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12.3 Sound Intensity and Sound Level
Chapter 12 Physics of Hearing
12.3 Sound Intensity and Sound Level
• Define intensity, sound intensity, and sound pressure level.
• Calculate sound intensity levels in decibels (dB).
In a quiet forest, you can sometimes hear a single leaf fall to the ground. After settling into bed, you may hear your blood pulsing through your ears. But when a passing motorist has his stereo
turned up, you cannot even hear what the person next to you in your car is saying. We are all very familiar with the loudness of sounds and aware that they are related to how energetically the source
is vibrating. In cartoons depicting a screaming person (or an animal making a loud noise), the cartoonist often shows an open mouth with a vibrating uvula, the hanging tissue at the back of the
mouth, to suggest a loud sound coming from the throat Figure 1. High noise exposure is hazardous to hearing, and it is common for musicians to have hearing losses that are sufficiently severe that
they interfere with the musicians’ abilities to perform. The relevant physical quantity is sound intensity, a concept that is valid for all sounds whether or not they are in the audible range.
Intensity is defined to be the power per unit area carried by a wave. Power is the rate at which energy is transferred by the wave. In equation form, intensity [latex]I[/latex]is
where[latex]P[/latex]is the power through an area[latex]A.[/latex]The SI unit for[latex]I[/latex]is[latex]\text{W/m}^2.[/latex]The intensity of a sound wave is related to its amplitude squared by the
following relationship:
Here [latex]\Delta{p}[/latex] is the pressure variation or pressure amplitude (half the difference between the maximum and minimum pressure in the sound wave) in units of pascals (Pa) or [latex]\text
{N/m}^2.[/latex] (We are using a lower case pp for pressure to distinguish it from power, denoted by [latex]P[/latex] above.) The energy (as kinetic energy [latex]\dfrac{mv^2}{2}[/latex] of an
oscillating element of air due to a traveling sound wave is proportional to its amplitude squared. In this equation, [latex]\rho[/latex] is the density of the material in which the sound wave
travels, in units of [latex]\text{kg/m}^3,[/latex] and [latex]v_{\text{w}}[/latex] is the speed of sound in the medium, in units of m/s. The pressure variation is proportional to the amplitude of the
oscillation, and so [latex]I[/latex] varies as [latex](\Delta{p})^2[/latex] (Figure 2). This relationship is consistent with the fact that the sound wave is produced by some vibration; the greater
its pressure amplitude, the more the air is compressed in the sound it creates.
Figure 1. Graphs of the gauge pressures in two sound waves of different intensities. The more intense sound is produced by a source that has larger-amplitude oscillations and has greater pressure
maxima and minima. Because pressures are higher in the greater-intensity sound, it can exert larger forces on the objects it encounters.
Sound intensity levels are quoted in decibels (dB) much more often than sound intensities in watts per meter squared. Decibels are the unit of choice in the scientific literature as well as in the
popular media. The reasons for this choice of units are related to how we perceive sounds. How our ears perceive sound can be more accurately described by the logarithm of the intensity rather than
directly to the intensity. The sound intensity level [latex]\beta[/latex] in decibels of a sound having an intensity [latex]I[/latex] in watts per meter squared is defined to be
where [latex]I_0=10^{-12}~\text{W/m}^2[/latex] is a reference intensity. In particular, [latex]I_0[/latex] is the lowest or threshold intensity of sound a person with normal hearing can perceive at a
frequency of 1000 Hz. Sound intensity level is not the same as intensity. Because [latex]\beta[/latex] is defined in terms of a ratio, it is a unitless quantity telling you the level of the sound
relative to a fixed standard ([latex]10^{-12}~\text{W/m}^2,[/latex] in this case). The units of decibels (dB) are used to indicate this ratio is multiplied by 10 in its definition. The bel, upon
which the decibel is based, is named for Alexander Graham Bell, the inventor of the telephone.
Sound intensity level β (dB) Intensity I(W/m^2) Example/effect
[latex]0[/latex] [latex]1\times10^{-12}[/latex] Threshold of hearing at 1000 Hz
[latex]10[/latex] [latex]1\times10^{-11}[/latex] Rustle of leaves
[latex]20[/latex] [latex]1\times10^{-10}[/latex] Whisper at 1 m distance
[latex]30[/latex] [latex]1\times10^{-9}[/latex] Quiet home
[latex]40[/latex] [latex]1\times10^{-8}[/latex] Average home
[latex]50[/latex] [latex]1\times10^{-7}[/latex] Average office, soft music
[latex]60[/latex] [latex]1\times10^{-6}[/latex] Normal conversation
[latex]70[/latex] [latex]1\times10^{-5}[/latex] Noisy office, busy traffic
[latex]80[/latex] [latex]1\times10^{-4}[/latex] Loud radio, classroom lecture
[latex]90[/latex] [latex]1\times10^{-3}[/latex] Inside a heavy truck; damage from prolonged exposure^1
[latex]100[/latex] [latex]1\times10^{-2}[/latex] Noisy factory, siren at 30 m; damage from 8 h per day exposure
[latex]110[/latex] [latex]1\times10^{-1}[/latex] Damage from 30 min per day exposure
[latex]120[/latex] [latex]1[/latex] Loud rock concert, pneumatic chipper at 2 m; threshold of pain
[latex]140[/latex] [latex]1\times10^2[/latex] Jet airplane at 30 m; severe pain, damage in seconds
[latex]160[/latex] [latex]1\times10^4[/latex] Bursting of eardrums
Table 2. Sound Intensity Levels and Intensities.
The decibel level of a sound having the threshold intensity of [latex]10^{-12}~\text{W/m}^2[/latex] is [latex]\beta=0\text{dB},[/latex] because [latex]\text{log}_{10}1=0.[/latex] That is, the
threshold of hearing is 0 decibels. Table 2 gives levels in decibels and intensities in watts per meter squared for some familiar sounds.
One of the more striking things about the intensities in Table 2 is that the intensity in watts per meter squared is quite small for most sounds. The ear is sensitive to as little as a trillionth of
a watt per meter squared—even more impressive when you realize that the area of the eardrum is only about [latex]1~\text{cm}^2[/latex], so that only [latex]10^{-16}[/latex] W falls on it at the
threshold of hearing! Air molecules in a sound wave of this intensity vibrate over a distance of less than one molecular diameter, and the gauge pressures involved are less than [latex]10^{-9}[/
latex] atm.
Another impressive feature of the sounds in Table 2 is their numerical range. Sound intensity varies by a factor of [latex]10^{12}[/latex] from threshold to a sound that causes damage in seconds. You
are unaware of this tremendous range in sound intensity because how your ears respond can be described approximately as the logarithm of intensity. Thus, sound intensity levels in decibels fit your
experience better than intensities in watts per meter squared. The decibel scale is also easier to relate to because most people are more accustomed to dealing with numbers such as 0, 53, or 120 than
numbers such as [latex]1.00\times10^{-11}[/latex].
One more observation readily verified by examining Table 2 or using [latex]I=\dfrac{(\Delta{p})^2}{2\rho{v}_{\text{w}}}[/latex] is that each factor of 10 in intensity corresponds to 10 dB. For
example, a 90 dB sound compared with a 60 dB sound is 30 dB greater, or three factors of 10 (that is, [latex]10^3[/latex] times) as intense. Another example is that if one sound is [latex]10^7[/
latex] as intense as another, it is 70 dB higher. See Table 3.
[latex]I_2/I_1[/latex] [latex]\beta_1-\beta_2[/latex]
2.0 3.0 dB
5.0 7.0 dB
10.0 10.0 dB
Table 3. Ratios of Intensities and Corresponding Differences in Sound Intensity Levels.
Example 1: Calculating Sound Intensity Levels: Sound Waves
Calculate the sound intensity level in decibels for a sound wave traveling in air at[latex]0^{\circ}\text{C}[/latex]and having a pressure amplitude of 0.656 Pa.
We are given [latex]\Delta{p},[/latex]so we can calculate [latex]I[/latex] using the equation [latex]I=(\Delta{p})^2/(2\rho{v}_{\text{w}})^2.[/latex] Using [latex]I,[/latex] we can calculate [latex]\
beta[/latex] straight from its definition in [latex]\beta\text{(dB)}=10~\text{log}_{10}(I/I_0).[/latex]
(1) Identify knowns:
Sound travels at 331 m/s in air at[latex]0^{\circ}\text{C}.[/latex]
Air has a density of [latex]1.29~\text{kg/m}^3[/latex] at atmospheric pressure and [latex]0^{\circ}\text{C}.[/latex]
(2) Enter these values and the pressure amplitude into [latex]I=(\Delta{p})^2/(2\rho{v}_{\text{w}})[/latex]
(3) Enter the value for[latex]I[/latex]and the known value for [latex]I_0[/latex] into [latex]\beta\text{(dB)}=10~\text{log}_{10}(I/I_0).[/latex] Calculate to find the sound intensity level in
This 87 dB sound has an intensity five times as great as an 80 dB sound. So a factor of five in intensity corresponds to a difference of 7 dB in sound intensity level. This value is true for any
intensities differing by a factor of five.
Example 2: Change Intensity Levels of a Sound: What Happens to the Decibel Level?
Show that if one sound is twice as intense as another, it has a sound level about 3 dB higher.
You are given that the ratio of two intensities is 2 to 1, and are then asked to find the difference in their sound levels in decibels. You can solve this problem using of the properties of
(1) Identify knowns:
The ratio of the two intensities is 2 to 1, or:
We wish to show that the difference in sound levels is about 3 dB. That is, we want to show:
Note that:
(2) Use the definition of[latex]\beta[/latex]to get:
This means that the two sound intensity levels differ by 3.01 dB, or about 3 dB, as advertised. Note that because only the ratio[latex]I_2/I_1[/latex]is given (and not the actual intensities), this
result is true for any intensities that differ by a factor of two. For example, a 56.0 dB sound is twice as intense as a 53.0 dB sound, a 97.0 dB sound is half as intense as a 100 dB sound, and so
It should be noted at this point that there is another decibel scale in use, called the sound pressure level, based on the ratio of the pressure amplitude to a reference pressure. This scale is used
particularly in applications where sound travels in water. It is beyond the scope of most introductory texts to treat this scale because it is not commonly used for sounds in air, but it is important
to note that very different decibel levels may be encountered when sound pressure levels are quoted. For example, ocean noise pollution produced by ships may be as great as 200 dB expressed in the
sound pressure level, where the more familiar sound intensity level we use here would be something under 140 dB for the same sound.
Take-Home Investigation: Feeling Sound
Find a CD player and a CD that has rock music. Place the player on a light table, insert the CD into the player, and start playing the CD. Place your hand gently on the table next to the speakers.
Increase the volume and note the level when the table just begins to vibrate as the rock music plays. Increase the reading on the volume control until it doubles. What has happened to the vibrations?
Check Your Understanding 1
Describe how amplitude is related to the loudness of a sound.
Check Your Understanding 2
Identify common sounds at the levels of 10 dB, 50 dB, and 100 dB.
Section Summary
• Intensity is the same for a sound wave as was defined for all waves; it is
where [latex]P[/latex] is the power crossing area [latex]A[/latex]. The SI unit for [latex]I[/latex] is watts per meter squared. The intensity of a sound wave is also related to the pressure
amplitude [latex]\Delta{p}[/latex]
where [latex]\rho[/latex] is the density of the medium in which the sound wave travels and [latex]v_{\text{w}}[/latex] is the speed of sound in the medium.
• Sound intensity level in units of decibels (dB) is
where [latex]I_0=10^{-12}~\text{W/m}^2[/latex] is the threshold intensity of hearing.
the power per unit area carried by a wave
sound intensity level
a unitless quantity telling you the level of the sound relative to a fixed standard
sound pressure level
the ratio of the pressure amplitude to a reference pressure
Conceptual Questions
1: Six members of a synchronized swim team wear earplugs to protect themselves against water pressure at depths, but they can still hear the music and perform the combinations in the water perfectly.
One day, they were asked to leave the pool so the dive team could practice a few dives, and they tried to practice on a mat, but seemed to have a lot more difficulty. Why might this be?
2: A community is concerned about a plan to bring train service to their downtown from the town’s outskirts. The current sound intensity level, even though the rail yard is blocks away, is 70 dB
downtown. The mayor assures the public that there will be a difference of only 30 dB in sound in the downtown area. Should the townspeople be concerned? Why?
Problems & Exercises
1: What is the intensity in watts per meter squared of 85.0-dB sound?
2: The warning tag on a lawn mower states that it produces noise at a level of 91.0 dB. What is this in watts per meter squared?
3: A sound wave traveling in [latex]20^{\circ}\text{C}[/latex] air has a pressure amplitude of 0.5 Pa. What is the intensity of the wave?
4: What intensity level does the sound in the preceding problem correspond to?
5: What sound intensity level in dB is produced by earphones that create an intensity of [latex]4.00\times10^{-2}~\text{W/m}^2[/latex]
6: Show that an intensity of [latex]10^{-12}~\text{W/m}^2[/latex] is the same as [latex]10^{-16}\text{W/cm}^2.[/latex]
7: (a) What is the decibel level of a sound that is twice as intense as a 90.0-dB sound? (b) What is the decibel level of a sound that is one-fifth as intense as a 90.0-dB sound?
8: (a) What is the intensity of a sound that has a level 7.00 dB lower than a [latex]4.00\times10^{-9}~\text{W/m}^2[/latex]sound? (b) What is the intensity of a sound that is 3.00 dB higher than a
9: (a) How much more intense is a sound that has a level 17.0 dB higher than another? (b) If one sound has a level 23.0 dB less than another, what is the ratio of their intensities?
10: People with good hearing can perceive sounds as low in level as[latex]-8.00~\text{dB}[/latex]at a frequency of 3000 Hz. What is the intensity of this sound in watts per meter squared?
11: If a large housefly 3.0 m away from you makes a noise of 40.0 dB, what is the noise level of 1000 flies at that distance, assuming interference has a negligible effect?
12: Ten cars in a circle at a boom box competition produce a 120-dB sound intensity level at the center of the circle. What is the average sound intensity level produced there by each stereo,
assuming interference effects can be neglected?
13: The amplitude of a sound wave is measured in terms of its maximum gauge pressure. By what factor does the amplitude of a sound wave increase if the sound intensity level goes up by 40.0 dB?
14: If a sound intensity level of 0 dB at 1000 Hz corresponds to a maximum gauge pressure (sound amplitude) of[latex]10^{-9}~\text{atm},[/latex]what is the maximum gauge pressure in a 60-dB sound?
What is the maximum gauge pressure in a 120-dB sound?
15: An 8-hour exposure to a sound intensity level of 90.0 dB may cause hearing damage. What energy in joules falls on a 0.800-cm-diameter eardrum so exposed?
16: (a) Ear trumpets were never very common, but they did aid people with hearing losses by gathering sound over a large area and concentrating it on the smaller area of the eardrum. What decibel
increase does an ear trumpet produce if its sound gathering area is[latex]900~\text{cm}^2[/latex]and the area of the eardrum is[latex]0.500~\text{cm}^2,[/latex]but the trumpet only has an efficiency
of 5.00% in transmitting the sound to the eardrum? (b) Comment on the usefulness of the decibel increase found in part (a).
17: Sound is more effectively transmitted into a stethoscope by direct contact than through the air, and it is further intensified by being concentrated on the smaller area of the eardrum. It is
reasonable to assume that sound is transmitted into a stethoscope 100 times as effectively compared with transmission though the air. What, then, is the gain in decibels produced by a stethoscope
that has a sound gathering area of[latex]15.0~\text{cm}^2,[/latex]and concentrates the sound onto two eardrums with a total area of[latex]0.900~\text{cm}^2[/latex]with an efficiency of 40.0%?
18: Loudspeakers can produce intense sounds with surprisingly small energy input in spite of their low efficiencies. Calculate the power input needed to produce a 90.0-dB sound intensity level for a
12.0-cm-diameter speaker that has an efficiency of 1.00%. (This value is the sound intensity level right at the speaker.)
1. 1 Several government agencies and health-related professional associations recommend that 85 dB not be exceeded for 8-hour daily exposures in the absence of hearing protection.
Check Your Understanding 1
Amplitude is directly proportional to the experience of loudness. As amplitude increases, loudness increases.
Check Your Understanding 2
10 dB: Running fingers through your hair.
50 dB: Inside a quiet home with no television or radio.
100 dB: Take-off of a jet plane.
Problems & Exercises
1: [latex]3.16\times10^{-4}~\text{W/m}^2[/latex]
3: [latex]3.04\times10^{-4}~\text{W/m}^2[/latex]
5: 106 dB
7: (a) 93 dB (b) 83 dB
9: (a) 50.1 (b)[latex]5.01\times10^{-3}[/latex] or [latex]\dfrac{1}{200}[/latex]
11: 70.0 dB
13: 100
15: [latex]1.45\times10^{-3}~\text{J}[/latex]
17: 28.2 dB | {"url":"https://pressbooks.bccampus.ca/douglasphys1107/chapter/12-3-sound-intensity-and-sound-level-2/","timestamp":"2024-11-08T04:04:29Z","content_type":"text/html","content_length":"108467","record_id":"<urn:uuid:83be364b-1f90-4317-b63a-f4102ce9b5fb>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00743.warc.gz"} |
Ohms Law Current Calculator - Online Calculation of Electric Current
Ohms Law Current Calculator
Calculate Current with Power and Voltage :
I = P/E
Calculate Current with Power and Resistance :
I = √P/R
Calculate Current with Voltage and Resistance :
I = E/R
Ohms Law states that the current passing or flowing through the conductor between two points is directly proportional to the potential difference that is applied across the two points. The electric
current on the other hand is inversely proportional to the resistance between the points. As the resistance in a circuit increases, the current generated in the circuit decreases.
The advanced online Ohms Law Current Calculator is used to calculate the electric current between the points flowing through a given conductor in various forms. Ohm's law can be used to solve simple
problems with electrical circuits of current.
Calculate the Electric Current as per the Ohms law for the given power and voltage.
Power (P) = 25 W
Voltage (E) = 20 V
Apply Formula:
I = P/E
I = 25/20
I = 1.25 A
• What is Ohm’s law? | Fluke. (2023, December 6).
• Uses of ohms law | Tenny. (2023, May 22). | {"url":"https://www.meracalculator.com/physics/electromagnetism/ohms-law-current.php","timestamp":"2024-11-13T22:39:52Z","content_type":"text/html","content_length":"42014","record_id":"<urn:uuid:f741d390-101c-46f9-b1d2-04139e768715>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00300.warc.gz"} |
seminars - Moments of Margulis functions and Quantitative Oppenheim Conjecture 2
The Oppenheim conjecture, proved by Margulis in 1986, states that for a non-degenerate indefinite irrational quadratic form Q in n ≥ 3 variables, the image set Q(Zn) of integral vectors is a dense
subset of the real line. Determining the distribution of values of an indefinite quadratic form at integral points asymptotically is referred to as quantitative Oppenheim conjecture. The quantitative
Oppenheim conjecture was established by Eskin, Margulis, and Mozes for quadratic forms in n ≥ 4 variables. In this talk, we discuss the quantitative Oppenheim conjecture for ternary quadratic forms
(n=3). The main ingredient of the proof is a uniform boundedness result for the moments of the Margulis function over expanding translates of a unipotent orbit in the space of 3-dimensional lattices,
under suitable Diophantine conditions of the initial unipotent orbit. | {"url":"http://www.math.snu.ac.kr/board/index.php?mid=seminars&sort_index=room&order_type=asc&page=77&l=en&document_srl=1243594","timestamp":"2024-11-05T03:22:50Z","content_type":"text/html","content_length":"45870","record_id":"<urn:uuid:59714f19-bce7-4c04-b1d1-8d3e93d85e30>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00653.warc.gz"} |
Comparative study of model updating methods using frequency response function data
Constructing accurate finite element models for engineering structures plays a key role in structural dynamic design and analysis. Finite element model updating using frequency response function data
arises great attention. In this paper, a comparison of two model updating approaches by using frequency response function data is investigated. The first method is based on sensitivity analysis,
which has been regarded as one of the most successful approaches in model updating. The second one is based on the representation of modeling errors as linear combinations of the individual element
matrices, which can be used for both error locating and model updating. The basic formulations of these two methods are introduced and the possible solution strategies are discussed. Numerical
simulations are conducted to compare the two model updating methods employing the GARTEUR Truss, two aspects effect on the updating solution including magnitude of initial modeling errors and the
completeness of measured coordinates are studied. At last, an experimental cantilever beam is updated by adopting the sensitivity method with tested frequency response function, it is shown that the
sensitivity method is effective even when the test data are extremely incomplete.
1. Introduction
Due to the high costs of the experiments, numerical simulations are often adopted to predict the dynamic behavior of engineering structures for reducing the costs in both design and analysis [1, 2].
An accurate numerical model is important in complex structures design, dynamics response computation and safety assessment. Since limited measurement data can be included in the testing model and
errors generally exist in the initial finite element model, an accurate numerical model for the purpose of dynamic analysis cannot be obtained by either using test modeling or the finite element
modeling. Model updating, a classical inverse problem, is often adopted to reduce the differences between predictions of the finite element model and observations from the dynamics tests, which is an
effective way to improve the accuracy of the initial finite element model. It draws great attention from researchers since this technique was proposed, although applying model updating requires a
good physical understanding of the structure. Rough finite element models were successfully updated with modal data obtained from experiments in many fields. Recently, Finite element model updating
using Frequency Response Functions (FRFs) data draws great attention to researchers owing to several advantages. (1) The FRFs are very sensitive to the damping of structures at resonance peaks, and
damping must be included in the finite element model; (2) no modal analysis is required in the updating thus the error in system identification can be avoided; (3) the updating problem is
over-determined due to the availability of FRF data at numerous excitation and observation points. All of the above characteristics make the approach with great prospects.
Finite element model updating using frequency response functions has been investigated in recent decades. Surveys on the model updating methods proposed in recent years were discussed by Imregun and
Visser [3], and also by Mottershed and Friswell [4]. Definitions on the two different types of error, namely the equation error and the output error, have been proposed by Natke [5] with the weighted
least square method to solve the problem in which the damping was not considered. With incomplete measurement data, Foster [6] and Link [7] employed the static and dynamic reduction techniques to
condense the system matrix. The truncated modal technique was employed to reduce the system matrix in the state space by Friswell [8]. Imregun and Ewins [9, 10] introduced the basic theory of model
updating using frequency response data comprehensively in which the stable solution of the algebraic equations was analyzed without consideration of the measurement noise. A correlation criterion has
been explored by Grafe [11] to derivate the formulas of model updating. Attempts had also been made to construct the governing equations for identifying the system matrices with modal expansion and
model reduction methods. The modeling and identification of the connections between the two models by using the FRFs data has been investigated by Visser [12].
Amongst the updating methods using FRFs data, the sensitivity method and the system matrix error method are which the researchers pay more attentions to. The former is based on the linearization of a
generally non-linear relationship between the selected parameters and the measurement [13]; in the latter method, the errors of the global system matrix are expressed by a linear relation of the
element matrix. In this paper, comparison of the two model updating approaches is conducted based on measured data from the GARTEUR Truss. The comparison aims to discuss the application perspectives
of the two methods and to provide references for engineers.
2. Basic theory
In this section, basic formulations of the two methods are presented, and then the rules for selecting the frequency points as well as the solution strategies of the updating problem will be briefly
2.1. The sensitivity method
Starting from the motion equations without considering damping in the frequency domain:
$\left[-{\omega }^{2}\mathbf{M}+\mathbf{K}\right]\mathbf{x}=\mathbf{f}\left(\omega \right).$
$\mathbf{Z}\left(\omega \right)=-{\omega }^{2}\mathbf{M}+\mathbf{K}.$
Eq. (1) can be transformed to:
$\mathbf{Z}\left(\omega \right)\mathbf{}\mathbf{x}\left(\omega \right)=\mathbf{f}\left(\omega \right),$
where $\mathbf{x}\left(\omega \right)$ and $\mathbf{f}\left(\omega \right)$ are the displacement of the structure and the input force in frequency domain, respectively. $\mathbf{Z}\left(\omega \
right)$ is the dynamic stiffness matrix in frequency domain and the inverse of which is the frequency response function $\mathbf{H}\left(\omega \right)$, the relationship between $\mathbf{Z}\left(\
omega \right)$ and $\mathrm{}\mathbf{H}\left(\omega \right)$ gives:
$\mathbf{H}\left(\omega \right)\mathbf{Z}\left(\omega \right)=\mathbf{I},$
in which:
$\mathbf{H}\left(\omega \right)={\left(-{\omega }^{2}\mathbf{M}+\mathbf{K}\right)}^{-1}.$
The displacement vector can be expressed by:
$\mathbf{x}\left(\omega \right)=\mathbf{H}\left(\omega \right)\mathbf{f}\left(\omega \right).$
When a unit force is applied on the $j$th degree-of-freedom (dof) of the structure, on the selected $m$ dofs, Eq. (6) can be simplified as:
$\mathbf{x}\left(\omega \right)=\left({\mathbf{H}}_{1j},{\mathbf{H}}_{2j},{\mathbf{H}}_{3j},\dots ,{\mathbf{H}}_{mj}\right)={\mathbf{H}}_{j},$
where ${\mathbf{H}}_{mj}$ is the element at the $m$th row and the $j$th column in the frequency response matrix, the vector ${\mathbf{H}}_{j}$ denotes the $j$th column of the frequency response
The error is defined as the difference between the measured and computed response given by:
where $\mathbf{p}$ is the parameter vector. ${\mathbf{x}}^{E}$ and ${\mathbf{x}}^{A}$ is respectively the experimental and the analytical response. Substituting Eq. (6) and (7) into Eq. (8), the
residual term gives:
$\mathbf{R}\left(\mathbf{p}\right)={\mathbf{H}}_{j}^{E}\left(\omega \right)-{\mathbf{H}}_{j}^{A}\left(\omega ,\mathbf{p}\right).$
The superscript $E$ and $A$ denote the FE model and the experimental model respectively. The sensitivity method is developed from a Taylor series expansion in which the high-order terms are
truncated, after the truncation, the $j$th column of frequency response function matrix ${\mathbf{H}}_{j}^{A}\left(\omega ,\mathbf{p}\right)$ with respect to ${\mathbf{p}}^{0}$ gives:
${\mathbf{H}}_{j}^{A}\left(\omega ,\mathbf{p}\right)={\mathbf{H}}_{j}^{A}\left(\omega ,{\mathbf{p}}^{0}\right)+\mathbf{S}\mathrm{\Delta }\mathbf{p},$
where $\mathbf{S}$ is the sensitivity matrix that can be expressed as:
$\mathbf{S}=\left[\begin{array}{ccc}\frac{\partial {\mathbf{H}}_{1j}^{A}\left(\omega ,{\mathbf{p}}^{0}\right)}{\partial {p}_{1}}& \cdots & \frac{\partial {\mathbf{H}}_{1j}^{A}\left(\omega ,{\mathbf
{p}}^{0}\right)}{\partial {p}_{{N}_{p}}}\\ ⋮& \ddots & ⋮\\ \frac{\partial {\mathbf{H}}_{mj}^{A}\left(\omega ,{\mathbf{p}}^{0}\right)}{\partial {p}_{1}}& \cdots & \frac{\partial {\mathbf{H}}_{mj}^{A}\
left(\omega ,{\mathbf{p}}^{0}\right)}{\partial {p}_{{N}_{p}}}\end{array}\right].$
$\mathrm{\Delta }\mathbf{p}$ and ${N}_{p}$ are the vector of correction and the number of the parameters need to be updated:
$\mathrm{\Delta }\mathbf{p}={\left(∆{p}_{1},∆{p}_{2},\dots ,∆{p}_{{N}_{p}}\right)}^{T}.$
Substituting Eq. (10) into Eq. (9), the residual Eq. (9) can be expressed in the following matrix form:
$\mathbf{R}\left(\mathbf{p}\right)={\mathbf{H}}_{j}^{E}\left(\omega \right)-{\mathbf{H}}_{j}^{A}\left(\omega ,{\mathbf{p}}^{0}\right)-\mathbf{S}\mathrm{\Delta }\mathbf{p}.$
${\mathbf{b}=\mathbf{H}}_{j}^{E}\left(\omega \right)-{\mathbf{H}}_{j}^{A}\left(\omega ,{\mathbf{p}}^{0}\right),$
where $\mathbf{b}={\left({b}_{1},{b}_{2},\dots ,{b}_{m}\right)}^{T}$ is the error vector between the experimental and analytical response. After $n$th iterations, if the correlation between the
computational and the experimental response is close to 1, the norm of the residual $‖\mathbf{R}\left(\mathbf{p}\right)‖$ will converged to zero, and the parameter vector ${\mathbf{p}}^{0}$ is
updated to be ${\mathbf{p}}^{n}$. Then:
$‖\mathbf{R}\left(\mathbf{p}\right)‖⟶0\mathrm{}\mathrm{}\mathrm{}\text{is equivalent to}\mathrm{}\mathrm{}\mathrm{}{\mathbf{H}}_{j}^{E}\left(\omega \right)-{\mathbf{H}}_{j}^{A}\left(\omega ,{\mathbf
{p}}^{n}\right)-\mathbf{S}\mathrm{\Delta }\mathbf{p}⟶0.$
Model updating can be transformed to an iteration problem; the following equations should be solved exactly in every iteration step:
$\mathbf{S}\mathrm{\Delta }\mathbf{p}=\mathbf{b}.$
It is noted that the sensitivity matrix is varying during the iteration procedure, and the sensitivity matrix must be calculated at each step until the norm of $\mathrm{\Delta }\mathbf{p}$ converged
to a small quantity.
The sensitivity method is based on the truncated Taylor expansion, and the solution of Eq. (16) can only identify the error direction of the parameters in a local constraint; however, due to the
truncated errors, when the initial parameters contains large errors, the finite element model may not be updated with this method.
2.2. The system matrix error method
Derivation of the system matrix error method begins with the measured frequency response functions, which is briefly introduced in this section.
According to Eq. (1), when a unit force is applied as excitation, the Equations of motion of the experimental model and the analytical model of the system can respectively be written as:
$\left[-{\omega }^{2}{\mathbf{M}}^{A}+{\mathbf{K}}^{A}\right]{\mathbf{H}}^{A}=\mathbf{I},$
$\left[-{\omega }^{2}{\mathbf{M}}^{E}+{\mathbf{K}}^{E}\right]{\mathbf{H}}^{E}=\mathbf{I}.$
The superscripts $A$ and $E$ denote FE model and experimental model, respectively.
Assuming that:
${\mathbf{K}}^{E}=\mathrm{\Delta }\mathbf{K}+{\mathbf{K}}^{A},\mathbit{}\mathbit{}\mathbit{}{\mathbf{M}}^{E}=\mathrm{\Delta }\mathbf{M}+{\mathbf{M}}^{A},\mathbit{}\mathbit{}\mathbit{}{\mathbf{H}}^{E}
=\mathrm{\Delta }\mathbf{H}+{\mathbf{H}}^{A},$
where $\mathrm{\Delta }\mathbf{K}$, $\mathrm{\Delta }\mathbf{M}$, $\mathrm{\Delta }\mathbf{H}$ are the stiffness, mass and frequency response function errors in the finite element model.
Substituting Eqs. (19) into (18), we have:
$\left[\left(\mathrm{\Delta }\mathbf{K}+{\mathbf{K}}^{A}\right)-{\omega }^{2}\left(\mathrm{\Delta }\mathbf{M}+{\mathbf{M}}^{A}\right)\right]\left(\mathrm{\Delta }\mathbf{H}+{\mathbf{H}}^{A}\right)=\
Rearranging Eq. (20), it gives:
$\left({\mathbf{K}}^{A}-{\omega }^{2}{\mathbf{M}}^{A}\right){\mathbf{H}}^{A}+\left({\mathbf{K}}^{A}-{\omega }^{2}{\mathbf{M}}^{A}\right)\mathbf{}\mathrm{\Delta }\mathbf{H}+\left(\mathrm{\Delta }\
mathbf{K}-{\omega }^{2}\mathrm{\Delta }\mathbf{M}\right)\left(\mathrm{\Delta }\mathbf{H}+{\mathbf{H}}^{A}\right)=\mathbf{I}.$
Substituting Eq. (17) into (21), we have:
$\left({\mathbf{K}}^{A}-{\omega }^{2}{\mathbf{M}}^{A}\right)\mathrm{\Delta }\mathbf{H}+\left(\mathrm{\Delta }\mathbf{K}-{\omega }^{2}\mathrm{\Delta }\mathbf{M}\right){\mathbf{H}}^{E}=0,$
$\left(\mathrm{\Delta }\mathbf{K}-{\omega }^{2}\mathrm{\Delta }\mathbf{M}\right){\mathbf{H}}^{E}=-\left({\mathbf{K}}^{A}-{\omega }^{2}{\mathbf{M}}^{A}\right)\mathbf{}\mathrm{\Delta }\mathbf{H}.$
Eq. (23) can be employed to solve for the unknown modeling errors. However, before this can be achieved, some parameterization is needed. To parameterize the modeling errors, it is assumed that the
error mass and stiffness matrices can be expressed as linear combinations of elemental mass and stiffness matrices, respectively:
$\begin{array}{c}\mathrm{\Delta }\mathbf{M}=\sum _{i=1}^{\mathbf{N}}{a}_{i}{\left[{\mathbf{M}}^{A}\right]}_{i}^{ele},\end{array}$$\mathrm{\Delta }\mathbf{K}=\sum _{i=1}^{\mathbf{N}}{b}_{i}{\left[{\
where ${\left[{\mathbf{M}}^{A}\right]}_{i}^{ele}$ and ${\left[{\mathbf{K}}^{A}\right]}_{i}^{ele}$ are the $i$th element mass and stiffness matrices, respectively. ${a}_{i}$ and ${b}_{i}$ are the
perturbation in design parameters corresponding to the $i\text{t}\text{h}$ element. Notation $\mathrm{\Sigma }$ represents the summation. Substituting Eq. (24) into Eq. (23), we can obtain the
following equation:
$\left[\sum _{i=1}^{N}{b}_{i}{\left[{\mathbf{K}}^{A}\right]}_{i}^{ele}-{\omega }^{2}\sum _{i=1}^{N}{a}_{i}{\left[{\mathbf{M}}^{A}\right]}_{i}^{ele}\right]{\mathbf{H}}^{E}=-\left({\mathbf{K}}^{A}-{\
omega }^{2}{\mathbf{M}}^{A}\right)\mathbf{}\text{Δ}\mathbf{H}.$
Eq. (25) can be transformed into a set of linear algebraic equations in terms of unknown perturbations in design parameters which denoted as ${a}_{i}$ ($i=\text{1,}\text{}\text{2,…,}\text{}N$) and $
{b}_{i}$ ($i=\text{1,}\text{}\text{2,…,}\text{}N$), respectively:
$\left[\begin{array}{ccc}{s}_{1}^{a}& \cdots & {s}_{N}^{a}\end{array}\begin{array}{ccc}{s}_{1}^{b}& \cdots & {s}_{N}^{b}\end{array}\right]\left\{\begin{array}{c}\mathbf{a}\\ \mathbf{b}\end{array}\
right\}=-\left({\mathbf{K}}^{A}-{\omega }^{2}{\mathbf{M}}^{A}\right)\mathrm{\Delta }\mathbf{H},$
where ${\mathrm{s}}_{i}^{a}$, ${\mathrm{s}}_{i}^{b}$, $\mathbf{a}$ and $\mathbf{b}$ are:
$\left\{\begin{array}{l}\begin{array}{c}{\mathrm{s}}_{i}^{a}=-{\omega }^{2}{\left[{\mathbf{M}}^{A}\right]}_{i}^{ele}{\mathbf{H}}^{E},\end{array}\\ {\mathrm{s}}_{i}^{b}={\left[{\mathbf{K}}^{A}\right]}
_{i}^{ele}{\mathbf{H}}^{E},\\ \mathbf{a}={\left(\begin{array}{ccc}{a}_{1}& \cdots & {a}_{N}\end{array}\right)}^{T},\\ \mathbf{b}={\left(\begin{array}{ccc}{b}_{1}& \cdots & {b}_{N}\end{array}\right)}^
Eq. (26) is formulated based on measured response function data with one measurement frequency selected. In practical vibration test, response function data are measured at many different measurement
frequencies. When response function data at sufficient number $n$ of measurement frequencies are used, Eq. (26) can be transformed as a set of over-determined algebraic equations:
$\mathbf{S}=\left[\begin{array}{ccc}{s}_{1}^{a}\left({\omega }_{1}\right)& \cdots & {s}_{N}^{a}\left({\omega }_{1}\right)\\ ⋮& \ddots & ⋮\mathbit{}\\ {s}_{1}^{a}\left({\omega }_{n}\right)& \cdots &
{s}_{N}^{a}\left({\omega }_{n}\right)\end{array}\mathbit{}\mathbit{}\mathbit{}\mathbit{}\begin{array}{ccc}{s}_{1}^{b}\left({\omega }_{1}\right)& \cdots & {s}_{N}^{b}\left({\omega }_{1}\right)\\ ⋮& \
ddots & ⋮\\ {s}_{1}^{b}\left({\omega }_{n}\right)& \cdots & {s}_{N}^{b}\left({\omega }_{n}\right)\end{array}\right],$
$\mathbf{p}=\left\{\begin{array}{c}\mathbf{a}\\ \mathbf{b}\end{array}\right\},\mathrm{}\mathrm{}\mathrm{}\mathbf{q}=\left\{\begin{array}{c}-\left({\mathbf{K}}^{A}-{\omega }_{1}^{2}{\mathbf{M}}^{A}\
right)\mathbf{}\mathrm{\Delta }\mathbf{H}\left({\omega }_{1}\right)\\ ⋮\\ -\left({\mathbf{K}}^{A}-{\omega }_{n}^{2}{\mathbf{M}}^{A}\right)\mathbf{}\mathrm{\Delta }\mathbf{H}\left({\omega }_{n}\right)
The coefficient matrix $\mathbf{S}$ and the vector $\mathbf{q}$ can be calculated in advance which is formed using the analytical model and the measured response function data. Eq. (28) can be solved
for $\mathbf{p}$ using linear least-square method and then the solution $\mathbf{p}$ is used to reconstruct the updated analytical model together with the original analytical model. However, when the
test data is incomplete and contaminated with noise, Eq. (28) is needed to be transformed and constraints have to be added.
2.3. Frequency point selection
Frequency response function are very sensitivity to damping properties at resonance peaks, local modes influence is include and no modal analysis is required, so that it contains much more
information than modal data, at each frequency point, the major system information are included. In the updating procedure, because of the huge number of frequency points, it is necessary to select
frequency point for improving the convergence speed and the precision, the selected frequencies should ensure the rank of the coefficient matrix in Eqs. (16) or (28) is not less than the number of
the parameters which need to be updated, and when the experimental FRF data is contaminated with measurement noises, frequency points selection seems more important.
There are two main principles of frequency are given in references [3, 12]:
1) Choose updating frequencies at the foot of experimental resonances peaks, and
2) Avoid frequencies between corresponding analytical and experimental resonances.
The first rule assures the coefficient matrix in Eqs. (16) or (28) to be well-posed. The second one avoids the problem of large residuals, as shown in Fig. 1, if frequencies between corresponding
analytical and experimental resonances are selected, in the process of the analytical data approaching to the experimental frequency response function, the analytical resonances will definitely
moving to the select frequency, and the amplitude at resonance is much bigger than the other positions, therefore, the right hand side (Residual 2) may become unbalance easily during the iterations,
this will cause the difficulties in solving linear algebraic equations. At each iteration step, it is necessary to ensure the chosen frequencies comply with the rules.
Fig. 1Iteration problem if frequencies are selected between corresponding analytical and experimental resonances
2.4. Solution strategies
Finite element model updating can often be transformed into optimization problem, for accurately reflecting the practical situations and the physical significances of the updated model, and the
solution of the optimum solution is often subject to a certain constraint. Due to the above reasons, based on the derivation of the updating theory, the linear algebraic Eq. (16) or (28) can be
transformed to an optimization problem with constrains:
$\left\{\begin{array}{l}\begin{array}{c}\text{M}\text{i}\text{n}{‖\mathbf{A}\mathbf{x}-\mathbf{b}‖}_{2}^{2},\end{array}\\ \text{s.t.}\mathrm{}\mathrm{}\mathrm{}\mathbf{V}\mathbf{L}\mathbf{B}\le \
mathbf{x}\le \mathbf{V}\mathbf{U}\mathbf{B},\end{array}\right\$
where $\mathbf{V}\mathbf{L}\mathbf{B}$ and $\mathbf{V}\mathbf{U}\mathbf{B}$ are respectively the lower limit and the upper limit of the parameter vector $\mathbf{x}$. Solving this problem by use of
mathematical method to get the estimated vector $\mathrm{\Delta }\mathbf{p}$, the iteration procedure will continue until the norm of $\mathrm{\Delta }\mathbf{p}$ converged to a small quantity.
3. Numerical case study
3.1. GARTEUR Truss
The simulation study using a GARTEUR Truss [14] is shown in Fig. 2, which is a cantilever truss. The numerical model of the Truss consists of 78 2-D beam elements, 74 nodes. Each node of the beam
element has three dofs (two translations and one rotation) and hence, the total number of dofs in the FE model is 222. The material properties are used in the FE model: Young’s modulus is assumed to
be $E=\text{0.75×}{\text{10}}^{\text{11}}\text{}\text{N/}{\text{m}}^{\text{2}}$ and density $\rho =\text{2800}\text{}\text{kg/}{\text{m}}^{\text{3}}$. The cross-sectional areas are ${A}_{h}=\text
{0.004}\text{}{\text{m}}^{\text{2}}$ (horizontal), ${A}_{v}=0\text{.006}\text{}{\text{m}}^{\text{2}}$ (vertical), and ${A}_{d}=\text{0.003}\text{}{\text{m}}^{\text{2}}$ (diagonal), the bending moment
is assumed to be the same for all the elements and is assumed to be $I=\text{0.0756}\text{}{\text{m}}^{\text{4}}$.
Fig. 2GARTEUR Truss model
Two simulated cases are carried out based on the GARTEUR Truss model. For Case-I, in order to generate the experimental data, stiffness modeling errors are introduced in the elements of the
analytical model of the structure by changing the bending stiffness of elements as shown in Table 1, assuming that the stiffness of element shown in Table 1 are all overrated 20 % in the FE model.
While in Case-II, the stiffness of elements in the FE model is overrated or underestimated have apparent discrepancies, details are shown in Table 2.
Table 1Stiffness modeling errors location (case 1)
Element no. 1 2 28 29 43 44 45 46 47 48 49 50
Error (%) –20 –20 –20 –20 –20 –20 –20 –20 –20 –20 –20 –20
Table 2Stiffness modeling errors location (case 2)
Element no. 1 2 28 29 43 44 45 46 47 48 49 50
Error (%) –90 –90 100 100 100 100 10 –90 –90 –90 –90 –90
The two cases are employed to investigate the two methods introduced in Section 2, the updated results of coordinates incomplete and coordinates complete are discussed. Considering the small modeling
error firstly, and then the large modeling errors.
3.2. Updated results evaluation criterions
To assess the progress of iterations and to compare different updated models certain model quality indices have been constructed. Percentage average error in natural frequencies (AENF) is calculated
using following expressions:
$\mathrm{A}\mathrm{E}\mathrm{N}\mathrm{F}=\frac{100}{m}\sum _{i=1}^{m}\mathrm{a}\mathrm{b}\mathrm{s}\left(\frac{{f}^{A}-{f}^{E}}{{f}^{E}}\right),$
where ${f}^{A}$ and ${f}^{E}$ are the natural frequencies, $m$ is the number of the considering modes. The updated model should have the abilities of reflect the test results in the considering
frequency domain, outside of which the test results predicted.
Percentage average error in parameters (AEP) is calculated as error in the predicted parameters as a percentage of the known exact parameters:
$\mathrm{A}\mathrm{E}\mathrm{P}=100×\sum _{i}^{np}\frac{\mathrm{a}\mathrm{b}\mathrm{s}\left({p}_{exact}^{i}-{p}_{updated}^{i}\right)}{\mathrm{a}\mathrm{b}\mathrm{s}\left({p}_{exact}^{i}\right)}/np,$
where $np$ is the number of the parameters, ${p}_{exact}^{i}$ and $\mathrm{}{p}_{updated}^{i}$ are the value of the parameters.
3.3. Results and discussions
In order to demonstrate their practical application, the two model updating methods using frequency response function of GARTEUR Truss model which is introduced in Section 3.1. For comparative
investigation on these methods, two cases including small modeling error and large modeling error are studied. In each case, the measured coordinates complete and incomplete are investigated, the
complete case means the response of all coordinates of the structure can be tested, and the incomplete case meaning only translation degree of freedoms can be obtained.
3.3.1. Case-I small modeling error
For the sensitivity method, calculation of the frequency response function matrix ${\mathbf{H}}_{j}^{A}\left(\omega ,\mathbf{p}\right)$ with respect to the parameter vector $\mathbf{p}$ is very
time-consuming, and a more important question is that the sensitivity matrix $\mathbf{S}$ will easily become ill-posed, therefore, the selection of parameters to be updated is also a very important
aspect. In this case study, at first, we select the stiffness of all elements in the structure as parameters, the equation is an ill-posed problem, it is hard to get a credible solution. However,
assume the position of the errors are known in advance, and select appropriate frequency points, iteration process is required during updating. The identification of element modeling errors is shown
in Fig. 3. It can be seen that all the introduced modeling errors are well identified after convergence. The response function curves of the experimental, analytical and updated models are shown in
Fig. 4.
Fig. 3Comparison of the exact and identified modeling errors (method 1, coordinates incomplete)
Fig. 4Comparison of the analytical, the experimental and the updated response function curves (method 1) (—o— updated, —— experimental, —*— analytical)
With regard to the second method, for the case of coordinates complete, it is need not to consider many limitations, the frequency point can be selected randomly, the identification of element
modeling errors is shown in Fig. 5, and it is found that the agreement between the exact error and the identified error is excellent. But, in vibration test, it is not realistic that all coordinates
which are specified in the analytical model have been measured. Therefore, the effect of coordinate incompleteness upon the updating procedure should be assessed. For the case of incomplete data, the
model condensation and coordinates expansion technique should be used to solve this problem. In this work, those unmeasured elements of the response function vector are replaced by their analytical
counterparts. The results for the identification of element modeling errors are shown in Fig. 6. The response function curves of the experimental, analytical and updated models are shown in Fig. 7.
From this figure, there are a few differences between the regenerated and experimental response function data.
Fig. 5Comparison of the exact and identified modeling errors (method-2, coordinates complete)
Table 3 is the error indices comparison for the Case-I of small modeling error, from the table it is can be found that, under coordinates complete condition, method-2 can get an excellent solution.
Fig. 6Comparison of the exact and identified modeling errors (method-2, coordinates incomplete)
Fig. 7Comparison of the analytical, the experimental and the updated response function curves (method-2) (—o— updated, —— experimental, —*— analytical)
Table 3Error indices comparison for the Case-I of small modeling error (%)
Percentage average error in natural frequencies (AENF) Percentage average error in parameters (AEP)
Experimental coordinates Updated Updated
Before updating Before updating
Method-1 Method-2 Method-1 Method-2
Coordinates complete 2.82 – 0 20 – 0
Coordinates incomplete 2.82 0.01 0.18 20 0.08 0.59
3.3.2. Case-II large modeling error
For Case-II, the modeling error is much larger. It is hard to get a convergent solution by using of the sensitivity method; the reason is introduced in Section 2.1. But for the system matrix error
method, in the case the coordinates is completely can be tested, it is also easy to get a reliable solution, the results can be seen in Fig. 8 and Fig. 9, it is shown that the solution is close to
the exact error. However, for the case of incomplete coordinates, it is not easy get a convergent updating solution. But if we know the stiffness error position in advance, under this condition, the
identification of stiffness error can be obtained. The comparison of the exact and identified modeling errors is shown in Fig. 10.
Fig. 8Comparison of the exact and identified modeling errors (method-2, coordinates complete)
Fig. 9Comparison of the analytical, the experimental and the updated response function curves (method-2) (—o— updated, —— experimental, —*— analytical)
Fig. 10Comparison of the exact and identified modeling errors (method-2, known error position in advance)
4. Experimental case study
A cantilever beam is updated by using the sensitivity method with experimental frequency response function. Geometric dimensions, including cross section and length of the rectangle tube, arrangement
of excitation and accelerometer in experimentation are shown in Fig. 11 and Fig. 12. The tube is made of steel, the density and Young’s module of which are respectively $\rho =\text{7900}\text{}\text
{kg/}{\text{m}}^{\text{3}}$ and $E=\text{2.0×}{\text{10}}^{\text{11}}\text{}\text{N/}{\text{m}}^{\text{2}}$. In this case, the experimental FRF data is obtained by using impact testing with single
input single output method; and the measuring direction is along with $z$ axis as shown in Fig. 12. The range of sample frequency is 0-2000 Hz.
The cantilever beam is modeled using beam element, three kinds of attributes (${E}_{1}$-${E}_{3}$) are assigned to six elements (element ①-⑥) which near the fixed end, ${E}_{4}$ is assigned to rest
twenty elements of the beam, the full finite element model consist of 27 nodes, 26 beam elements, and 156 active degrees of freedom.
Fig. 11Experimental setup of the impact testing
Fig. 12The cantilever beam (Unit: mm)
Fig. 13Comparison of the analytical, the experimental and the updated response function curves (—o— updated, —— experimental, —*— analytical)
Table 4Variation of parameters after model updating
Parameters Initial values Updated Variation (%)
${E}_{1}$ (N/m^2) 2.0×10^11 1.86×10^11 –7.0
${E}_{2}$ (N/m^2) 2.0×10^11 1.73×10^11 –13.5
${E}_{3}$ (N/m^2) 2.0×10^11 1.72×10^11 –14.0
${E}_{4}$ (N/m^2) 2.0×10^11 1.83×10^11 –8.5
${E}_{1}\text{}{E}_{4}$ are selected parameters to be updated in the experimental case study. Comparison of the analytical, experimental, and updated response functions are shown in Fig. 12. The
updated parameters are shown in Table 4. It is shown that, after model updating, the finite element model can more accurately reflect the dynamic characteristics in the frequency domain of 0-1000 Hz.
5. Conclusions
Two model updating methods using frequency response function data have been investigated. On the basis of results from the comparative study, the following conclusions are drawn.
1) When the position of modeling errors are known in advance, the sensitivity method is good for solve the case with small errors. And it is difficult to get a convergent solution by using of the
sensitivity method when the initial error is large, this indicated that the initial value of parameters to be updated and the optimization constraints are important.
2) With regard to the system matrix error method. Under the condition of coordinates are measured completely, the frequency point can be selected randomly, and it is found that the agreement between
the exact error and the identified error is excellent even when large modeling error exists; this method can also be used for localizing the inaccurate parameters. For the case of incomplete
coordinates, it is not easy get a convergent updating solution. However, if we know the stiffness error position in advance, the identification of stiffness error can be obtained.
3) The experimental cantilever beam is updated by adopting the sensitivity method with tested frequency response function, it is shown that the sensitivity method is effective even when the test data
are extremely incomplete.
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About this article
model updating
sensitivity method
error of system matrix
frequency response function
The work described in this paper was supported by a Program for New Century Excellent Talents in University a research grant (NCET-11-0086), a research grant by the National Natural Science
Foundation of China (10902024), a Doctoral Program of Higher Education of China (20130092120039), and a project funded by the Priority Academic Program Development of Jiangsu Higher Education
Institutions (PAPD-1105007001).
Copyright © 2014 JVE International Ltd.
This is an open access article distributed under the
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Discrete and Algorithmic Mathematics
Ramon Llull Prize
The Ramon Llull Prize in Discrete Mathematics is an academic award for outstanding doctoral thesis in Discrete Mathematics. It is awarded every two years by the Spanish Red de Matemática Discreta y
Algorítmica (MDA). The prize winner is announced during the biennial edition of the Discrete Mathematics Days (DMD). Next edition of DMD will be held in July 3-5 2024 in Alcalá de Henares:
The awardee will give a prize talk at the conference. Travel within Europe and accommodation expenses will be covered by the Red MDA. The composition of the international jury of the prize will be
announced at the meeting.
The topics of the prize cover all areas related to Discrete Mathematics, including but not limited to
• Coding Theory and Cryptography
• Combinatorial Number Theory
• Combinatorics
• Discrete and Computational Geometry
• Discrete Optimization
• Graph Theory
• Theoretical Computer Science
Eligibility of the candidates:
• Candidates must have defended their doctoral thesis from 1st January 2019 to 31st December 2023.
• Candidates must either have defended the doctoral thesis in a Spanish university or be of Spanish nationality.
• Candidates must have written the doctoral thesis in English.
• Candidates must present their candidacy by sending to the President of the Jury, Prof. Gabor Lugosi at gabor.lugosi@gmail.com the following documents:
□ CV.
□ Copy of the doctoral thesis and publications arising from it.
□ Two recommendation letters to be sent directly by the recommenders to the President of the Jury.
The deadline for the first edition of the Prize is 31 of January, 2024.
The list of the prize winners is the following.
1st Edition
Jury: Gábor Lugosi (UPF-ICREA, chair), Frédéric Havet (CRNS-INRIA, Sophia Antipolis), Peter Keevash (Oxford)
Alberto Espuny Díaz (Heidelberg)
The 1st Ramon Llull Prize in Discrete Mathematics is awarded to Alberto Espuny Díaz for his PhD dissertation "Hamiltonicity problems in random graphs". The thesis contains several excellent results,
most notably the solution of a 40-year old conjecture of Bollobás regarding the threshold for the existence of a Hamilton cycle in the percolated hypercube, a truly outstanding achievement.
Slides of the ceremony.
Who was Ramon Llull?
Ramon Llull (1232-1316) was a medieval philosopher born in Mallorca in 1232. Professor at the University of Paris, indefatigable traveler, he wrote more than 200 books on religion, phylosophy,
science, moral and social order. He introduced a method named Ars combinatoria, a cosmogony integrating science, philosophy and religion in a common combinatorial framework, a complex devise
synthesized in his Combinatorial Wheel, which intended to unify all knowledge and solve all problems, creating a vast network that connected men with God. Apart from his contribution to
Combinatorics, Llull is considered to be the inventor of Condorcet voting method, 500 years before it was introduced by the Marquis of Condorcet. The Lullian Art had a notable influence on the
development of Mathematics, in particular in the work of Leibniz on formal logic Dissertio de Arte Combinatoria.
• John Fauvel and Robin J. Wilson, “The Lull before the storm: Combinatorics and religion in the Renaissance”, Bulletin of the Institute of Combinatorics and its Applications (ICA) , 11 (1994), pp.
• Martin Gardner, Logic Machines and Diagrams, 2nd edition, University of Chicago Press, 1982
• G. Hägele and F. Pukelsheim (2001). “Llull’s writings on electoral systems”. Studia Lulliana. 41: 3–38. Archived from the original on 2006-02-07. | {"url":"https://dam-network.github.io/ramon-llull/","timestamp":"2024-11-09T01:25:15Z","content_type":"text/html","content_length":"11641","record_id":"<urn:uuid:22f0e80a-a608-4afb-bdb4-0db2ba08e8b3>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00401.warc.gz"} |
Z-Score Price Breakout Strategy
1. Z-Score Price Breakout Strategy
Z-Score Price Breakout Strategy
, Date: 2023-12-07 15:17:43
The Z-Score price breakout strategy uses the z-score indicator of price to determine whether the current price is in an abnormal state, so as to generate trading signals. When the z-score of price is
higher or lower than a threshold, it means the price has entered an abnormal state, at which point long or short positions can be taken.
Strategy Principle
The core indicator of this strategy is the z-score of price, calculated as follows:
Z_score = (C - SMA(n)) / StdDev(C,n)
Where C is the closing price, SMA(n) is the simple moving average of n periods, and StdDev(C,n) is the standard deviation of closing price for n periods.
The z-score reflects the degree of deviation of the current price from the average price. When the price z-score is greater than a certain positive threshold (e.g. +2), it means the current price is
above the average price by 2 standard deviations, which is a relatively high level. When it is less than a certain negative threshold (e.g. -2), it means the current price is below the average price
by 2 standard deviations, which is a relatively low level.
This strategy first calculates the z-score of price, then sets a positive and negative threshold (e.g. 0 and 0). When the z-score is higher than the positive threshold, it generates a buy signal.
When lower than the negative threshold, it generates a sell signal.
Advantage Analysis
• Using price z-score to judge price anomalies is a common and effective quantitative method
• Easily achieve both long and short trading
• Flexible parameter settings, adjustable cycle, threshold, etc.
• Can be combined with other indicators to form a trading system
Risk Analysis
• The z-score strategy is crude and prone to false signals
• Need to set appropriate parameters like cycle and threshold
• Need to consider stop loss strategies to control risk
Optimization Directions
• Optimize cycle parameters to find the best cycle
• Optimize positive and negative thresholds to reduce false signals
• Add filter conditions, combine with other indicators
• Add stop loss strategies
The z-score price breakout strategy judges whether the current price is in an abnormal state, and trades according to the positive and negative of the price z-score. This strategy is simple and easy
to implement, allows two-way trading, but also has some risks. By optimizing parameters, adding stop loss and combining with other indicators, this strategy can be enhanced to form a complete
quantitative trading system.
start: 2023-11-29 00:00:00
end: 2023-12-04 19:00:00
period: 15m
basePeriod: 5m
exchanges: [{"eid":"Futures_Binance","currency":"BTC_USDT"}]
// Copyright by HPotter v1.0 18/01/2017
// The author of this indicator is Veronique Valcu. The z-score (z) for a data
// item x measures the distance (in standard deviations StdDev) and direction
// of the item from its mean (U):
// z = (x-StdDev) / U
// A value of zero indicates that the data item x is equal to the mean U, while
// positive or negative values show that the data item is above (x>U) or below
// (x Values of +2 and -2 show that the data item is two standard deviations
// above or below the chosen mean, respectively, and over 95.5% of all data
// items are contained within these two horizontal references (see Figure 1).
// We substitute x with the closing price C, the mean U with simple moving
// average (SMA) of n periods (n), and StdDev with the standard deviation of
// closing prices for n periods, the above formula becomes:
// Z_score = (C - SMA(n)) / StdDev(C,n)
// The z-score indicator is not new, but its use can be seen as a supplement to
// Bollinger bands. It offers a simple way to assess the position of the price
// vis-a-vis its resistance and support levels expressed by the Bollinger Bands.
// In addition, crossings of z-score averages may signal the start or the end of
// a tradable trend. Traders may take a step further and look for stronger signals
// by identifying common crossing points of z-score, its average, and average of average.
// You can change long to short in the Input Settings
// Please, use it only for learning or paper trading. Do not for real trading.
strategy(title="Z-Score Strategy", shorttitle="Z-Score Strategy")
Period = input(20, minval=1)
Trigger = input(0)
reverse = input(false, title="Trade reverse")
hline(Trigger, color=purple, linestyle=line)
xStdDev = stdev(close, Period)
xMA = sma(close, Period)
nRes = (close - xMA) / xStdDev
pos = iff(nRes > Trigger, 1,
iff(nRes < Trigger, -1, nz(pos[1], 0)))
possig = iff(reverse and pos == 1, -1,
iff(reverse and pos == -1, 1, pos))
if (possig == 1)
strategy.entry("Long", strategy.long)
if (possig == -1)
strategy.entry("Short", strategy.short)
barcolor(possig == -1 ? red: possig == 1 ? green : blue )
plot(nRes, color=blue, title="Z-Score") | {"url":"https://www.fmz.com/strategy/434552","timestamp":"2024-11-02T01:57:36Z","content_type":"text/html","content_length":"13767","record_id":"<urn:uuid:d0909371-7596-4c90-ae76-6e4413658c0f>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00781.warc.gz"} |
Solution: Altius
Answer: PEAKCCOLO
Written by Jonathan
With the help of the answers we already have and the skyscraper grids, we note that each skyscraper grid has 6 sets of 6 ?s, and there are 6 answers. This suggests we can associate each answer with 6
numbers from 1-6 somehow. Given the answers teams have, one thing to try would simply be to convert the letters to numbers A1Z26. We note that each string we already have has 6 digits in them, which
fits the skyscrapers.
Now we can start associating answers to skyscrapers. The easiest break in comes from ABBEY, because of the direction of the given arrows and the 1 in ABBEY, it can only fit at the bottom of grid 1.
VOW becomes the only word that can pair at the top of grid 1 with ABBEY. Of the remaining 4 words, CUBED is the only word that can belong to the top of Grid 3, and then it has only one match in BLEW.
For grid 2, the 2 words are fixed because of the 4 given at the bottom of C2 (otherwise 6 has to be in R3C2 which contradicts the 5 in R3), so COVE belongs at the top and WELD on the left. We can
then solve each of the 3 skyscrapers.
Extracting and ordering as given gives the meta answer PEAKCCOLO.
Author's Notes
In this meta, teams could backsolve the puzzles in the meta without having solved the meta. I had a list of 2 3-letter words, 13 4-letter words, and 15 5-letter words to choose from, and many were
pretty close to each other. The constraints include the fact that the answers can only contain the letters ABCDEFKLMNOPUVWXYZ, with a 1 occuring exactly once (so exactly one of AKLMNOPU appears in
the answer). Given 2 answers per grid, a lot of the words with many 5s and 6s could not fit either. It was a pretty fun and cool process constructing this meta! | {"url":"https://www.puzzlerojak.xyz/solution/altius.html","timestamp":"2024-11-09T13:50:10Z","content_type":"text/html","content_length":"6193","record_id":"<urn:uuid:47eb399e-3524-4d38-8fdd-cf1081f8ff35>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00655.warc.gz"} |
Comparing Floyd's Heapsort to Williams Implementations
Heaps are ubiquitous in computer science. If you need the minimum, or maximum of a collection, few choices are better than a heap for getting them. Heap-forming algorithms are the foundation of many
different algorithms and data structures. From heapsort to priority queues, to huffman coding, heaps really are everywhere.
A max heap
Heaps make use of two fundamental operations: siftup and siftdown, which take an index of a position in a heap in place it in a heap ordered position. When an entire collection is placed in heap
order, its called heapifying the collection.
template <class T>
void siftup(T a[], int n, int k) {
T v = a[k];
while (k > 0 && a[k/2] <= v) {
a[k] = a[k/2];
k = k/2;
a[k] = v;
template <class T>
void siftdown(T a[], int n, int k) {
T v = a[k];
while (2*k < n) {
int j = k+k;
if (j < n && a[j] < a[j+1]) j++;
if (v >= a[j]) break;
a[k] = a[j]; k = j;
a[k] = v;
When the a binary heap is used as an explicit data structure, as is the case with a priority queue, it is often built using siftup when inserting a value and siftdown upon removing the root. And so
when heapsort was first published, it followed suit.
To siftDown or siftUp?
J. Williams utilized both siftup and siftdown in his original heapsort implementation. It is Robert Floyds "Tree-Sort" that was published a few months later that we know commonly today as heapsort.
Heapsort can generally be separated into two phases:
1. The initial "heapifying "of the input array when the collection to be sorted is initially transformed into heap order
2. The "pop and maintain" phase which continuously removes the root and placing it in it's sorted position, followed by the re-heapifying of the unsorted portion of the collection.
void heapsort(int a[], int l, int r) {
int n = r-l;
int *pq = a+l;
//phase 1
makeHeap(pq, n);
//phase 2
while (n > 0) {
exch(pq, 0, n--);
siftdown(pq, n, 0);
It is the first phase of the algorithm which differs between the two approaches. Williams original implementation builds the heap using siftUp():
void makeHeap_williams(int a[], int n) {
for (int i = 1; i < n; i++)
siftup(a, n, i);
Floyds version on the other hand makes uses siftDown(). Notice the difference in both array bounds, and direction of iteration between the two versions.
void makeHeap_floyd(int a[], int l, int r) {
for (int i = n/2; i >= 0; i--)
siftdown(a, n, i);
It seems at first glance to be a subtle difference of little importance: they both transform the provided array into heap order, but one goes up and one goes down. The truth of the matter though, is
that Floyd's heapify runs closer to O(n) while Williams version is O(n log n) and that is not a subtle difference.
//The heaps produced in the first phase of both algorithms, and the total //running time of both - Floyd's is more than 2x faster!!
Williams heapify:
Heap Sort completed succesfully in 0.04622
Floyds heapify:
Heap Sort completed succesfully in 0.02129
As you can see, both algorithms build valid heaps, but they don't build the same heaps, and Floyds version does it much faster. Even at small values of N, its apparent that building the heap bottom
up is the clear winner. Why?
Further Reading
[1] http://math0.wvstateu.edu/~baker/cs405/code/HeapSort.pdf
[2] https://en.wikipedia.org/wiki/Heapsort
[3] My templated C++ implementation of all algorithms discusses in this post is available on my github | {"url":"http://maxgcoding.com/another-look-at-heapifying","timestamp":"2024-11-03T05:59:59Z","content_type":"text/html","content_length":"14443","record_id":"<urn:uuid:1ad5debe-0988-408d-8dfe-2c90581bb37a>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00434.warc.gz"} |
Example: Optimal Auctions for Beta Distributions
This example considers the problem of selling of 2 items to a single additive buyer whose values for the items are distributed independently according to Beta distributions. Note that in the special
case with Beta(1,1), the value for each item is distributed uniformly in [0,1].
The figure shows how the framework in the papers "Mechanism Design via Optimal Transport" and "Strong Duality for a Multiple-Good Monopolist" can be applied to compute the optimal mechanism that
maximizes the seller's expected revenue:
• The square represents the region [0,1]^2 where the measure lies, the dark shaded area is where the measure is negative while the light shaded area is where it is positive.
• The red dashed line gives the position of the first 0 when integrating from right to left, while the blue dashed line gives the position of the first 0 when integrating from top to bottom.
• The thick black line gives the optimal price for the grand-bundle of both items.
• The solid black, blue and red lines partition the square in at most 4 regions:
1. The region where the buyer gets both items with probability 1.
2. The region where the buyer gets no items
3. The region where the buyer gets item 1 with probability 1 and item 2 with probability strictly less than 1.
4. The region where the buyer gets item 2 with probability 1 and item 1 with probability strictly less than 1.
In the regions where the probability for an item is between 0 and 1, the probability is given by the slope of the corresponding curve at that point.
• The buyer's utility is 0 below the thick solid curve. Above the thick solid curve, it is given by the L[1] distance to the curve. | {"url":"https://tzamos.com/betas/","timestamp":"2024-11-04T21:57:23Z","content_type":"text/html","content_length":"14045","record_id":"<urn:uuid:5e0cac32-519e-41e5-af6c-f36e1dd1a599>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00746.warc.gz"} |
Incidences Between Points and Lines on Two- and Three-Dimensional Varieties
Let P be a set of m points and L a set of n lines in R^4, such that the points of P lie on an algebraic three-dimensional variety of degree D that does not contain hyperplane or quadric components (a
quadric is an algebraic variety of degree two), and no 2-flat contains more than s lines of L. We show that the number of incidences between P and L is (Formula Presented.) for some absolute constant
of proportionality. This significantly improves the bound of the authors (Sharir, Solomon, Incidences between points and lines in R^4. Discrete Comput Geom 57(3), 702–756, 2017), for arbitrary sets
of points and lines in R^4, when D is not too large. Moreover, when D and s are constant, we get a linear bound. The same bound holds when the three-dimensional surface is embedded in any
higher-dimensional space. The bound extends (with a slight deterioration, when D is large) to the complex field too. For a complex three-dimensional variety, of degree D, embedded in C^4 (or in any
higher-dimensional C^d), under the same assumptions as above, we have (Formula Presented.). For the proof of these bounds, we revisit certain parts of [36], combined with the following new incidence
bound, for which we present a direct and fairly simple proof. Going back to the real case, let P be a set of m points and L a set of n lines in R^d, for d≥ 3 , which lie in a common two-dimensional
algebraic surface of degree D that does not contain any 2-flat, so that no 2-flat contains more than s lines of L (here we require that the lines of L also be contained in the surface). Then the
number of incidences between P and L is (Formula Presented.) When d= 3 , this improves the bound of Guth and Katz (On the Erdős distinct distances problem in the plane. Ann Math 181(1), 155–190,
2015) for this special case, when D≪ n^1 / 2. Moreover, the bound does not involve the term O(nD). This term arises in most standard approaches, and its removal is a significant aspect of our result.
Again, the bound is linear when D= O(1). This bound too extends (with a slight deterioration, when D is large) to the complex field. For a complex two-dimensional variety, of degree D, when the
ambient space is C^3 (or any higher-dimensional C^d), under the same assumptions as above, we have (Formula Presented.). These new incidence bounds are among the very few bounds, known so far, that
hold over the complex field. The bound for two-dimensional (resp., three-dimensional) varieties coincides with the bound in the real case when D= O(m^1 / 3) (resp., D= O(m^1 / 6)).
Funders Funder number
Blavatnik Computer Science Research Fund
Hermann Minkowski-MINERVA Center for Geometry
United States-Israel Binational Science Foundation
Israel Science Foundation 2012/229
Tel Aviv University
Israeli Centers for Research Excellence 4/11
• Algebraic techniques for discrete geometry
• Geometric incidences
• Lines on varieties
• Polynomial partitioning
• Ruled surfaces
Dive into the research topics of 'Incidences Between Points and Lines on Two- and Three-Dimensional Varieties'. Together they form a unique fingerprint. | {"url":"https://cris.tau.ac.il/en/publications/incidences-between-points-and-lines-on-two-and-three-dimensional-","timestamp":"2024-11-09T06:22:11Z","content_type":"text/html","content_length":"57659","record_id":"<urn:uuid:404ba626-fadd-438c-ab2e-2483623ad624>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00644.warc.gz"} |
Title : Alternation, Degree and Sensitivity : New Bounds and Super-linear Gaps
Speaker : Krishnamoorthy Dinesh (IITM)
Details : Tue, 29 Aug, 2017 2:00 PM @ Alan Turing Hall
Abstract: : Often while solving computational problems, we would want to understand how ``sensitive'' the solution
is to changes in the input instance. Abstracting the computational problem as a Boolean function, the
sensitivity (denoted ​s​(f)) is defined to be the number of input bits (maximized over all input
settings) which if flipped, changes the value of the function. If instead of a single bit, we allow a
block of bits to be flipped and we want to count the maximally disjoint such blocks, we get to a notion
called block sensitivity (denoted bs​(​f)). Nisan and Szegedy (1992) conjectured that for every
Boolean function $f$, the block sensitivity is upper bounded polynomially by sensitivity. Attempts to
resolve this conjecture, popularly called as the ​​sensitivity ​conjecture, has led to the
discovery of connections between several other parameters of Boolean functions.
In this ​work, we show new relations and establish gaps between the Boolean function parameters --
alternation, sensitivity, degree and F_2-degree of Boolean function f denoted alt(f), s(f), deg(f) and
F_2-def(f) respectively. A motivation for this study is the recent result of Lin and Zhang (2016) who
showed that the Sensitivity Conjecture holds for Boolean functions whose alternation is upper bounded by
polynomial in sensitivity.
​(1) We show that there exists a family of Boolean functions for which alt(f) is super-linear in s(f),
alt(f) is super-linear in deg(f) and s(f) is super-linear in F_2-deg(f). The main tool used is a bound
on alternation of composed Boolean functions.​ ​En-route the above proof, we show that for any
Boolean function f, there exists a linear transform L such that alt(f) is at most 2s(g)+1 where g(x) is
​(2) We ​use​ th​e the arguments in Lin and Zhang (2016) implies that for any Boolean function
f, deg(f) is at most alt(f)​ x F_2-deg(f)^2. As two quick applications, we answer a question of
Kulkarni and Santha (2013) on asymptotically bounding logarithm of sparsity of monotone Boolean
functions by its degree and we improve a bound on influence of Boolean functions by Guo and Komargodski | {"url":"https://cse.iitm.ac.in/seminar_details.php?arg=MjUw","timestamp":"2024-11-07T02:51:56Z","content_type":"application/xhtml+xml","content_length":"15232","record_id":"<urn:uuid:0187868a-d6e2-4648-a5ae-f892140369a0>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00137.warc.gz"} |
SSC CGL Set 33
Submitted by Atanu Chaudhuri on Sun, 03/07/2016 - 13:44
Solutions to quantitative algebra questions SSC CGL Set 33
Learn to solve quantitative questions on algebra. Know how to solve 10 Algebra questions in SSC CGL set 33 in 12 minutes. Learn algebra problem solving.
For best results take the test first at,
33rd SSC CGL question set and 9th on Algebra.
Learn to solve quantitative questions on algebra - answering time was 12 minutes
Q1. If $x = 2.361$, $y=3.263$, and $z=5.624$, then the value of $x^3 + y^3 - z^3 + 3xyz$ is,
a. 35.621
b. 1
c. 0
d. 19.277
Solution - Problem analysis
The target expression being a well-known cube expression in three variables, we remember vaguely an expression similar to $x+y+z = 0$ associated with it so that the value of the target expression is
very much simplified.
With this goal in view we try to discover an additive-subtractive relationship between $x$, $y$ and $z$ and quickly find that,
$x + y = z$.
Solution - Simplifying actions
We have $x + y = z$
Or, $x^3 + y^3 + 3xy(x+y) = z^3$
Or, $x^3 + y^3 - z^3 + 3xyz = 0$.
Answer: Option c: 0.
Key concepts used: First establishing simple relationship between $x$, $y$, and $z$ -- using cube of sums expression in two variables -- basic algebra concepts.
Q2. If $6 + \displaystyle\frac{1}{x}=x$, then the values of $x^4 + \displaystyle\frac{1}{x^4}$ is,
a. 1444
b. 1442
c. 1448
d. 1446
Solution - Problem analysis:
The target expression being in the form of sum of inverses the given expression also needs to be transformed into a sum of inverses in single power of $x$.
$6 + \displaystyle\frac{1}{x}=x$,
Or, $x - \displaystyle\frac{1}{x} = 6$.
We will now derive higher powers of sum of inverses in two stages to arrive at the sum of inverses in fourth power of $x$.
Solution - Simplifying actions
$x - \displaystyle\frac{1}{x} = 6$,
Or squaring both sides,
$x^2 - 2 + \displaystyle\frac{1}{x^2} = 36$
Or, $x^2 + \displaystyle\frac{1}{x^2} = 38$.
Squaring both sides again,
$x^4 + 2 + \displaystyle\frac{1}{x^4} = 38^2=1444$,
Or, $x^4 + \displaystyle\frac{1}{x^4} = 1444 - 2=1442$.
Answer: Option b : 1442.
Key concepts used: Observing that the target expression is in the form of sum of inverses, the given expression is also transformed to a sum (subtraction) of inverses and in two stages of squaring
the value of the target expression is reached -- sum of inverses concept.
You may refer to the article on principle of inverses for more details on how effectively the concept on sum of inverses can be used.
Q3. If $x^2 + \displaystyle\frac{1}{x^2} = 66$, then the value of $\displaystyle\frac{x^2 - 1 + 2x}{x}$ is,
a. $\pm{8}$
b. $6, -10$
c. $10, -6$
d. $\pm{4}$
Solution - Problem analysis:
Usually we have to derive value of higher powers of sum of inverses when lower power of sum of inverses is given. In this problem, the situation is opposite and from higher powers, we have to derive
lower powers of sum of inverses. But first we need to verify the nature of the target expression to see whether it is really a sum of inverses or not.
$\displaystyle\frac{x^2 - 1 + 2x}{x} = x - \displaystyle\frac{1}{x} + 2$.
So the target expression is really in the form of sum of inverses.
Solution - Simplifying actions:
Given expression,
$x^2 + \displaystyle\frac{1}{x^2} = 66$,
Or, $x^2 - 2 + \displaystyle\frac{1}{x^2} = 64$,
Or, $\left(x - \displaystyle\frac{1}{x}\right)^2 = 8^2$
Or, $x - \displaystyle\frac{1}{x} = \pm{8}$
So the target expression when evaluated is,
$x - \displaystyle\frac{1}{x} + 2 = \pm{8} + 2 = 10, -6$.
Answer: Option c: $10, -6$.
Key concepts used: From higher powers of sum of inverses value of lower power of sum of inverses derived after confirming that the target expression is really a sum of inverses -- principle of
inverses -- working backwards approach.
Q4. Find the minimum value of $2x^2 - (x - 3)(x + 5)$, where $x$ is real,
a. 20
b. 14
c. -12
d. 8
Solution - Problem analysis
As this is a minimization of quadratic expression problem, the sign of $x^2$ term should be positive. But before any significant action can be taken on the target expression, it needs to be converted
in proper form consisting of a square term in $x$ where all terms of $x$ are absorbed leaving only a numeric term outside the square term.
Let us transform the given expression in that specific form and then apply the reasoning of minimization.
Solution - Problem simplification
$2x^2 - (x - 3)(x + 5)$
$=2x^2 -(x^2 + 5x - 3x - 15)$
$=2x^2 - x^2 - 2x + 15$
$=x^2 -2x +15$
$=(x-1)^2 + 14$.
We can now apply our reasoning.
As $x$ is real, the square in $x$, that is, $(x-1)^2$ will always have a positive or zero value and the target expression will have the minimum value only when the square expression is zero at $x
In that case the minimum value of the target expression is 14.
Answer: Option b: 14.
Key concepts used: Maxima minima technique -- Transforming the given expression in terms of an expression containing a square term in $x$ absorbing all terms in $x$ so that the remaining numeric term
can be identified as the minimum value of the expression.
Q5. If $x+y=7$ then the value of $x^3 + y^3 + 21xy$ is,
a. 243
b. 143
c. 443
d. 343
Solution - Problem analysis:
To use the value of the given expression, the target expression needs to be expressed as a cube of sum expression.
Solution - Simplifying actions
$=x^3 + y^3 + 3xy\times{7}$
$=x^3 + y^3 + 21xy$,
Or, $7^3 = x^3 + y^3 + 21xy$,
Or, $x^3 + y^3 + 21xy = 343$.
Answer: Option d: 343.
Key concepts used: Using the cube of sum expression and the value of given expression to transform the target expression suitably -- basic algebra concepts.
Q6. If $x=\displaystyle\frac{\sqrt{3}}{2}$ then the value of $\displaystyle\frac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} - \sqrt{1-x}}$ will be,
a. $-\sqrt{3}$
b. $1$
c. $\sqrt{3}$
d. $-1$
Solution 1: Quick solution by principle of target expression simplification first and pattern discovery
Following the golden principle of algebraic expression simplification, we first try to see whether the target expression can be simplified by itself before substitution of the surd value of $x$.
We find it easily possible by multiplying both numerator and denominator with $\sqrt{1+x}+\sqrt{1-x}$. This is the key pattern discovery.
Denominator reduces to just $2x$,
And the numerator reduces to a numeric value,
Result is,
Answer: Option c : $\sqrt{3}$.
This is a quick solution all in mind without the complexity of double square root surd simplification.
Solution 2: By double square root surd simplification: Problem analysis:
As the value of $x$ contains a square root and the target expression is in terms of square root of $x$, it is apparently a case of double square root surd simplification.
To free $\sqrt{1+x}$ and $\sqrt{1-x}$ of the double square roots, we need to transform each of the expressions under the square roots in terms of a square of surd expression.
Let us try to transform the square root expressions to the desired form.
Solution 2: Simplifying steps
$\sqrt{1 +x} = \sqrt{1 + \displaystyle\frac{\sqrt{3}}{2}} $
$= \sqrt{\displaystyle\frac{2 + \sqrt{3}}{2}} $
$= \sqrt{\displaystyle\frac{4+2\sqrt{3}}{4}}$
$=\displaystyle\frac{1}{2}\sqrt{(\sqrt{3} + 1)^2}$
$=\displaystyle\frac{1}{2}(\sqrt{3} + 1)$.
$\sqrt{1 - x} = \displaystyle\frac{1}{2}(\sqrt{3} - 1)$.
So we have the the target expressoion as,
$\displaystyle\frac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} - \sqrt{1-x}}$
$=\displaystyle\frac{\sqrt{3} + 1 + \sqrt{3} - 1}{\sqrt{3} + 1 - \sqrt{3} + 1}$, the $\frac{1}{2}$ canceled out.
Answer: Option c : $\sqrt{3}$.
Key concepts used: Freeing a surd expression out of a square root by expressing a two term surd expression as a three term expanded square of surd expression -- simplification.
Before going into the complexity of double square root surd simplification, you should try to look for any feasible way of avoiding the complexity. This would save valuable solving time.
Q7. If $p^3 + 3p^2 + 3p = 7$ then the value of $p^2 + 2p$ is,
a. 3
b. 4
c. 5
d. 6
Solution - Problem analysis:
Identifying the given expression as a part of a cube of sum expression we examine the target expression to find it also as a part of square expression in a sum of $x$.
Solution - Simplifying actions
$p^3 + 3p^2 + 3p = 7$,
Or, $p^3 + 3p^2 + 3p + 1 = 8$,
Or, $(p+1)^3 = 2^3$,
Or, $p+1 = 2$,
Or, $(p+1)^2 = p^2 + 2p + 1 = 4$.
So finally,
$p^2 + 2p = 3$.
Answer: Option a: 3.
Key concepts used: Identifying the given expression as a part of cube of sum expression as well as the target expression as a part of square of sum expression -- basic algebra concepts.
Q8. If $\left(x + \displaystyle\frac{1}{x}\right)^2 = 3$ then the value of $(x^{72} + x^{66} + x^{54} + x^{36} + x^{24} + x^6 + 1)$ is,
a. 4
b. 2
c. 3
d. 1
Solution: Quick solution by pattern discovery based on difference in powers of consecutive terms and mathematical reasoning
The target expression has 7 terms and highest power of $x$ is 72. Seems to be a complex problem. There must be some pattern in the expression, and this usually is a relation between two consecutive
terms. We are using mathematical reasoning.
And when we look for it, we find an interesting pattern between each of the three pairs of terms—the power difference between the first two terms is 6 and for the next two pairs of terms it is 18.
For two terms in $x$ with power difference of 6 you can get a factor $x^3+\displaystyle\frac{1}{x^3}$ out of the sum of two terms.
Let's show you,
In the same way when the power difference between two terms in $x$ is 18, you get a factor of $x^9+\displaystyle\frac{1}{x^9}$ out of the sum of two terms.
$x^{54}+x^{36}=x^{45}\left(x^9+\displaystyle\frac{1}{x^9}\right)$, and
The job is now to get the value of $x^3+\displaystyle\frac{1}{x^3}$ and then of $x^9+\displaystyle\frac{1}{x^9}$ from the given expression.
At this point itself we are sure from the simple numeric choice values that both $x^3+\displaystyle\frac{1}{x^3}$ and $x^9+\displaystyle\frac{1}{x^9}$ are zero and answer will be option d: 1. There
cannot be any term in $x$ left over.
Still let us check whether our reasoning is true.
Given expression is,
Or, $x^2-1+\displaystyle\frac{1}{x^2}=0$.
This is the second factor in the two-factor expansion of sum of inverses of cubes,
And so, $x^9+\displaystyle\frac{1}{x^9}$ also is 0 as it has the first factor in its two-factor expansion of sum of cubes as $x^3+\displaystyle\frac{1}{x^3}$, which is 0.
The first six terms then add to 0 leaving 1 as the final result.
Answer: Option d: 1.
Key concepts used: Mathematical reasoning -- Pattern discovery based on difference in power of two consecutive terms -- Two-factor expansion of sum of cubes.
Q9. If $a^2 + b^2 + c^2 = 2(a -b -c) -3$, then $4a - 3b + 5c$ is,
a. 3
b. 2
c. 5
d. 6
Solution - Problem analysis:
As the target expression is asymmetric and not easily relatable to the given expression, it is expected that actual values need to be found out for $a$, $b$ and $c$ for substitution and final
Solution - Simplifying actions
We analyze the given expression and gather friendly terms on the LHS,
$a^2 + b^2 + c^2 = 2(a -b -c) -3$,
Or, $(a-1)^2 + (b+1)^2 + (c+1)^2 = 0$
As the sum of squares is 0, each of the squares must be 0.
So, $a = 1$, $b=-1$ and $c=-1$.
Thus the target expression is,
$4a - 3b + 5c = 4 + 3 - 5 = 2$.
Answer: Option b: 2.
Key concepts used: Using principle of collection of friendly terms for expressing the given expression as a sum of squares equal to 0 -- resulting each of the square to become 0 -- evaluation of the
variables and the target expression value.
Q10. If $3x + \displaystyle\frac{1}{2x} = 5$, then the value of $8x^3 + \displaystyle\frac{1}{27x^3}$ is,
a. $118\frac{1}{2}$
b. $0$
c. $30\frac{10}{27}$
d. $1$
Solution - Problem analysis:
The given expression though is a sum of inverse its coefficients do not exactly correspond with those of the target inverse expression. We need to first make the coefficients of the corresponding
terms of the given and the target expression conform to each other. Effectively we have to apply coefficient conformation technique between the given and the target expressions.
Solution - Simplifying actions
The target expression is a sum of inverses in power of cube of terms, $2x$ direct and $3x$ inverse. We will transform the given expression in that form.
$3x + \displaystyle\frac{1}{2x} = 5$,
Multiplying both sides by $\displaystyle\frac{2}{3}$ for making the coefficients between the given and the target expressions conform we have,
$2x + \displaystyle\frac{1}{3x}=\displaystyle\frac{10}{3}$.
So by the sum of cubes expression,
$(2x)^3 + \left(\displaystyle\frac{1}{3x}\right)^3 $
$= \left(2x + \displaystyle\frac{1}{3x}\right)\left(\left(2x + \displaystyle\frac{1}{3x}\right)^2 - 3\times{2x}\times{\displaystyle\frac{1}{3x}}\right)$
$=\displaystyle\frac{10}{3}\left(\left(\displaystyle\frac{10}{3}\right)^2 - 2\right)$
Answer: Option c: $30\frac{10}{27}$.
Key concepts used: Input transformation to conform the given coefficients to the coefficients of similar terms of the target expression -- principle of inverses -- cube of sum expression --
coefficient conformation technique.
Guided help on Algebra in Suresolv
To get the best results out of the extensive range of articles of tutorials, questions and solutions on Algebra in Suresolv, follow the guide,
Suresolv Algebra Reading and Practice Guide for SSC CHSL, SSC CGL, SSC CGL Tier II and Other Competitive exams.
The guide list of articles includes ALL articles on Algebra in Suresolv and is up-to-date. | {"url":"https://suresolv.com/ssc-cgl/ssc-cgl-level-solution-set-33-algebra","timestamp":"2024-11-05T11:54:15Z","content_type":"text/html","content_length":"46027","record_id":"<urn:uuid:237eac4d-0497-460e-adb7-9bde0861cfab>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00800.warc.gz"} |
KS Editor Mod
Hey everybody. I've been working on a modification of the KS Editor.
Level Editor Modification by Looki, 2009
New features:
* Open a world by double-clicking it.
* Open a world by searching for it in the input box above the list.
* Hold down the arrow keys when navigating through the world.
* Custom objects are displayed on-screen if "Screen->Show custom objects" is set.
* In-built INI Editor with syntax highlighting:
---Press F3 or "Script" left to the tileset settings. Reload the original file by "Edit->Reload World.ini".
---(NOT WORKING) You can press on a room's group, e.g. [x1000y1000], to jump to the room.
---The font can be changed in Worlds\EditorSettings.temp.
* Tile replacing, Press Shift+Right click. Like Flood Fill, but not only for connected tiles.
* Rectangular selection: Select multiple tiles at once.
---You can right-click a tile to keep the current selection rectangle. You can press F to reset it.
* Rectangular filling: Hold down Ctrl while placing tiles. You can select multiple tiles to fill using a pattern.
* Undo, Press Ctrl+Z or "Edit->Undo". Undo only works for object and tile placement.
* Flood Fill, Press Shift+Click. Use can select multiple tiles to use a pattern. Only for tiles (single layer).
* Minor fix for German users by setting "Qwertz" in EditorSettings.temp to 1 (Z gets Y, bank-changing shortcut).
* The executable to use for testing is read from "Executable" in EditorSettings.temp.
* Press Ctrl+G or "Go to>Coordinates" to jump to a screen by entering its coordinates.
The Worlds folder only contains a EditorSettings.temp file which has to be updated in order to be able to test your level.
One note, undo only works per screen. If you switch the screen, all the undo information is lost.
I hope you like it as much as I do!
Get it | {"url":"https://nifflas.lp1.nl/index.php?topic=347.msg3037","timestamp":"2024-11-07T17:06:26Z","content_type":"application/xhtml+xml","content_length":"63324","record_id":"<urn:uuid:96a261ef-0c98-4ee8-8b4e-47bdf1c73a3e>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00728.warc.gz"} |
Time Value of Money Formula | Calculator (Excel template)
Updated July 31, 2023
Time Value of Money Formula (Table of Contents)
Time Value of Money Formula
The time value of money is a very important concept for each individual and for making important business decisions. Companies will consider the time value of money while deciding whether to
acquire new business equipment or to invest in new product development or facilities and establishing the credit terms for selling their services or products.
There are two aspects of the time value of money the first one is the future value of money, and the second one is the present value of money. In the future, the future value of money will represent
the worth of money invested today. Conversely, the present value of money will indicate the current value of an amount that will be received or paid in the future.
The formula to calculate the future value of money allows comparison with the present value of the money:
FV = PV * [ 1 + ( i / n ) ]^ (n * t)
PV = FV / [ (1 + i/n) ]^(n * t)
• FV = Future Value of Money
• PV = Present Value of Money
• i = Rate of interest
• t = number of years
• n = number of compounding periods per year
Examples of Time Value of Money Formula (With Excel Template)
Let’s take an example to understand the calculation of the Time Value of Money formula in a better manner.
Example #1
Let us Assume that a sum of money, say $100,000 is invested for two years at 8% interest. What will be the future value of the sum invested?
The formula to calculate Future Value is as below:
FV = PV * [ 1 + ( i / n ) ]^ (n * t)
• FV = 100,000 * [ 1 + ( 8% / 1 ) ] ^( 1 * 2 )
• FV = 116,640
Example #2
Below is the extract from the standard chartered bank deposit rate (recurring deposit) available for various periods. Now let’s assume that you decide to invest $100,000, say for period of 6 months,
then what is the value you would expect to receive?
Let’s now try to calculate the future value of money:
Here, PV is $100,000, the Rate of interest for 6 months applicable is 3.50% p.a., the number of years is 0.5 (1/2), and the number of compounding per period will be 2.
The formula to calculate Future Value is as below:
FV = PV * [ 1 + ( i / n ) ]^ (n * t)
• FV = 100,000 * [ 1 + ( 3.50% / 2 ) ] ^( 2 * 0.5 )
• FV = 1,01,750
Example #3
Below is again an extract of loan details from a standard chartered bank where the bank will lend, say 100,000 to its client at a rate of interest 10.99% and say the term is 2 years.
In this case, the bank will calculate the present value of the principal only, that is, the present value of the money which the bank would receive in the future.
Here the FV is 100,000, i is 10.99%, t is 2 years, and n is 1 year.
The formula to calculate the Present Value of the principal amount is as below:
PV = FV / [ (1 + i/n) ]^(n * t)
• PV = 100,000 / [ (1+10.99/1)]^(2*1)
• PV = 81,176.86913
The Time Value of Money concept will indicate that the money earned today will be more valuable than its fair value or intrinsic value in the future. This will be due to its earning capacity, which
will be the potential of the given amount. The time Value of Money (i.e. TVM) can also be called Discounted present value. Money deposited in the savings bank account will earn a certain interest
rate as it must compensate for keeping the amount of money invested by a client and away from them at the current period. Therefore, if any bank holder deposits $200 in the bank account, then the
customer will expect to receive more than $200 after 1 year.
Relevance and Uses
The time value of money is a wider concept and can also be related to purchasing power and inflation. Consider both factors along with the rate of return that investing the amount of money may
Why is this factor so much important? The reason is inflation constantly erodes the value of money and, henceforth, the purchasing power of the money. This can be best exemplified by the value or the
prices of commodities such as food or gas. Let’s say. For example, you were given a certificate for $150 of free gasoline in 1991. Then you could have purchased a lot more gallons of gasoline than
you could have in hand if you were given $150 of free gas a decade, i.e., 10 years later.
The time value of money formula can be used in many financial decision-making:
• Capital budgeting
• Valuation of companies
• Loan amount and EMI calculation
• Annuity Calculation
• Insurance premium calculation
Time Value of Money Formula Calculator
You can use the following Time Value of Money Calculator.
Future Value of Money =
Future Value of Money = PV * (1 + (i / n))^(n * t)
0 * (1 + (0 / 0))^(0 * 0) = 0
Recommended Articles
This has been a guide to the Time Value of Money formula. Here we discuss How to Calculate the Time Value of Money using FV Formula and practical examples. We also provide a Time Value of Money
Calculator with a downloadable excel template. You may also look at the following articles to learn more – | {"url":"https://www.educba.com/time-value-of-money-formula/","timestamp":"2024-11-09T17:46:17Z","content_type":"text/html","content_length":"343604","record_id":"<urn:uuid:7785f28a-d9dc-43ce-8f59-5bfbf177a343>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00465.warc.gz"} |
Add calculations to category groups in Numbers on iPad
For each group or subgroup in a categorized table, you can add a function (a predefined calculation such as sum, count, or average) to summarize the data in any column. For example, in a table with
shirt sales data categorized by month, you could add a function to count the number of shirts sold each month. You can add these functions to the summary row in each category in your table.
Add a function to a summary row cell
You can add a calculation, such as count, subtotal, or average, for any column of data in any group’s summary row. The same calculation is automatically added to all groups at the same level in the
category’s hierarchy.
• Tap an empty cell in the summary row, tap then choose a function.
A label row automatically appears above the summary row. These labels can’t be edited, but you can change their text style and hide them.
You can also sort groups by summary row values. See Sort table data in Numbers on iPad.
Remove a function from a summary row cell
• Tap the cell in the summary row with the function you want to remove, tap then tap No Summary.
Show or hide the label row
A label row is an optional row located above a summary row. It shows the name of the category and for any cell with a function added, the name of the function (count, subtotal, average, and so on).
• Select the summary row, tap then choose Show Label Row or Hide Label Row.
You can also chart the results of the calculations in a summary row column. Select the column containing the summary calculations you want to chart, tap tap Create New Chart, then select a chart | {"url":"https://support.apple.com/en-asia/guide/numbers-ipad/tan3ec8281cf/ipados","timestamp":"2024-11-04T15:59:00Z","content_type":"text/html","content_length":"188937","record_id":"<urn:uuid:333c7be9-7712-40cd-a5cf-3d4657563bc0>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00086.warc.gz"} |
The time required for a citizen to complete the 2000 U.S. - College School Essays
The time required for a citizen to complete the 2000 U.S.
Question 1
The time required for a citizen to complete the 2000 U.S. Census “long” form is normally distributed with a mean of 40 minutes and a standard deviation of 10 minutes. What proportion of the citizens
will require less than one hour?
Question 2
If the random variable Z has a standard normal distribution, then P(Z ≤ -1.37) is
Question 3
Bob’s z-score for the last exam was 1.52 in Prof. Axolotl’s class BIO 417 “Life Cycle of the Ornithorhynchus”. Bob said, “Oh, good, I’m in the top 10%.” Is he right?
Question 4
If arrivals occur at a mean rate of 1.6 events per minute, the exponential probability of waiting less than 1 minute for the next arrival is
Question 5
The lengths of brook trout caught in a certain Colorado stream have a mean of 14 inches and a standard deviation of 3 inches. The first quartile for the lengths of brook trout would be
Question 6
Assume that X is normally distributed with a mean μ= $64. Given that P(X ≥$75) = 0.2981, we can calculate that the standard deviation of X is approximately
Question 7
If arrivals follow a Poisson distribution with mean 1.2 arrivals per minute, find the 75th percentile of waiting times (i.e., 75 percent below).
Question 8
The MPG (miles per gallon) for a certain compact car is normally distributed with a mean of 31 and a standard deviation of 0.8. What is the probability that the MPG for a randomly selected compact
car would be less than 32?
Question 9
A letter is mailed to a sample of 500 homeowners. Based on past experience, the probability of an undeliverable letter is 0.06. The normal approximation to the probability of 40 or more undeliverable
letters is
Which model best describes your waiting time until you get the next non-working web URL (“This page cannot be displayed”) as you click on web sites for Florida condo rentals?
If you need assistance with writing your assignment, essay, our professional assignments / essay writing service is here to help!
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OptCirClust | CRAN/E
Circular, Periodic, or Framed Data Clustering
CRAN Package
Fast, optimal, and reproducible clustering algorithms for circular, periodic, or framed data. The algorithms introduced here are based on a core algorithm for optimal framed clustering the authors
have developed (Debnath & Song 2021) . The runtime of these algorithms is O(K N log^2 N), where K is the number of clusters and N is the number of circular data points. On a desktop computer using a
single processor core, millions of data points can be grouped into a few clusters within seconds. One can apply the algorithms to characterize events along circular DNA molecules, circular RNA
molecules, and circular genomes of bacteria, chloroplast, and mitochondria. One can also cluster climate data along any given longitude or latitude. Periodic data clustering can be formulated as
circular clustering. The algorithms offer a general high-performance solution to circular, periodic, or framed data clustering.
• Version0.0.4
• R versionunknown
• Needs compilation?Yes
• Last release07/28/2021
Last 30 days
Last 365 days
The following line graph shows the downloads per day. You can hover over the graph to see the exact number of downloads per day.
Data provided by cranlogs
• Imports7 packages
• Suggests5 packages
• Linking To1 package
• Reverse Imports1 package | {"url":"https://cran-e.com/package/OptCirClust","timestamp":"2024-11-03T10:28:17Z","content_type":"text/html","content_length":"70275","record_id":"<urn:uuid:2542c92a-7676-463e-b239-757a0b466b80>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00283.warc.gz"} |
Mastering Bilinear Interpolation with Numpy and Scipy - Adventures in Machine Learning
Bilinear Interpolation: An Introduction
In the world of computing and mathematics, interpolation refers to the estimation of values that lie between known data points. In other words, it’s the process of filling in the gaps.
Bilinear interpolation is a particular type of interpolation that is used to estimate the value of an unknown point in a two-dimensional space. It’s essentially a weighted average of the four nearest
known data points to the unknown point.
Bilinear interpolation is widely used in image processing, data analysis, computer vision, and digital terrain modeling to name a few.
Applications of Bilinear Interpolation
1. Image Processing
In digital image processing, bilinear interpolation is used to increase or decrease the size of an image. When an image is resized, the new pixel values have to be calculated based on the existing
pixel values.
Bilinear interpolation is ideal for this task because it can produce smooth and visually pleasing results.
2. Data Analysis
In data analysis, bilinear interpolation is used to estimate missing values in data sets.
For example, if you have a set of data points with some missing values, bilinear interpolation can be used to estimate the missing values.
3. Computer Vision
Bilinear interpolation is also widely used in computer vision to estimate the position, orientation, and scale of objects in an image.
This is particularly useful in object recognition and tracking.
4. Digital Terrain Modeling
In digital terrain modeling, bilinear interpolation is used to generate continuous elevation models by estimating the height of terrain points based on the height of known data points.
Bilinear Interpolation with Numpy
Numpy is a popular Python library used for scientific computing. It provides a fast and efficient array computing capability, which makes it ideal for tasks such as bilinear interpolation.
1. Manual Computation of Bilinear Interpolation
To manually compute bilinear interpolation with Numpy, we need to first break the process down into steps.
Given four known data points, we can compute the weighted average to estimate the value of an unknown point as follows:
1. Compute the linear interpolation along the x-axis for the two data points above the unknown point, and the two data points below the unknown point.
2. Compute the linear interpolation along the y-axis between the two provisional values from step 1, giving us the final estimated value.
2. Implementation of Bilinear Interpolation with Numpy
To implement bilinear interpolation with Numpy, we can write a function called bi_interpolation that takes in four known data points and the coordinates of the unknown point.
The function should then perform the steps outlined above and return the estimated value of the unknown point. For visualization purposes, we can create a grid of data points and plot the estimated
values using Numpy’s meshgrid and imshow functions.
This gives us a visual representation of how bilinear interpolation works.
In conclusion, bilinear interpolation is a handy tool in data analysis, computer vision, digital terrain modeling, and image processing. It can be implemented manually using Numpy or using built-in
functions that calculate the weighted averages much faster.
It behooves professionals in scientific computing to be well-versed in bilinear interpolation as it can be a valuable asset in exploring datasets, detecting outliers, and smoothing sharp edges.
Bilinear Interpolation with Scipy
Scipy is another popular Python library used for scientific computing and provides functions for interpolation, integration, optimization, and other mathematical operations. Scipy provides the
griddata function, which can be used to perform bilinear interpolation.
Using Scipy for Bilinear Interpolation
Interpolation requires a set of known data points to estimate new values. In Scipy, these known data points are called coordinate points.
The griddata function can then be used to interpolate these points and estimate the values at new points. When using griddata, the interpolation method used for each grid cell is determined by the
points and their distance to the new grid cell.
Scipy uses linear interpolation by default, but other methods, including bilinear interpolation, can be selected.
Implementation of Bilinear Interpolation with Scipy
To implement bilinear interpolation with Scipy, we first need to define a set of coordinate points.
These points should cover the entire region we want to interpolate and have a regular spacing. Once we have defined the coordinate points, we can use the griddata function to interpolate new values.
The griddata function takes three arguments: the coordinate points, the values at those points, and a new set of points where the interpolation is to be performed. In the case of bilinear
interpolation, we can set the interpolation method to ‘linear’, and Scipy will automatically perform bilinear interpolation wherever necessary.
To visualize the results of our interpolation, we can create a grid of points and plot the interpolated values using Scipy’s imshow function. Overall, bilinear interpolation with Scipy is
straightforward and easy to implement.
The griddata function provides a fast and efficient way to perform bilinear interpolation on large datasets without the need for manual computation.
In conclusion, bilinear interpolation is a powerful tool that can be used to estimate unknown values in a two-dimensional space. It is widely used in image processing, data analysis, computer vision,
and digital terrain modeling.
Numpy and Scipy are two popular Python libraries used for scientific computing, and both provide functions for bilinear interpolation. Numpy’s manual computation provides flexibility while Scipy’s
griddata function offers an efficient and user-friendly method for interpolation.
By understanding bilinear interpolation and its implementation with Numpy and Scipy, professionals in scientific computing can perform efficient data analysis, conduct object recognition and
tracking, and accurately model elevation data. Bilinear interpolation is an essential tool that can be used to estimate unknown values in a two-dimensional space and is used in a variety of fields
such as image processing, data analysis, computer vision, and digital terrain modeling.
Both Numpy and Scipy are popular Python libraries used for scientific computing and provide functions for bilinear interpolation. Numpy’s manual computation provides flexibility, while Scipy’s
griddata function offers an efficient and user-friendly method for interpolation.
Understanding bilinear interpolation and its implementation with Numpy and Scipy is crucial for data analysis, object recognition, and accurately modeling elevation data. Scientists and engineers
need to master bilinear interpolation to develop more precise models, detect outliers, and analyze large datasets efficiently. | {"url":"https://www.adventuresinmachinelearning.com/mastering-bilinear-interpolation-with-numpy-and-scipy/","timestamp":"2024-11-11T05:03:24Z","content_type":"text/html","content_length":"69541","record_id":"<urn:uuid:8371775a-0bce-4ff8-95c5-b7a8e4ae1ab7>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00561.warc.gz"} |
Fast Step Graph
Fast Step Graph is an optimized implementation in R of the Stepwise approach used for discovering high-dimensional Gaussian Graphical Models. It aims to accurately estimate the \(\mathbf{\Omega}\)
precision matrix when dealing with datasets where the number of features is significantly larger than the number of samples (\(p >> n\)), such as in genomics.
This implementation builds upon the original code available at link, which accompanies the associated research paper. Fast Step Graph enhances the computational efficiency of the original code to
handle much larger graph structures than previously reported, while reducing the training time. Several improvements have been made, including the elimination of redundant code, utilization of
column-wise data structures (better for R), avoid list creation, manipulation and expansion within loops, and the integration of a faster subroutine for regression. Additionally, this implementation
addresses a bug introduced in the original code.
Despite these enhancements, the primary bottleneck of Fast Step Graph lies in the requirement of substantial memory resources \(\text{Memory} \propto \Big(\frac{p(p-1)}{2}\Big)\) for storing the
entire graph, particularly when \(p\) grows.
Clone this repository or simply download the .zip file and follow the instructions in this link to see an example.
How to cite this repository? | {"url":"https://cran-r.c3sl.ufpr.br/web/packages/FastStepGraph/readme/README.html","timestamp":"2024-11-07T09:09:25Z","content_type":"application/xhtml+xml","content_length":"3302","record_id":"<urn:uuid:03ae0732-40c9-49b8-9178-5cfaddc3456c>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00706.warc.gz"} |
Inertial navigation systems use the angular rate and acceleration measurements from gyroscopes and accelerometers to determine the position, velocity, and attitude of a system by integrating this
data over time. The non-linearities inherent to attitude require these integrations to occur at very high rates. To minimize the computational requirements of the user system, most inertial
navigation systems output what are known as coning and sculling integrals which are integrated internally and can then be used at lower rates for full state integration.
Coning and Sculling Integrals
The coning and sculling integrals are integration processes that properly account for the coning and sculling motion and are valid despite the non-linearities inherent to real-world motion.
Typically, the coning and sculling integrals are performed at higher rates, which allows the integration of the velocity and angular rate outputs to be performed at much lower speeds, thus reducing
the amount of bandwidth needed to process the data. The coning integral provides a principal rotation vector known as Delta-Theta, $\Delta\boldsymbol{\theta}$, while the sculling integral generates a
Delta-Velocity, $\Delta\boldsymbol{v}$, over specified amount of time, $\Delta t$.
These techniques have the advantage of providing the change in orientation and change in velocity over an arbitrary amount of time with higher accuracy as compared to averaging the accelerations or
angular rates over longer time steps. In addition, the coning and sculling integrals provide the benefit of lower computational complexity as compared to other algorithms, such as the quaternion
attitude update.
Attitude Integration
The Delta-Theta output from the coning integral is easily combined with quaternions to produce a continuously updated attitude estimate. An updated quaternion value ($\boldsymbol{q}_{k+1}$) is
computed from the previous quaternion value ($\boldsymbol{q}_{k}$) using Equation \ref{eq:aiqu}. This equation assumes the scalar term of the quaternion is $q_4$ and that $\Delta\boldsymbol{\theta}$
is provided in radians.
$$\label{eq:aiqu} \boldsymbol{q}_{k+1} = \begin{bmatrix} \cos\gamma[I_{3\times3}] - \left[\boldsymbol{\Psi}^\times\right] & \boldsymbol{\Psi} \\ -\boldsymbol{\Psi}^\intercal & \cos\gamma \end
\begin{equation*} \gamma = \frac{\|\Delta\boldsymbol{\theta}\|}{2}\qquad \boldsymbol{\Psi} = \begin{cases}\frac{\sin\gamma}{2\gamma}\Delta\boldsymbol{\theta}& \gamma\ge1\mathrm{e}{-5}\\ \frac{1}{2}\
Delta\boldsymbol{\theta} & \gamma<1\mathrm{e}{-5} \end{cases} \qquad \left[\boldsymbol{\Psi}^\times\right] = \begin{bmatrix} 0 & -\Psi_3 & \Psi_2 \\ \Psi_3 & 0 &{-\Psi}_1 \\ {-\Psi_2} & \Psi_1 & 0 \
Because no computation can be achieved with perfect numerical precision, it is recommended that the updated quaternion is normalized per Equation \ref{eq:aiqun} to ensure that this updated quaternion
value remains unit length.
$$\label{eq:aiqun} \hat{\boldsymbol{q}}_{k+1} = \frac{\boldsymbol{q}_{k+1}}{\|\boldsymbol{q}_{k+1}\|}$$
Once the quaternion has been calculated from the Delta-Theta, this orientation can then be converted into the desired attitude representation. For more information about the quaternion and different
attitude representations, refer to Sections 2.3 and 2.4.
Position and Velocity Integration
Information about an object's position can be obtained by integrating the velocity solution over a discrete period of time. Given a Delta-Velocity output in the body frame (${^B}\Delta\boldsymbol{v}
$), the attitude at the start of the integration step is used to transform it into the inertial frame (typically NED):
The inertial frame Delta-Velocity must then be corrected for gravity ($\boldsymbol{g}$) and the Coriolis term arising from Earth's angular rate ($\boldsymbol{\omega}_\oplus$) and the current velocity
estimate ($\boldsymbol{v}_k$), each in the inertial frame.
$$\begin{split} \Delta\boldsymbol{v}_{g/cor}&=\Delta t\left(\boldsymbol{g}-2\boldsymbol{\omega}_{\oplus} \times \boldsymbol{v}_k\right) \\ \Delta\boldsymbol{v}_{c}&={^I}\Delta\boldsymbol{v}+\Delta\
boldsymbol{v}_{g/cor} \end{split}$$
VectorNav sensors can be configured to output the term $\Delta\boldsymbol{v}_c$ directly, utilizing the onboard Kalman filter attitude estimates, eliminating these steps. Once the corrected
Delta-Velocity is available, the position and velocity can be easily updated via Equation \ref{eq:intposvel}.
$$\begin{split} \boldsymbol{v}_{k+1}&=\boldsymbol{v}_{k}+\Delta\boldsymbol{v}_{c}\\ \boldsymbol{p}_{k+1}&=\boldsymbol{p}_{k}+\Delta t\boldsymbol{v}_{k} + \frac{\Delta t}{2} \Delta\boldsymbol{v}_{c}\ | {"url":"https://www.vectornav.com/resources/inertial-navigation-primer/math-fundamentals/math-coning","timestamp":"2024-11-06T17:38:02Z","content_type":"text/html","content_length":"41131","record_id":"<urn:uuid:a720ae0b-c640-40e6-bf2a-1283aa9cafa3>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00572.warc.gz"} |
Logarithmic derivative of wave function
Next: Ghost states Up: Generation of pseudopotential Previous: How the MBK scheme   Contents
To check the transferability of generated pseudopotentials, a useful measure is to compare logarithmic derivatives of wave functions [16]. If the logarithmic derivative of pseudopotential is
comparable to that by the all electron calculation through a wide range of energy, then the pseudopotential would possess a good transferability. In Fig. 4 shows the logarithmic derivatives in a
carbon atom, indicating a good transferability of the pseudopotential. The keywords concerned to the calculations of the logarithmic derivative are as follows:
• log.deri.RadF.calc
When the logarithmic derivatives are calculated, then ON, otherwise, OFF.
• log.deri.MinE
The lower bound of energy (Hartree) used in the calculation of logarithmic derivatives of radial wave functions.
• log.deri.MaxE
The upper bound of energy (Hartree) used in the calculation of logarithmic derivatives of radial wave functions.
• log.deri.R
Radius at which the logarithmic derivatives of radial wave functions are evaluated.
You can find details for these keyword in the section, Input file. In case of log.deri.RadF.calc=ON, calculated logarithmic derivatives are output in files *.ld#, where * is the file name that you
specified by the keyword, System.Name, and # is the angular momentum number. If the fully relativistic calculation is performed as 'eq.type=dirac', the file name is *.ld%_#, where % runs 0 to 1
corresponding to
Figure: Logarithmic derivatives of radial wave functions under the all electron potential, semi-local pseudopotential, and fully separable pseudopotential of a carbon atoms
Next: Ghost states Up: Generation of pseudopotential Previous: How the MBK scheme   Contents 2011-09-28 | {"url":"https://openmx-square.org/adpack_man2.2/node15.html","timestamp":"2024-11-03T06:26:33Z","content_type":"text/html","content_length":"5601","record_id":"<urn:uuid:08eac704-29b3-4bfc-a4b3-ca52553fc054>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00127.warc.gz"} |
ciin - Intuitionistic Logic Explorer
Description: Extend class notation to include indexed intersection. Note: Historically (prior to 21-Oct-2005), set.mm used the notation ∩ 𝑥 ∈ 𝐴𝐵, with the same intersection symbol as cint 3821.
Although that syntax was unambiguous, it did not allow for LALR parsing of the syntax constructions in set.mm. The new syntax uses a distinguished symbol ∩ instead of ∩ and does allow LALR parsing.
Thanks to Peter Backes for suggesting this change. | {"url":"https://us.metamath.org/ileuni/ciin.html","timestamp":"2024-11-13T23:12:30Z","content_type":"text/html","content_length":"6461","record_id":"<urn:uuid:45c27f06-033c-4d4b-a9c8-8cece193ac3e>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00629.warc.gz"} |
Selection sort in JavaScript | Guide to Selection sort in JavaScript
Updated April 10, 2023
Introduction to Selection Sort in JavaScript
Selection sort in JavaScript is one of the most simplest and intuitive sorting algorithms. It works to repeatedly find minimum elements in the unsorted, considering ascending order, and move to the
start of the array. In the Computer field, there are some tools that are often used and based on Algorithms internally. As programmers, we move through data that is built with modern programming
languages available in one way or another. Using these built-in sorting algorithms, work becomes easier for a programmer but it is still necessary to understand what is going on underneath and what
are all the different types of sorting algorithms. So today, we shall deep dive into the Selection sort algorithm, which is one fundamental way of sorting algorithms.
For such kinds of Sorting techniques, we have Algorithms or a Flowchart.
Algorithm for Selection sort:
Step 1: Consider an array of elements, out of which we set the first element as a minimum.
Step 2: Compare the minimum value with the second. And if the second element is less than the minimum element, the second element is assigned as a minimum element.
Step 3: So Step 2 gets repeated by comparing the minimum element with the third element. If the third element in less than the minimum element, the third element will be assigned as the minimum
element else nothing. And the likewise process will repeat till the last element.
Step 4: After each change, the minimum element is placed at the start of the unsorted element.
Step 5: For each iteration, the index starts with 0 i.e. first unsorted element. So all the above steps are repeated until the elements are placed at their correct positions i.e. in Ascending order.
How does Selection Sort work?
We shall see How Selection Sort works with few examples,
It is a comparison-based algorithm, which divides into two parts. One is the sorted part and the other being the unsorted part. Initially, sorted part is empty. The first element from the unsorted
part is compared and swapped with the before element which actually adds to the sorted list, and the process goes on.
Example #1
<!DOCTYPE html>
<h2>Selection Sort Example</h2>
<p>Selection sort for array of elements in JavaScript</p>
<p id="demo"></p>
function selectionSorting(inputArr) {
for (var i = 0; i < inputArr.length; i++) {
var tempEle = inputArr[i];
for (var j = i + 1; j < inputArr.length; j++) {
if (tempEle > inputArr[j]) {
tempEle = inputArr[j];
var index = inputArr.indexOf(tempEle);
var tempVal = inputArr[i];
inputArr[i] = tempEle;
inputArr[index] = tempVal;
var inputArr = [2, 7, 9, 1, 8];
document.write("Here is the input array: ", inputArr + "<br>");
document.write("Here is the sorted array: ", inputArr);
So here we see the input array and the sorted array, which is sorted with the Selection sort Algorithm. Let us explain this in a simpler manner as to How this above input array has been sorted.
inputArr = [2, 7, 9, 1, 8];
1. As per the Algorithm, the minimum element will be 2 i.e. first element.
2. Compare ‘2’ with ‘7’. Since 2 < 7, no swapping
3. Compare ‘2’ with ‘9’, no swapping
4. Compare ‘2’ with ‘1’, which will be swapped. inputArr = [1, 7, 9, 2, 8]
5. Now, the index of minimum element changes to 1 i.e. 7.
6. So, as the left part is sorted part, 7 is now compared with 9. No swapping
7. ‘7’ is now compared with ‘2’, swaps. indexArr = [1, 2, 9, 7, 8].
8. Now, index of the minimum element changes to 2 i.e. 9.
9. Compare ‘9’ with ‘7’, swaps. indexArr = [1, 2, 7, 9, 8].
10. Index of minimum element changes to 3 i.e. 9.
11. Compare ‘9’ with ‘8’, swaps. indexArr = [1, 2, 7, 8, 9].
12. Array is sorted.
Example #2
Selection Sort ( maintaining a copy of the array elements since objects are passed by reference)
<!DOCTYPE html>
<h2>Selection Sort Example</h2>
<p>Selection sort for array of elements in JavaScript</p>
<p id="demo"></p>
const selectionSorting = (originalArr) => {
const listArr = [...originalArr]
const lengthOfArr = listArr.length
for (let i = 0; i < lengthOfArr; i++) {
let minimum = i
for (let j = i + 1; j < lengthOfArr; j++) {
if (listArr[minimum] > listArr[j]) {
minimum = j
if (minimum !== i) {
;[listArr[i], listArr[minimum]] = [listArr[minimum], listArr[i]]
return listArr
const inputArr = [29, 72, 13, 98, 76, 32, 12, 10, 89]
document.write("Here is the sorted array:" ,selectionSorting(inputArr)) //[1,3,4,5,6]
So this is how we have used a different way of writing the logic for Selection sort and have got a sorted array.
With this, we can conclude our topic ‘Selection sort in JavaScript’. We have seen what Selection sorting is and How is it implemented by looking at the Algorithm above. We have also seen few examples
here which will give an insight on the working model of Selection sorting. Once the concept is understood, it can be coded in any manner or in any kind of programmatic logic to achieve the same.
There are also many other Sorting Algorithms coming our way. Thanks! Happy Learning!!
Recommended Articles
This is a guide to Selection sort in JavaScript. Here we discuss definition, syntax, and parameters, working of the Selection sort in JavaScript examples with code implementation. You may also have a
look at the following articles to learn more – | {"url":"https://www.educba.com/selection-sort-in-javascript/","timestamp":"2024-11-14T11:28:59Z","content_type":"text/html","content_length":"315518","record_id":"<urn:uuid:5e6b69b8-9e31-4b41-ab6f-09cbebe159e8>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00175.warc.gz"} |
About Blaise Goutéraux
I am a theorist working at the interface between high energy theory and condensed matter.
Strongly-correlated quantum phases of nature, such as the high Tc superconductors, fall outside the realm of the quasiparticle theory developed by Landau, so-called Fermi liquids. They display
unconventional transport properties (most famously, a resistivity linear in temperature, as well as others) which cannot be accounted for by quasiparticles and Boltzmann transport. On the other hand,
strong interactions make it very difficult tackling these systems using conventional approaches, which are often based around perturbation theory in a small coupling. Dualities, such as the gauge/
gravity duality, or effective field theories, offer calculable self-consistent frameworks to think about strongly-correlated phases of matter.
The holographic principle (aka gauge/gravity duality or ads/cft) is one of the most important insights coming out of string theory in the twentieth century. It lets us ask fundamental questions about
gravity and the emergence of spacetime, but also about strongly-correlated quantum phases of matter in general, by mapping the dynamics of some strongly-interacting quantum field theories to theories
of gravity in anti de Sitter spacetimes. Anti de Sitter is a solution of Einstein's equations with a negative cosmological constant (in contrast to our Universe, which appears to be well-described by
a positive, albeit very small, cosmological constant).
More generally, I am also interested in hydrodynamics and effective field theory approaches, in the presence of broken symmetries. These approaches rest on the assumption that at long distances and
late times, collective behavior emerges as a result of the conservation of a few fundamental quantities, such as charge, energy or momentum. The low energy dynamics is captured by the conservation
equations as well as by a number of constitutive relations for the conserved currents. These involve an expansion in small gradients (of time and space). At the end of the day, the dynamics depends
on a small, finite set of transport coefficients, which can be computed efficiently in microscopic models. In these sense, hydrodynamics allows to describe the universal sector of strongly-coupled
phases of matter.
These symmetry-based approaches naturally connect with gauge/gravity duality, since the symmetries of the problem can be encoded in the gravity dual. By perturbing away from thermal equilibrium, all
the holographic transport coefficients can be efficiently computed. This is particularly useful once symmetries are weakly broken (eg weak breaking of translations by disorder) and in the presence of
quasi-conserved quantities.
You can find more details on what I have been up to recently by browing my list of publications on arXiv, Inspire or Google Scholar. | {"url":"https://www.cpht.polytechnique.fr/?q=en/node/490","timestamp":"2024-11-05T09:49:45Z","content_type":"text/html","content_length":"16290","record_id":"<urn:uuid:a1c454e3-ec87-4198-b1ad-4662f0933f56>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00628.warc.gz"} |
Integral Method for Enclosures | Ansys Courses
This lesson covers the integral equation approach to radiative transfer in enclosures. It explains the radiosity method and how it is used to solve problems involving radiative transfer enclosures.
The lesson also introduces the concept of the integral equation approach, which is not as popular as the matrix method due to its tedious nature. However, it is essential for studying gas radiation.
The lesson further explains how to set up the problem and solve it using the integral equation approach. It also discusses the importance of the extinction coefficient in determining the mean free
path of a photon. The lesson concludes by explaining how to estimate the intensity emission by the gas using Kirchhoff’s law.
Video Highlights
01:16 - Discussion on the integral equation approach
12:35 - Introduction to the concept of the exponential kernel approximation
21:32 - Explanation of the Leibnitz rule
41:59 - Explanation of the extinction coefficient and absorption coefficient of a gas
53:38 - Discussion on Kirchhoff’s law and the concept of absorptivity
Key Takeaways
- The radiosity method is commonly used to solve problems involving radiative transfer enclosures.
- The integral equation approach, though more tedious, is essential for studying gas radiation.
- The extinction coefficient is crucial in determining the mean free path of a photon.
- Kirchhoff’s law can be used to estimate the intensity emission by the gas.
- The integral equation approach can provide a complete analytic solution for radiosity, temperature, and other properties. | {"url":"https://innovationspace.ansys.com/courses/courses/shape-factor-enclosures-in-radiative-heat-transfer/lessons/integral-method-for-enclosures-lesson-9/","timestamp":"2024-11-10T00:12:58Z","content_type":"text/html","content_length":"180193","record_id":"<urn:uuid:85e0c9a8-e971-498d-9e5b-fbf40cae1c71>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00752.warc.gz"} |
Averages From Frequency Table Worksheet
Averages from frequency table at a glance
Given a table of data, we can find the mean, median, mode and range of the data set. For example, this table shows the number of goals scored in 21 football matches:
Number of GoalsΒ Β Β Β Β Β Β Β Frequency
0Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β 5
1Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β 4
2Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β 7
3Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β 4
• The mode is the most common number of goals scored. The number of goals with the highest frequency is 2 and so the mode = 2.
• The median is the middle value. In this case it is the 11th value, which would be 2.Β
• To find the mean number of goals, we multiply the number of goals by their frequency, add these totals (30) and divide by the total number of games (21). The mean is therefore 1.4.
• The range is the difference between the highest and lowest number of goals. 3 subtract 0 is equal to 3.Β
We can also find averages from a grouped frequency table, where data is arranged into groups. The class intervals for grouped data are often denoted using inequalities. For example, the classes of
data about the shoe size, s, of a number of people could be 1β €s<4, 4β €s<7, and 7β €s<10.
Β For grouped data:Β
• The modal class is the class with the highest frequency.
• The class which contains the median which is the class in which the middle value lies.
• To find the estimated mean, we multiply the midpoint of each class by its frequency, add these together and divide by the total number of items included in the table.Β
You may be asked to find missing numbers in a frequency table using averages.
Looking forward, students can then progress to additional Statistics worksheets, for example a mean, median, mode and range worksheet or pie chart worksheet.
For more teaching and learning support on Statistics our GCSE maths lessons provide step by step support for all GCSE maths concepts. | {"url":"https://thirdspacelearning.com/secondary-resources/gcse-maths-worksheet-averages-from-frequency-table/","timestamp":"2024-11-07T16:28:02Z","content_type":"text/html","content_length":"164143","record_id":"<urn:uuid:2356ed01-24ed-440b-b447-7d9e6ecb4f48>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00500.warc.gz"} |
Moving Average Convergence Divergence - ThirdBrainFxMoving Average Convergence Divergence
Moving Average Convergence Divergence
Moving Average Convergence Divergence
Moving average convergence divergence (MACD) is a trend-following momentum indicator that shows the relationship between two moving averages of a security’s price.
The MACD is calculated by subtracting the 26-period exponential moving average (EMA) from the 12-period EMA.
The result of that calculation is the MACD line. A nine-day EMA of the MACD called the “signal line,” is then plotted on top of the MACD line, which can function as a trigger for buy and sell
signals. Traders may buy the security when the MACD crosses above its signal line and sell—or short—the security when the MACD crosses below the signal line. Moving average convergence divergence
(MACD) indicators can be interpreted in several ways, but the more common methods are crossovers, divergences, and rapid rises/falls.
The Formula for MACD is:
MACD = 12-Period EMA − 26-Period EMA
MACD is calculated by subtracting the long-term EMA (26 periods) from the short-term EMA (12 periods). An exponential moving average (EMA) is a type of moving average (MA) that places a greater
weight and significance on the most recent data points.
The exponential moving average is also referred to as the exponentially weighted moving average. An exponentially weighted moving average reacts more significantly to recent price changes than a
simple moving average (SMA), which applies an equal weight to all observations in the period.
The MACD has a positive value whenever the 12-period EMA is above the 26-period EMA (the blue line in the price chart) and a negative value when the 12-period EMA is below the 26-period EMA. The more
distant the MACD is above or below its baseline indicates that more the distance between the two EMAs is growing.
Key Features
• Moving average convergence divergence (MACD) is calculated by subtracting the 26-period exponential moving average (EMA) from the 12-period EMA.
• MACD triggers technical signals when it crosses above (to buy) or below (to sell) its signal line.
• The speed of crossovers is also taken as a signal of a market is overbought or oversold.
• MACD helps investors understand whether the bullish or bearish movement in the price is strengthening or weakening. | {"url":"https://www.thirdbrainfx.com/indicator/indicators-of-trading-macd/","timestamp":"2024-11-07T22:06:40Z","content_type":"text/html","content_length":"32168","record_id":"<urn:uuid:cceff259-3b77-4864-8495-52f69000795f>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00493.warc.gz"} |
When you’re shopping around for a loan, you may be tempted to compare offers based on the interest rate quoted by the lender. But the interest rate is only a slice of the full picture. A more
comprehensive metric you can use is the APR. In this article, we’ll look at the differences between APR vs. interest rate and how knowing the difference can help you find better loan deals.
What is an interest rate?
The interest rate is the amount of interest a borrower will pay over one year. For example, if you borrowed $10,000 with a 10 percent interest rate and had to pay the whole thing back after one year,
then you’d owe $11,000 (the original $10,000 plus $1,000 in interest).
Many loans use fixed interest rates when calculating the payment amount (such as a conventional 30-year mortgage). However, other loans are designed to have interest rates that can fluctuate over
time, such as adjustable-rate mortgages or ARMs. Note that the interest rate used by credit cards is also subject to change.
What is an APR?
The annual percentage rate, or APR, is how much the borrower is truly paying to finance their loan. Although it often gets misinterpreted as the interest rate, they are not the same thing. APR takes
into consideration the total amount of interest expense you’ll pay plus any other applicable expenses, including:
• Fees (such as origination fees or some closing costs)
• Private mortgage insurance (PMI)
• Mortgage points (also called prepaid interest)
When broadening the picture to include these additional expenses, it becomes more transparent to the borrower that the “true” cost of the loan is much higher than just the interest alone. So, the
advertised APR will be higher than the interest rate.
The Federal Truth in Lending Act requires lenders in all covered loans to disclose the APR before an agreement is signed. This is useful to the borrower because it gives them a good starting point
for similar types of loans. However, you still need to be careful and understand that not all APRs are necessarily the same. For example, the rules used to calculate APR are slightly different for
open-ended credit (i.e. a credit card or line of credit) than they are for closed-end credit (e.g. a fixed-amount personal loan or standard purchase-price mortgage).
How to calculate APR and interest rate
When you apply for a loan, the interest rate you’ll be offered will be based on several criteria. The main factor will be the Federal Reserve’s interest rate and how this has affected what the
industry calls the “prime rate”. This is the starting point used by lenders to calculate the interest rates for all other loans.
Other factors that can alter your interest rate will include things like the type of loan being applied for and your credit score. Borrowers with a higher credit score may qualify for a better
interest rate.
Once the interest rate is known, APR can be estimated as follows:
1. Take the total interest paid plus any other fees charged by the lender to the borrower as a condition of giving the loan (e.g. an origination fee, or points).
2. Divide this value by the total principal (amount being borrowed).
3. Divide the result by the total number of days in the loan.
4. Multiply the result by 365 to adjust this figure relative to one year.
5. Multiply this result by 100, to obtain the percentage.
There are some variations, based on the specific type of loan, in the types of fees which must be included in the APR and the exact method lenders are required to use for calculation of the APR
however, so this method of calculating APR may not yield the exact rate reflected by the lender.
The bottom line
If you’re thinking about getting a loan, don’t compare offers based on the interest rate alone. Remember to use APR since it will include the interest expense plus fees. Although APR won’t be the
same for all loans, it will give you a much better starting point for deciding which offer is right for you. | {"url":"https://business.mammothtimes.com/mammothtimes/article/icrowdnewswire-2022-12-21-how-to-calculate-apr-vs-interest-rate","timestamp":"2024-11-11T10:56:23Z","content_type":"text/html","content_length":"106333","record_id":"<urn:uuid:253df0cd-6746-49c1-88d9-50a25417abf7>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00331.warc.gz"} |
MathFiction: Once Upon a Wardrobe (Patti Callahan)
Megs is a student at Oxford University in 1950 whose eight year old brother is so ill that he is unlikely to live another year. While Megs loves equations, her brother George loves the new book "The
Lion, the Witch, and the Wardrobe" by C.S. Lewis and he convinces his sister to meet the author to learn about the book's origins. Through those conversations, she comes to better appreciate fiction.
The little that is said of mathematics in the book makes me think the author does not really know much about it. When she first meets Lewis, Megs says "I attend Somerville College reading maths".
However, besides vague references to "figures" and "sums", the only math that is ever mentioned are Einstein and Newton's physical theories. None of that is what a student of mathematics was likely
to be learning at Oxford in the mid-twentieth century.
If I was to quibble about this mathematical detail, however, I would be missing the point (and probably confirming the author's prejudices about mathematicians). The point, which the book bluntly
states in this conversation between Megs and her friend Padraig, concerns the importance of stories:
(quoted from Once Upon a Wardrobe)
"Surely you know physics is more important than fiction?" This is an absurd discussion. I can rightly enjoy a good story, but thinking novels are the same as Einstein and Newton's theories is absurd.
"I think they are neither more important nor less important," he says to my surprise. "No. Not one bit."
"But we cannot understand our world without the genius of the mathematicians. It's a language of the universe," I say.
"You are ignoring imagination; you need it for your work too. But I can't really understand my life without stories. They offer me... they offer all of us the truth in their myths mysteries, and
"I'm sorry. I'm being a dolt. I just can't figure out how to help George, and this whole thing is frustrating me. At least with a math problem I can work on it until the right answer shows up."
"Shows up?" Padraig sits again also, then scoots even closer so our knees are touching. "Like a character." [...] I think what I was trying to say is that when my fictional characters show up, or the
ones you're reading about in that book, they have a place they're going. A journey. A math problem does too. I've seen Father spend years on one equation until it shows him the way it is meant to go.
That's what a story does with me. I'm not trying to convince you, Megs with the flashing blue eyes, that my work is more important than yours, but maybe it's just as important."
I certainly would not argue with the idea that fiction is important and has a role to play in our understanding of the world. (That, in part, is why I maintain this website about mathematical
fiction.) However, at the risk of again being a stereotypical mathematician, let me also say that the book goes a bit too far for my tastes when it suggests that mathematics is nothing more than
another kind of story telling. Similarly, although I agree that mathematics also requires imagination (a true statement that the book supports with a quote from Albert Einstein), that doesn't mean
that mathematicians are making use of stories when they utilize it. Fiction and mathematics are just two of many human endeavors which require imagination.
Callahan, who also wrote "Becoming Mrs. Lewis", clearly has an affinity for and much knowledge about the author C.S. Lewis. And, if there really is anyone who fails to see that fiction is more than
just a form of entertainment, then reading this bittersweet tale could help them see the error of their ways. However, I did not particularly enjoy it or get much from it. (The only thing I can
really say I learned from it, assuming it is true, is the interesting fact that Florence Augusta Lewis, mother of the author of the Narnia books, had a degree in mathematics. I have not been able to
confirm this, but I have no particular reason to doubt that she did. If so, it is indeed a very rare achievement for a British woman of that time!) | {"url":"https://kasmana.people.charleston.edu/MATHFICT/mfview.php?callnumber=mf1646","timestamp":"2024-11-09T12:45:02Z","content_type":"text/html","content_length":"12216","record_id":"<urn:uuid:7649e650-1df9-4906-9af7-e4f58472a12f>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00735.warc.gz"} |
Who can help me with this math homework question?
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Who can help me with this math homework question?
Which equation represents a nonlinear realationship between x and y? A) y= 4x B) y= 3x + 14 C) y= x over 3 (as in a fraction, the x is the top #) - 2 D) y= 2x exponent2(the little 2 on thge top
corner of 2x) -1
What do you think the answer is
Answer #1
I know it is D because linear equation should not have an exponent
Ask an advisor one-on-one! | {"url":"https://funadvice.com/q/who_can_help_me_with_this_math_homework","timestamp":"2024-11-05T18:35:06Z","content_type":"text/html","content_length":"21937","record_id":"<urn:uuid:6dfd7822-20fb-4d14-b70a-e87b884b6b96>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00215.warc.gz"} |
Why Math Is the Most Common Subject for Tutoring Services - Grade Potential Cleveland, OH
Why Mathematics Is the Most Common Topic for Tutoring Services
We've all been there. There is consistently that particular subject we all difficulty with, and just working on homework makes your head hurt. For some, it may be English, Biology, Chemistry, or
Physics, while others might stumble over historical dates. But for most of students, that arduous topic is arithmetic.
Is arithmetic too difficult? Are you just not built for that type of mental work? Well, we disagree. You'd be surprised to discover that many students have difficulty with the similar things as you.
The majority of the time, you just require a different point of view.
Read on and learn what tools you have at your disposal to conquer this mental obstruction and be the best of your class.
Common Grounds Why Students Struggle with Arithmetics
Commonly, mathematics makes student anxious because as you study further, it keeps getting more complex. It begins with more straightforward theories, but they build on each other and proceed to more
complicated theories. For example, you must be knowledgeable regarding addition and subtraction before tackling multiplication and division.
Finally, mathematics is inaccessible for some since they don’t understand that arithmetic is cumulative. It demands time, loads of work, and calmness. Now and then, the assigned time during the
school year for the subject isn't adequate for learners to comprehend all of the material that the plan is required to cover.
There is moreover the problem of fully comprehending the subject matter. To completely learn a lesson, you must digest it and easily apply it to math problems, which also requires a lot of emphasis
and care.
Mathematics has a prestige for being difficult. This is mostly from improper learning approaches for students, a absence of logical methods|system, and misconceptions regarding intelligence. It is
essential to know that passing a standardized test does not mean that you can implement the same concept on different subject area.
So, what can you do if you (or your son or daughter) struggle with mathematics?
Is a Math Tutor Ideal for You (or Your Learner)?
The most apparent answer is looking for guidance. If you are struggling with mathematics, why not get a tutoring service to provide a new viewpoint? It is common to search for outside assistance to
make up for confusing classes.
The ideal way to make arithmetic more comprehensible is to turn back and go through older lessons to understand those concepts fully. Once that is completed, you can create upon them to progress to
more complicated topics and improve your test scores.
In sizeable classes with many students asking for attention from the teacher, there is short time to tackle each query and confusion. Likewise, classes are generally organized to use the time
accessible to complete the program rather than focusing on the attention that each child needs to enhance problem-solving skills.
This technique is not the math teacher's fault since the time during the semester is restricted. Still, it does pose a challenge for students. Unless your grade comprise of a small group, it is hard
to get individual focus.
Using learning centers, you will have the time to study through the syllabus, repeat former lessons, and ultimately overcome the problems you face with repetition, patience, and practice.
High school students and college students may get discouraged when they are told that they should review elementary school lessons. Luckily, going over these math concepts takes less time than you'd
believe! With high-quality private tutors and a little discipline, you'll be going from middle school math to advanced computer science in no time!
Do you Choose Online or In-Person Mathematics Tutoring?
So, you've determined to go for an online tutoring service. Luckily, with the unprecedented rise of online platforms during the pandemic, there are multiple choices for online and in-person tutoring
services. Which one should you select?
To assist you comprehend the benefits of each, here is a comprehensive guide on what each can give in terms of teaching.
What Are the Benefits of Online Learning?
Online tutoring consists of online sessions using applications that takes advantage of existing platforms to further learning.
There is a lot to know utilizing this method. Online tutoring is beneficial for students with a very tight timetable. In-between the teacher and the student, they can set up a time to connect, no
matter how late or early.
Additionally, without commuting, it could be less expensive and more appropriate since you don’t need take into account for conveyance to a teaching center. You can do test prep, revise math
concepts, and receive homework help with online math tutoring from your bedroom. That's exceptional!
What Are the Benefits of In-Person Tutoring?
In-person tutoring follows a equivalent structure. You and your instructor will set a time and location for a tutoring session. During these sessions, you can revise your math classes, clear your
queries, or solve exercises to better your math skills.
If you can afford the commute time, in-person tutoring is ideal for hands-on work! Since the instructor is next-to-next, he can read you and guide whenever he feels you are baffled about any math
concept throughout the session. Body language makes all the difference!
Likewise, assume you have a short attention span. In that instance, this technique is desirable. It is simpler to help you keep you involved in the lesson with face-to-face lessons since the tutor
may have tools and props to make math problems more interesting.
Finally, choosing among in-person and online tutoring will depend on your learning style and transportation restrictions.
Grade Potential Can Guide You Succeed in Mathematics
The key to comprehending arithmetics is to prevent being afraid of failure. If you need help understanding trigonometry or passing pre-algebra, reach out for help.
You will get the care required with Grade Potential regardless of your pick! Online and in-person tutoring programs are one phone call away. With a tutoring company, you will have highly qualified
teachers at a moment's notice.
Contact us if you require help with arithmetic classes or you know your child is having problems. We will reply to all your questions! | {"url":"https://www.clevelandinhometutors.com/blog/why-math-is-the-most-common-subject-for-tutoring-services","timestamp":"2024-11-05T13:43:10Z","content_type":"text/html","content_length":"78451","record_id":"<urn:uuid:6147b262-32bb-4c28-b9e8-801f6cf2a900>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00358.warc.gz"} |
Centrifugal Force Calculator | How to Calculate Centrifugal Force? - physicsCalculatorPro.com
Centrifugal Force Calculator
The Centrifugal Force Calculator is used to determine the amount of force applied on a rotating body depending on its mass, velocity, and rotational radius. To receive the centrifugal force, angular
velocity, and angular acceleration values in the output section, simply enter the inputs in the available fields and press the calculate button.
What is Centrifugal Force?
The inertia force that emerges in each rotating object is known as centrifugal force. It's only required in a rotating reference frame, or when we're looking at the system from the perspective of a
moving object.
Newton's first law states that if no force is applied to an object, it will move in a straight line. A centrifugal force, acting outwards from the centre of rotation, is required for rotation to
Centrifugal Force Formula
The following are the centrifugal force formulas
• F = mv²/r
• v = ωr
• ω = v/r
• a = F/m
• r = mv²/F
• m = Fr/v²
Where, F = centrifugal force, m = mass of the object, v = velocity, r = radius, ω = angular velocity, a = centrifugal acceleration
How to Compute Centrifugal Force?
The steps for calculating angular acceleration and centrifugal force are described below
• Step 1: Calculate the object's mass, velocity, and radius.
• Step 2: Multiply the object mass by the squared velocity.
• Step 3: To get the centrifugal force, divide the product by the radius.
• Step 4: To determine the angular velocity, divide the velocity by the radius.
• Step 5: To calculate the centrifugal acceleration, divide the force by the mass of the object.
For more concepts check out physicscalculatorpro.com to get quick answers by using this free tool.
How to use Centrifugal Force Calculator?
The following is the procedure how to use the centrifugal force calculator
• Step 1: Input the unknown value's mass, radius, tangential velocity and x in the appropriate input fields.
• Step 2: To acquire the result, click the "Calculate the Unknown" button.
• Step 3: Finally, the output field will show the angular velocity, force, centrifugal acceleration.
Centrifugal Force Examples
Question 1: An object with a mass of 3 kg is moving at a speed of 7.2 m/s in a circular direction. The circular walkway has a radius of 3.5 metres. What are the centrifugal force and acceleration?
Given: Mass of the object m = 3 kgs
Velocity v = 7.2 m/s
Radius r = 3.5 m
Centrifugal force F = mv²/r
F = (3 x 7.2²)/3.5
= (3 x 51.84)/3.5
= 155.52/3.5
= 44.43
Angular velocity ω = v/r
= 7.2 / 3.5
= 2.05
Centrifugal Acceleration a = F/m
a = 44.43/3
= 14.81
As a result, the centrifugal force on the item is 44.43 N, the angular velocity is 2.05 rad/s, and the centrifugal acceleration is 14.81 m/ sec^2.
FAQs on Centrifugal Force Calculator
1. What is a centrifugal force with an example?
When viewed from a rotating frame of reference, centrifugal force acts on any object travelling in a circular path. For Example: At the poles and on the equator, the weight of an object is different.
2. What is centrifugal load, and how does it work?
The consequence of a centrifugal load is that the entire model spins around an axis you designate. Instead, the comparable forces that would arise from angular rotation and/or angular acceleration
are calculated and applied to each element's nodes.
3. What is the principle of centrifugal force?
The centrifugal force is a fictitious force that occurs when a particle moves in a circular route and has the same magnitude and dimensions as the centripetal force but points in the opposite
4. Is it true that gravity is a centrifugal force?
An object's gravity force is the centrifugal force perceived as pointing downwards towards the hull in the spinning frame of reference.
5. How do you remember centripetal and centrifugal?
The centripetal force attracts items to the centre. The word centrifugal has an F in it. Keep in mind that centrifugal forces cause items to move far away. Objects are drawn closer together by
centripetal forces, which are spelt with a P. | {"url":"https://physicscalculatorpro.com/centrifugal-force-calculator/","timestamp":"2024-11-05T02:23:18Z","content_type":"text/html","content_length":"34479","record_id":"<urn:uuid:fe3d570d-9682-4488-b83b-13a5f4e3e4f2>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00142.warc.gz"} |
Simple Balanced Parentheses
3.6. Simple Balanced Parentheses¶
We now turn our attention to using stacks to solve real computer science problems. You have no doubt written arithmetic expressions such as
where parentheses are used to order the performance of operations. You may also have some experience programming in a language such as Lisp or Scheme with constructs like:
(defun square(n)
(* n n))
This defines a function called square that will return the square of its argument n. Scheme and Lisp are both notorious for using lots and lots of parentheses.
In both of these examples, parentheses must appear in a balanced fashion. Balanced parentheses means that each opening symbol has a corresponding closing symbol and the pairs of parentheses are
properly nested. Consider the following correctly balanced strings of parentheses:
Compare those with the following, which are not balanced:
The ability to differentiate between parentheses that are correctly balanced and those that are unbalanced is an important part of recognizing many programming language structures.
The challenge then is to write an algorithm that will read a string of parentheses from left to right and decide whether the symbols are balanced. To solve this problem we need to make an important
observation. As you process symbols from left to right, the most recent opening parenthesis must match the next closing symbol (see Figure 4). Also, the first opening symbol processed may have to
wait until the very last symbol for its match. Closing symbols match opening symbols in the reverse order of their appearance; they match from the inside out. This is a clue that stacks can be used
to solve the problem.
Once you agree that a stack is the appropriate data structure for keeping the parentheses, the statement of the algorithm is straightforward. Starting with an empty stack, process the parenthesis
strings from left to right. If a symbol is an opening parenthesis, push it on the stack as a signal that a corresponding closing symbol needs to appear later. If, on the other hand, a symbol is a
closing parenthesis, pop the stack. As long as it is possible to pop the stack to match every closing symbol, the parentheses remain balanced. If at any time there is no opening symbol on the stack
to match a closing symbol, the string is not balanced properly. At the end of the string, when all symbols have been processed, the stack should be empty. The C++ and Python code to implement this
algorithm is shown in ActiveCode 1.
This function, parChecker, assumes that a Stack class is available and returns a Boolean result as to whether the string of parentheses is balanced. Note that the Boolean variable balanced is
initialized to true as there is no reason to assume otherwise at the (lines 15-16). Note also in line 22 that pop removes a symbol from the stack. At the end (lines 28-31), as long as the expression
is balanced and the stack has been completely emptied, the string represents a correctly balanced sequence of parentheses.
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Understanding and Managing Reinvestment Risk in Investments
4.3.5 Reinvestment Risk
Reinvestment risk is a critical concept in the realm of investments, particularly for those dealing with fixed-income securities such as bonds. It is the risk that arises from the possibility that
future cash flows, such as coupon payments or principal repayments, will have to be reinvested at a lower interest rate than the original investment. This risk can significantly impact the overall
returns of an investment portfolio, especially in a declining interest rate environment. In this section, we will delve into the nuances of reinvestment risk, its implications for investors, the
factors that influence it, and strategies to manage and mitigate this risk effectively.
Understanding Reinvestment Risk
Reinvestment risk is primarily associated with fixed-income securities, where investors receive periodic interest payments, known as coupons. The essence of reinvestment risk lies in the uncertainty
of the rate at which these coupon payments can be reinvested. When interest rates fall, the income generated from reinvesting these payments decreases, potentially leading to lower overall returns.
Key Characteristics of Reinvestment Risk
• Interest Rate Dependency: Reinvestment risk is heavily dependent on the prevailing interest rate environment. A decline in interest rates increases the risk, as future cash flows will be
reinvested at lower rates.
• Coupon Payments: Bonds with higher coupon rates are more susceptible to reinvestment risk because they generate more frequent and larger cash flows that need to be reinvested.
• Investment Horizon: Longer investment horizons expose investors to greater reinvestment risk, as there are more opportunities for interest rates to fluctuate over time.
Impact of Reinvestment Risk on Investors
The impact of reinvestment risk on investors can be profound, particularly for those relying on fixed-income investments for steady income. When interest rates decline, the reinvestment of coupon
payments at lower rates results in reduced income, affecting the investor’s cash flow and potentially their financial goals.
Example Scenario
Consider an investor who purchases a bond with a 6% coupon rate. If, after a year, the prevailing interest rates drop to 4%, the investor faces the challenge of reinvesting the coupon payments at
this lower rate. This reduction in reinvestment earnings can significantly impact the overall yield of the investment.
Factors Influencing Reinvestment Risk
Several factors can influence the degree of reinvestment risk an investor faces:
1. Coupon Rate: Bonds with higher coupon rates generate more frequent cash flows, increasing the exposure to reinvestment risk.
2. Maturity: Longer-term bonds are more susceptible to changes in interest rates over time, heightening reinvestment risk.
3. Interest Rate Volatility: Greater volatility in interest rates increases the uncertainty of future reinvestment rates.
4. Economic Conditions: Economic factors, such as inflation and central bank policies, can influence interest rate trends and, consequently, reinvestment risk.
Illustrating Reinvestment Risk
To better understand reinvestment risk, let’s explore a detailed scenario:
Scenario: Bond Investment in a Declining Rate Environment
An investor purchases a 10-year bond with a face value of $10,000 and a 6% annual coupon rate. The bond pays $600 in interest annually. Initially, the investor expects to reinvest these coupon
payments at the same 6% rate. However, after the first year, the central bank reduces interest rates, and the prevailing rate drops to 4%.
Impact on Reinvestment Earnings:
• Year 1: Coupon payment of $600 is reinvested at 6%, earning $36.
• Year 2: Coupon payment of $600 is reinvested at 4%, earning $24.
The reduction in reinvestment earnings from $36 to $24 illustrates the impact of declining interest rates on the investor’s income.
Strategies to Manage Reinvestment Risk
Investors can employ several strategies to manage and mitigate reinvestment risk effectively:
1. Zero-Coupon Bonds: These bonds do not pay periodic interest. Instead, they are issued at a discount and mature at face value. By eliminating periodic coupon payments, zero-coupon bonds reduce
reinvestment risk.
2. Laddered Portfolios: Creating a bond ladder involves purchasing bonds with staggered maturities. This strategy spreads reinvestment risk over time, as bonds mature at different intervals,
allowing the investor to reinvest at varying interest rates.
3. Diversification: Diversifying across different asset classes and securities can help mitigate the impact of reinvestment risk on a portfolio.
4. Matching Investment Strategy with Income Needs: Aligning the investment strategy with the investor’s income requirements can help manage reinvestment risk. For instance, selecting bonds with
maturities that coincide with anticipated cash flow needs can reduce the need for reinvestment.
5. Interest Rate Forecasting: Keeping abreast of economic indicators and interest rate forecasts can help investors anticipate changes in the interest rate environment and adjust their strategies
Reinvestment risk is an inherent challenge for investors in fixed-income securities. Understanding its implications and the factors that influence it is crucial for effective risk management. By
employing strategies such as zero-coupon bonds, laddered portfolios, and diversification, investors can mitigate the impact of reinvestment risk and align their investment strategies with their
financial goals. As the interest rate landscape continues to evolve, staying informed and adaptable is key to navigating the complexities of reinvestment risk.
Quiz Time!
📚✨ Quiz Time! ✨📚
### What is reinvestment risk? - [x] The uncertainty of reinvesting coupon payments at prevailing interest rates. - [ ] The risk of bond default. - [ ] The risk of inflation eroding returns. - [ ]
The risk of currency fluctuations affecting investment value. > **Explanation:** Reinvestment risk refers to the uncertainty of reinvesting coupon payments or principal repayments at prevailing
interest rates, which can affect overall returns. ### How does a decline in interest rates affect reinvestment risk? - [x] It increases reinvestment risk by reducing future income from reinvestments.
- [ ] It decreases reinvestment risk by increasing future income from reinvestments. - [ ] It has no effect on reinvestment risk. - [ ] It eliminates reinvestment risk entirely. > **Explanation:** A
decline in interest rates increases reinvestment risk because future cash flows will be reinvested at lower rates, reducing income. ### Which type of bond is most susceptible to reinvestment risk? -
[x] Bonds with higher coupon rates. - [ ] Zero-coupon bonds. - [ ] Short-term bonds. - [ ] Inflation-linked bonds. > **Explanation:** Bonds with higher coupon rates are more susceptible to
reinvestment risk because they generate more frequent and larger cash flows that need to be reinvested. ### What is a bond ladder? - [x] A strategy involving bonds with staggered maturities to spread
reinvestment risk. - [ ] A type of bond with adjustable interest rates. - [ ] A bond with a fixed maturity date. - [ ] A bond that pays interest only at maturity. > **Explanation:** A bond ladder is
a strategy that involves purchasing bonds with staggered maturities, allowing for reinvestment at varying interest rates and spreading reinvestment risk. ### How can zero-coupon bonds help manage
reinvestment risk? - [x] By eliminating periodic coupon payments, reducing reinvestment risk. - [ ] By providing higher coupon rates. - [ ] By offering adjustable interest rates. - [ ] By being
inflation-linked. > **Explanation:** Zero-coupon bonds do not pay periodic interest, thus eliminating the need to reinvest coupon payments and reducing reinvestment risk. ### What factor increases
exposure to reinvestment risk? - [x] Longer investment horizons. - [ ] Shorter investment horizons. - [ ] Stable interest rates. - [ ] High inflation rates. > **Explanation:** Longer investment
horizons increase exposure to reinvestment risk as there are more opportunities for interest rates to fluctuate over time. ### Which strategy involves aligning investment strategy with income needs?
- [x] Matching investment strategy with income needs. - [ ] Investing in high-yield bonds. - [ ] Using leverage to increase returns. - [ ] Investing solely in equities. > **Explanation:** Aligning
the investment strategy with the investor's income requirements can help manage reinvestment risk by selecting bonds with maturities that coincide with anticipated cash flow needs. ### What is the
effect of interest rate volatility on reinvestment risk? - [x] It increases the uncertainty of future reinvestment rates. - [ ] It decreases the uncertainty of future reinvestment rates. - [ ] It has
no effect on reinvestment risk. - [ ] It eliminates reinvestment risk. > **Explanation:** Greater volatility in interest rates increases the uncertainty of future reinvestment rates, thereby
increasing reinvestment risk. ### How does diversification help mitigate reinvestment risk? - [x] By spreading risk across different asset classes and securities. - [ ] By concentrating investments
in a single asset class. - [ ] By investing solely in fixed-income securities. - [ ] By focusing on short-term investments. > **Explanation:** Diversification helps mitigate reinvestment risk by
spreading risk across different asset classes and securities, reducing the impact of any single investment's reinvestment risk. ### True or False: Reinvestment risk is only a concern for equity
investors. - [ ] True - [x] False > **Explanation:** Reinvestment risk is primarily a concern for fixed-income investors, as it relates to the reinvestment of coupon payments or principal repayments
at prevailing interest rates. | {"url":"https://csccourse.ca/4/3/5/","timestamp":"2024-11-08T20:57:44Z","content_type":"text/html","content_length":"91405","record_id":"<urn:uuid:51e4c1ea-b664-448b-8f97-ccba320c1512>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00638.warc.gz"} |
How do you define a variable and write an expression for each phrase: the difference between a number and one-third of the number? | HIX Tutor
How do you define a variable and write an expression for each phrase: the difference between a number and one-third of the number?
Answer 1
We are working with an unknown number and we need to call it something.
#"Let the number be x"#
The word DIFFERENCE is an indication that we must subtract, to find out by how much one number is bigger than another number.
DIFFERENCE is always used with the word AND, telling us which two numbers we need to subtract.
The numbers we are working with are #x# and #"one third of x"#
"One third of" , means the number, divided by 3. It can be written as #1/3x, or (1x)/3, or x/3#
The expression is: # x - 1/3x#
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Answer 2
To define a variable for this phrase, let's denote the number as ( x ).
The difference between a number and one-third of the number can be expressed as:
[ x - \frac{x}{3} ]
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Answer from HIX Tutor
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
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What is the capacity of Flexrail1™?
Flexrail1™ is rated at 20 amps, therefore the capacity of the system is 2400 watts. We recommend de-rating to 80% so that no circuit exceeds 1920 watts. De-rating is also advised when multiple
connection points are being used.
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Transmission Lines in RF PCB Design
Over the last ten to twenty years, advances in IC fabrication technology have led to huge increases in both the operating frequencies and edge rates of switching signals in digital circuits. This has
really pushed digital circuit design into the frequency domains that were previously considered RF and/or microwave. As a result, it has become important to consider the transmission line parameters
of the connecting traces used for digital signals in circuit boards as much as they have always been considered for RF connections.
A transmission line is a two or more port network connecting a generator circuit at the driving end to a load at the receiving end. Transmission lines most commonly consist of two conductors
(although sometimes more). A three phase power transmission line for example uses three or more conductors. Examples of commonly used transmission lines are a simple two-wire line, a co-axial cable,
or a circuit board trace and return path as in the case of microstrip and stripline.
At DC and low frequencies, where wavelength is much longer than the circuit path, a transmission line was often thought of in terms of resistance only and early transmission line modeling was based
on resistive loss with voltage drop as the only real concern. Analysis could be done using traditional static circuit theory and lumped elements. An equivalent electrical circuit used for DC analysis
of low speed transmission lines could be created with resistive elements only, as follows:
Figure 1. Resistive Model Of A DC Or Low Speed Transmission Line
The voltage drop in each conductor is calculated using only the resistance, R, of the line and the current flowing in the line. The larger the cross sectional area of the conductor, and the higher
the conductivity of the material, the lower the resistance. The conductance, G, represents the leakage current through the dielectric material between the two conductors, at DC and low frequencies
this can be dismissed as negligible.
At higher frequencies however, the wavelength is now much shorter than the typical circuit size, the parasitic inductive and capacitive elements of the transmission line start to have an effect and
the behavior of the transmission line is considerably different. Analysis requires high frequency transmission line theory with distributed elements.
An equivalent electrical circuit for a high speed transmission line can be drawn with only passive components in a ladder network. The complete transmission line is made of cascaded sections of the
same equivalent circuit. For exact analysis we would need an infinite number of infinitely small segments of length x. The transmission line can then be represented as follows:
Figure 2. Complete Model Of A High Speed Transmission Line Showing Parasitic Elements
If we assume the transmission line is physically constant and the dielectric is uniform over the entire length, then usable results can be modeled by defining the characteristics of the transmission
line per unit length.
There is a wealth of information regarding the analysis of transmission line characteristics and the effect these characteristics have on high frequency signals. It is important that layout designers
understand that the traces they create in their designs are not simply copper connections to transfer a signal from one point to another, albeit that is what they actually do. These connections are
actually transmission lines, and need to be designed carefully if they are not to cause the design to function poorly or even not at all.
The higher the operating frequencies, and more importantly, the faster the edge rates of the signals, the more important it is to understand that transmission line design needs to be considered.
Higher signal speeds and edge rates means that what used to be considered a short trace is now actually a long trace in terms of the wavelength of the signal.
Fortunately, there are also many advances in both device technology and fabrication processes that actually work in the layout designer’s favor when designing circuit boards for these high
frequencies. Traces and spaces can be much smaller than was previously the case, dielectrics are much thinner and of higher quality materials, and components are becoming much smaller which allows us
to place them closer together. These factors all work to allow us to create complex, high density, high speed designs where the transmission length of the signals can be much shorter than was
previously the case.
Detailed transmission line analysis is an extremely complex topic, and is well beyond the scope of this guideline. In the next sections some common types of planar transmission lines used in circuit
board design are shown and the formulae for calculating the characteristic impedance and propagation delay. This is by no means a complete reference on any of these transmission line structures. One
of the most complete references available, and highly recommend to both layout designers and electrical engineers, is:
This post is a condensed snippet from A Practical Guide to RF and Mixed Technology Printed Circuit Board Layout, written by Optimum's Brendon Parise. Available for purchase by following this link.
If you think we’re the right fit for your next project, or just have a quick question, we’d love to hear from you. | {"url":"https://www.optimumdesign.com/blogs-and-articles/transmission-lines-pt1","timestamp":"2024-11-10T09:32:27Z","content_type":"text/html","content_length":"55995","record_id":"<urn:uuid:c6531709-0244-4132-94d4-ab6f064681bb>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00784.warc.gz"} |
Arrival-Time Detection in Wind-Speed Measurement: Wavelet Transform and Bayesian Information Criteria
School of Automation Engineering, University of Electronic Science and Technology of China, Chengdu 611731, China
Center for Information Geoscience, University of Electronic Science and Technology of China, Chengdu 611731, China
Author to whom correspondence should be addressed.
Submission received: 3 November 2019 / Revised: 29 December 2019 / Accepted: 1 January 2020 / Published: 2 January 2020
The time-difference method is a common one for measuring wind speed ultrasonically, and its core is the precise arrival-time determination of the ultrasonic echo signal. However, because of
background noise and different types of ultrasonic sensors, it is difficult to measure the arrival time of the echo signal accurately in practice. In this paper, a method based on the wavelet
transform (WT) and Bayesian information criteria (BIC) is proposed for determining the arrival time of the echo signal. First, the time-frequency distribution of the echo signal is obtained by using
the determined WT and rough arrival time. After setting up a time window around the rough arrival time point, the BIC function is calculated in the time window, and the arrival time is determined by
using the BIC function. The proposed method is tested in a wind tunnel with an ultrasonic anemometer. The experimental results show that, even in the low-signal-to-noise-ratio area, the deviation
between mostly measured values and preset standard values is mostly within 5 μs, and the standard deviation of measured wind speed is within 0.2 m/s.
1. Introduction
Accurate measurement of wind speed is of considerable significance in many fields [
]. In atmospheric science, accurate measurement of wind speed has a direct impact on accurate weather forecasting [
]. In agriculture, accurate wind-speed measurements contribute to crop cultivation and growth [
]. On an entire high-speed rail line, one anemometer is placed at every mile to measure wind speed and direction for the safety of the trains using that line [
]. In the military, it is necessary to measure wind speed accurately in a complex environment, as it is vitally essential for the precise targeting of weapons [
]. In the manufacturing process of wind tunnels, accurate wind-speed measurement is needed to calibrate wind speed in wind tunnels [
]. At present, prevailing wind-measuring instruments include mechanical anemometers [
], thermal anemometers [
], pitot tube anemometers [
], and ultrasonic wind-measuring instruments [
]. A mechanical anemometer has a rotating part that is easily damaged in use. The measuring range of wind speed is limited because of the design principle of the thermal anemometer. When utilizing a
pitot-tube anemometer, temperature must be measured, and its magnitude is greatly influenced by environmental factors, while the application conditions are too harsh. In contrast, ultrasonic
anemometers are widely utilized because they have no mechanical structure, no start-up wind speed limitation, a wide measurement range, and many other advantages [
Recently, there have been many methods utilized in ultrasonic anemometers, such as the phase-difference [
], Doppler [
], and time-difference [
] methods. The Doppler method must measure the current temperature and requires that the air in the measured wind field contains suspended particles, which requires more stringent testing conditions.
The phase-difference method is also difficult to implement due to its complex underlying principle. Li et al. proposed an array signal processing method to design an ultrasonic anemometer [
], but only simulation experiments were carried out. Its practicability must still be studied further.
The time-difference method is a mainstream approach adopted in ultrasonic anemometers that includes a direct time-difference method and a cross-correlation time-difference method [
]. As mentioned above, the key to the time-difference method is determining the arrival time of the ultrasonic echo signal accurately. The commonly used methods for measuring the arrival time of an
echo signal include the peak-value, information criteria [
], and Teager-Kaiser energy operator (TKEO) [
] methods. The peak-value method is widely applied in anemometers. It utilizes the time corresponding to the sampling point with the largest amplitude in the echo as the arrival time of the echo
signal, but this method is so subjective and open to being affected by noise. The information criteria function method, which was utilized to determine the location of acoustic emissions (AEs) from
concrete by Kurz et al. [
], finds the envelope of the signal first, and then determines the arrival time of the echo by calculating the Akaike information criterion (AIC) function of the envelope of the signal. Furthermore,
Liu et al. utilize the Hilbert-Huang Transform (HHT) and AIC function to determine the arrival time of the impact signal [
]. They divide the signal into two local stationary auto-regressive (AR) processes that correspond to the noise part before the signal arrives and the signal itself. After determining the order of
the AR model, the time corresponding to the minimum value of the AIC function is the arrival time of the echo signal. However, in the case of long-time series, the method needs a complex
multi-independent variables fitting model to increase the fitting accuracy, which aggravates the computational complexity. The TKEO method, which is adopted to determine the arrival time of P and S
waves in earthquakes by Ismail et al. [
], is based on a hybrid usage of empirical mode decomposition and TKEO algorithms.
For feature recognition of non-stationary signals, many signal processing methods have been advanced. The wavelet transform (WT) is a superior algorithm among them [
] which has multi-resolution characteristics and enables a target signal to be observed from coarse to fine. In this paper, according to the non-stationarity of an ultrasonic echo signal, we propose
a method for determining the arrival time of ultrasonic echo signals based on the WT and Bayesian information criteria (BIC). First, the WT is utilized for the time-frequency analysis of noisy
signals to obtain the time-frequency distribution. Then, the BIC function is calculated, and its sampling point with the minimum value of the function corresponds to the arrival time of the echo
signal. To verify the accuracy and stability of this method, a three-dimensional ultrasonic anemometer is designed to compare the method with three other methods in a wind tunnel. The experimental
results show the advantages of the proposed method.
2. Method
2.1. Time-Frequency Location Analysis
The ultrasonic echo signal is considered to be non-stationary. Its characteristic is that it has two parts: a non-signal part containing only noise before the signal arriving, and an effective part
after the signal arrives. Because the acoustic signal excited by the ultrasonic transducer has a single frequency, the echo signal is also of single frequency. A typical echo signal is depicted in
Figure 1
. Noise, which is considered to be random and uncorrelated to the echo signal, is sampled by sensors before the arrival of the echo signal. The sensor is based on the mechanical principle, and it
produces a tail vibration that makes it impossible for the echo to correspond with the transmitted acoustic wave after receiving the ultrasonic signal, as shown in
Figure 1
. Therefore, the best way to recognize the ultrasonic echo signal is to determine the arrival time of the first wave.
However, the frequency spectra of the noise and echo signal overlap significantly in frequency range. In a complex environment, noise sources, such as strong electromagnetic fields and high
temperature, inevitably interfere with the instrument [
Figure 2
shows a group of typical echo signals in a wind tunnel with electromagnetic interference. In
Figure 2
a, the amplitude of background noise has exceeded the amplitude of the first wave, and because of the existence of electromagnetic interference, the echo waveform exhibits some distortion. The
spectrum of background noise is uniformly distributed in the entire frequency domain, as is shown in
Figure 2
b, which demonstrates that the arrival time of the first wave cannot be clearly identified, and the filter cannot effectively eliminate the noise.
To accurately determine the arrival time of the first wave in noise, we must utilize parameters with distinct characteristics when the first wave arrives. As mentioned above, it can be seen that the
frequency of the signal changes significantly when the echo arrives, so the joint time-frequency analysis can be utilized to estimate the arrival time of the first wave. When the echo signal arrives,
the frequency of the signal increases rapidly with time, which means that when the first wave arrives, the joint time-frequency analysis requires good time resolution to provide accurate
time-frequency positioning. This can be achieved by pre-processing the signal using the WT. That involves using a signal scalogram to carry out time-frequency analysis and obtain the distribution
relationship between time, frequency, and energy.
The WT, proposed by Morlet [
], can provide a time-frequency window that changes with frequency. The definition of a continuous WT is expressed as
$W x ( a , b ) = 1 a ∫ − ∞ + ∞ x ( t ) ψ * ( t − b a ) d t ,$
is a scale factor associated with frequency and
is a displacement factor that describes wavelet movement in the time domain. A mother wavelet generates a function family
$ψ a , b ( t ) = 1 a ψ ( t − b a )$
by changing
$x ( t )$
is an echo signal that changes persistently and
$ψ ( t )$
is the mother wavelet that must satisfy the following conditions:
$∫ − ∞ + ∞ ψ ( t ) d t = 0 .$
In addition, a Fourier transform (FT)
$Ψ ( ω )$
of wavelet function
$ψ ( t )$
must satisfy the following conditions:
$C ψ = ∫ − ∞ + ∞ | Ψ ( ω ) | 2 | ω | d ω < ∞$
The selection of the wavelet function is essential in the WT, which is related to whether good resolution can be obtained in the time-frequency domain. The frequency of the echo signal received by
the sensor is a single one, which means a high temporal resolution is required for time localization of this single frequency. The Morlet WT (MWT), which utilizes the Morlet wavelet function, is an
ideal option [
], and its analytic formula is
$ψ ( t ) = e j ω 0 t e − t 2 2 , ω 0 ≥ 5 ,$
$ω 0$
is the wavelet central frequency and
an imaginary number unit. Therefore, Equation (1) can be rewritten as
$| W x ( a , b ) | 2 = | 1 a ∫ − ∞ + ∞ x ( t ) e − j w 0 t − b a e − ( t − b ) 2 2 a 2 d t | 2 ,$
$| W x ( a , b ) | 2$
is a scalogram of the echo signal that denotes the energy distribution by displacement factor
and scale factor
. The scale factor
is related to the frequency
as expressed by
$f S$
is the sampling frequency. Thus, Equation (5) can be rewritten as
$| W x ( f , b ) | 2 = | 2 π f ω 0 f S ∫ − ∞ + ∞ x ( t ) e − j 2 π f ( t − b ) f S e − 2 [ π f ( t − b ) ] 2 ( ω 0 f S ) 2 d t | 2 ,$
according to which the time-frequency relationship of the typical echo signal can be obtained and the time-frequency distribution of the echo signal drawn as in
Figure 3
Figure 3
shows the MWT of
Figure 2
a. In
Figure 3
, the time window clearly shows that the energy changes significantly with signal frequency when the echo signal arrives. Tests on numerous amounts of noise show the same regularity. Based on this
characteristic, sampling points that change with abrupt energy can be found on the echo signal, which can be utilized as a reference to find the precise arrival time of the echo signal.
2.2. Accurate Arrival-Time Determination of Echo Signal Based on BIC
The sampling points obtained by the WT in the time window contain the precise arrival time of the echo signal, and the real arrival time needs further precise positioning. As mentioned in the
Introduction, the BIC function is applied to the pre-processed result using a certain method that is defined as
$B I C = − 2 ln ( L ) + k ln n ,$
is a likelihood function,
the order of the model, and
the number of sampling points [
]. To distinguish noise from the echo signal, one ideal method combines the AR model and BIC to obtain accurate sampling points of first wave [
]. The
th-order AR model is expressed as [
$u t = ∑ i = 1 m a i u t − i + ε t ,$
$ε t$
is Gaussian white noise with zero mean, and its variance is
$σ 2$
$a i$
is the AR coefficient; and
$u t$
denotes the discrete ultrasonic echo signal acquired by an analog-to-digital converter.
$ε t$
also obeys the normal distribution, which means that the probability density function of it is [
$ρ ( ε t ) = 1 2 π σ 2 e − ε t 2 2 σ 2 .$
The rough sampling point $P$, which corresponds to the arrival time of the first wave and is obtained by the MWT, is defined as $S 2$. Before and after the point $P$, part of the echo signal is cut
out as the time window that contains sampling points, and its length is $N$. Thus, the starting point of the acquisition window, which is defined as $S 1$, is $P − 0.5 N$, and the ending point of the
window, which is defined as $S 3$, is $P + 0.5 N$. $M ( 1 )$ and $M ( 2 )$ denote the orders of the AR model before and after the $P$ point, respectively.
Therefore, according to Equations (9) and (10) and the definition of the likelihood function [
], the likelihood function of all sampling points in the entire time window is
$L ( a i , σ i 2 ) = ∏ i = 1 2 ( 1 2 π σ i 2 ) Δ N i 2 e x p ( − 1 2 σ i 2 ∑ t = h i s i ( x t − ∑ m = 1 M ( i ) a m i x t − m ) 2 ) ,$
$Δ N i$
is defined as
$S i − S i − 1$
$h 1$
$P − 0.5 N$
$h 2$
$P + 1$
$x t$
is the discrete ultrasonic echo signal, and
$σ i$
is the variance of the noise corresponding to the AR model of order
$M ( i )$
. The maximum value of Equation (11) is the maximum likelihood estimation of the echo signal [
], which is
$L ( σ i 2 ) = − N 2 ( 1 + ln 2 π ) − 1 2 ∑ i = 1 2 Δ N i σ i 2 ,$
$σ i 2$
$1 Δ N i ∑ t = S i K ( x t − ∑ m = 1 M ( i ) a m ( i ) x t − m ) 2$
when the derivative of Equation (11) is 0 [
Thus, according to Equations (8) and (12), the AR-BIC picker of the echo signal can be obtained as
$B I C ( K , M ) = N ( 1 + ln 2 π ) + Δ N 1 ln [ 1 Δ N 1 ∑ t = S 1 K ( x t − ∑ m = 1 M ( 1 ) a m ( 1 ) x t − m ) 2 ] + Δ N 2 ln [ 1 Δ N 2 ∑ t = k + 1 S 3 ( x t − ∑ m = 1 M ( 2 ) a m ( 2 ) x t − m ) 2
] + ( ∑ i = 1 2 M ( i ) ) ln N = N ( 1 + ln 2 π ) + Δ N 1 ln σ 1 2 + Δ N 2 ln σ 2 2 + ( ∑ i = 1 2 M ( i ) ) ln N ,$
is the range through all sampling signal points in the time window. According to references [
], the echo signal received by sensors can be considered as the pure echo signal with added Gaussian white noise, and the noise is uncorrelated with the echo signal. Thus, the variance of the noise,
$σ i$
, can be regarded as the variance of the echo signal received by sensors, and Equation (13) can be rewritten as
$B I C ( K , M ) = N ( 1 + ln 2 π ) + Δ N 1 ln [ v a r ( x ( S 1 , K ) ) ] + Δ N 2 ln [ v a r ( x ( K + 1 , S 2 ) ) ] + ( ∑ i = 1 2 M ( i ) ) ln N ,$
$v a r ( x ( S 1 , K ) )$
denotes the variance of the corresponding interval from
$S 1$
in the time series
$x ( t )$
. When Equation (14) takes minimum values, it determines the arrival time of the echo signal accurately, which is shown in
Figure 4
Figure 4
a shows the time window containing the precise arrival time of the echo signal and
Figure 4
b is the corresponding BIC function. The arrival time of the echo signal indicated by the dotted line is 0.124 ms, which is measured by the proposed method, while the actual determined time of the
echo signal without noise is 0.125 ms, which is indicated by a vertical straight line. It can be seen from
Figure 4
a that the noise has a major influence on the echo signal, so that the characteristics of the signal cannot be identified directly. However, the BIC function can effectively identify the original
characteristics of the signal from the noise and accurately find the arrival time of the echo signal, as shown in
Figure 4
b. The result indicates the satisfactory precision of our method.
Figure 5
summarizes the proposed method.
3. Experiments and Results
3.1. Experimental Platform
To test the proposed method, a three-dimensional ultrasonic anemometer was designed, consisting of a three-dimensional nonorthogonal ultrasonic sensor array, echo signal acquisition circuit, control
and computing core, ultrasonic sensor drive circuit, and host computer, which is shown in
Figure 6
c. The ultrasonic sensor array is composed of three pairs of ultrasonic sensors that emit the ultrasound in sequence and receive it from the corresponding sensors. All sensors are Airmar AT200 type
sensors with a working frequency of 200 kHz. As shown in
Figure 6
a, the experiment was carried out in the Low-Speed Wind Tunnel at the China Aerodynamics Research and Development Center.
3.2. Performance under Different Signal-To-Noise Ratios
To test the stability of the proposed method, we designed several actual tests. We fixed the distance of one pair of sensors and measured the standard arrival time of the echo signal in the noiseless
environment. Then, 500 tests were carried out under different signal-to-noise ratios (SNRs) to obtain the measured arrival times using the proposed method. For comparison, the Kurz, TKEO, and
high-order statistics methods were carried out in the same test environment.
Figure 7
Figure 8
Figure 9
show the distribution of deviation between measured time
and standard time
denoted by Δ
, which is obtained from the proposed, the Kurz, and the TKEO methods, as well as the high-order statistics method under experimental conditions of 10, 5, and 0 dB, respectively. The abscissa denotes
, and the ordinate denotes the number of each Δ
Figure 7
Figure 8
Figure 9
As shown in
Figure 7
Figure 8
Figure 9
, most of the deviations distribute close to zero and are less than ±5 μs using the proposed method, while the distribution is quite intensive. In addition, the distribution shows great stability as
the SNR decreases. In contrast, the distributions of Δ
are out of the range of ±15 μs using the Kurz, TKEO, and high-order statistics methods, which means that the measured results acquired from the three methods are more dispersive than those obtained
utilizing the proposed method, and are more likely to have a relatively larger error in determining the arrival time of the echo signal.
3.3. Stability and Accuracy Tests of Wind Speed in Wind Tunnel
To test the stability and accuracy of the proposed method in actual wind-speed measurements, three pre-set wind speeds were measured with the above-described ultrasonic anemometer in a wind tunnel.
The wind speeds were preset to four separate levels in the wind tunnel: 0, 10, 15, and 20 m/s. Eight tests were carried out at each wind speed, and 150 wind speed data continuously measured in each
test. The ambient temperature and relative humidity in the wind tunnel were 16.3 °C and 58%, respectively. The average and standard deviations of a series of wind speeds obtained from each continuous
test were calculated, as shown in
Figure 10
Table 1
Figure 10
a shows the experiment without opening the wind tunnel, in which the theoretical wind speed was 0 m/s. According to reference [
], when the wind tunnel is not opened, the measured wind speed within 0.1 m/s can be regarded as being 0 m/s. The results shown in
Figure 10
a indicate that this method can accurately measure the arrival time of the echo signal and has a strong anti-noise ability for background noise.
Figure 10
b–d show the results of testing the actual wind speed in the wind tunnel when the pre-set wind speeds were 10, 15, and 20 m/s, respectively. In each wind-speed test, seen from the viewpoints of the
average and standard deviation of those experiments, the deviation between the mean value and the wind speed produced by the wind tunnel is within 0.16 m/s, and the standard deviation is within 0.2 m
/s, which means that the proposed method has great accuracy and stability.
3.4. Discussion of Results of Wind-Speed Tests
It can be found from
Figure 10
that the standard deviation of test results increases with increasing wind speed in the wind tunnel. After carefully checking the anemometer, it is found that the bracket carrying the ultrasonic
sensors has installation tolerance. With an increasing wind speed, the bracket oscillates to a certain extent, which leads to waveform distortion when the sensor receives the ultrasonic echo signal.
Since the anemometer is a prototype instrument utilized to verify the method, we plan to redesign a new bracket with ultrasonic sensors installed to ensure that this phenomenon will not re-occur.
4. Conclusions
In this paper, a new method is proposed to determine the precise arrival time of an ultrasonic echo signal. The method is based on the WT and BIC. The time-frequency distribution of echo signals with
strong background noise is obtained by the WT, and the time corresponding to the beginning of frequency variation is determined as the rough position of arrival time. Then, based on the rough
position, a segment containing sampled points is cut out before and after the rough point as the time window. Finally, the BIC function of the echo signal in the time window is calculated, and the
time corresponding to the minimum BIC-function value is determined as the precise arrival time of the ultrasonic echo signal.
To verify the proposed method, we designed an ultrasonic anemometer and tested it in a wind tunnel. To demonstrate the stability and anti-noise ability of the proposed method, the Kurz, TKEO, and
high-order statistics methods were compared. After 500 experiments, the distribution of the deviation between the statistical measured value and pre-supposed standard value show that our method is
much more accurate in determining the arrival time. Moreover, the actual different wind speeds in the wind tunnel were measured by the proposed method, and the results also verify that the proposed
method performs satisfactorily in terms of stability and accuracy.
Author Contributions
W.Z. and Z.L. designed the method in this paper; Z.L. embedded the method in the anemometer; X.G. designed the structure of the anemometer; Y.L. built the hardware platform of the anemometer; and
Y.S. revised the paper. All authors have read and agreed to the published version of the manuscript.
This research is supported by the National Science Foundation of China (Grant No. 61201131) and the Fundamental Research Funds for the Central Universities (Grant No. ZYGX2016J104).
We thank Shiyuan Liu, Hu Sun, and Hengyu Liu for their valuable comments on the writing of this paper.
Conflicts of Interest
The authors declare no conflicts of interest.
Figure 2. Influence of noise on an echo signal: (a) an echo signal with noise; (b) the signal frequency spectrum.
Figure 6. Hardware comprising experimental platform: (a) anemometer in wind tunnel; (b) sensor positions; (c) overall structure of experimental anemometer.
Figure 7. Distribution of deviation in the environment of 10 dB: (a) proposed, (b) Kurz, (c) TKEO, and (d) high-order statistics methods.
Figure 8. Distribution of deviation in the environment of 5 dB: (a) proposed, (b) Kurz, (c) TKEO, and (d) high-order statistics methods.
Figure 9. Distribution of deviation in the environment of 0 dB: (a) proposed, (b) Kurz, (c) TKEO, and (d) high-order statistics methods.
Wind Speed in Wind Tunnel Mean Measured Value Standard Deviation
(m/s) (m/s) (m/s)
0 0.07 0.03
10 10.15 0.09
15 15.06 0.16
20 20.16 0.19
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Zhang, W.; Li, Z.; Gao, X.; Li, Y.; Shi, Y. Arrival-Time Detection in Wind-Speed Measurement: Wavelet Transform and Bayesian Information Criteria. Sensors 2020, 20, 269. https://doi.org/10.3390/
AMA Style
Zhang W, Li Z, Gao X, Li Y, Shi Y. Arrival-Time Detection in Wind-Speed Measurement: Wavelet Transform and Bayesian Information Criteria. Sensors. 2020; 20(1):269. https://doi.org/10.3390/s20010269
Chicago/Turabian Style
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KoulMde: Koul's Minimum Distance Estimation in Regression and Image Segmentation Problems
Many methods are developed to deal with two major statistical problems: image segmentation and nonparametric estimation in various regression models. Image segmentation is nowadays gaining a lot of
attention from various scientific subfields. Especially, image segmentation has been popular in medical research such as magnetic resonance imaging (MRI) analysis. When a patient suffers from some
brain diseases such as dementia and Parkinson's disease, those diseases can be easily diagnosed in brain MRI: the area affected by those diseases is brightly expressed in MRI, which is called a white
lesion. For the purpose of medical research, locating and segment those white lesions in MRI is a critical issue; it can be done manually. However, manual segmentation is very expensive in that it is
error-prone and demands a huge amount of time. Therefore, supervised machine learning has emerged as an alternative solution. Despite its powerful performance in a classification problem such as
hand-written digits, supervised machine learning has not shown the same satisfactory result in MRI analysis. Setting aside all issues of the supervised machine learning, it exposed a critical problem
when employed for MRI analysis: it requires time-consuming data labeling. Thus, there is a strong demand for an unsupervised approach, and this package - based on Hira L. Koul (1986) <doi:10.1214/aos
/1176350059> - proposes an efficient method for simple image segmentation - here, "simple" means that an image is black-and-white - which can easily be applied to MRI analysis. This package includes
a function GetSegImage(): when a black-and-white image is given as an input, GetSegImage() separates an area of white pixels - which corresponds to a white lesion in MRI - from the given image. For
the second problem, consider linear regression model and autoregressive model of order q where errors in the linear regression model and innovations in the autoregression model are independent and
symmetrically distributed. Hira L. Koul (1986) <doi:10.1214/aos/1176350059> proposed a nonparametric minimum distance estimation method by minimizing L2-type distance between certain weighted
residual empirical processes. He also proposed a simpler version of the loss function by using symmetry of the integrating measure in the distance. Kim (2018) <doi:10.1080/00949655.2017.1392527>
proposed a fast computational method which enables practitioners to compute the minimum distance estimator of the vector of general multiple regression parameters for several integrating measures.
This package contains three functions: KoulLrMde(), KoulArMde(), and Koul2StageMde(). The former two provide minimum distance estimators for linear regression model and autoregression model,
respectively, where both are based on Koul's method. These two functions take much less time for the computation than those based on parametric minimum distance estimation methods. Koul2StageMde()
provides estimators for regression and autoregressive coefficients of linear regression model with autoregressive errors through minimum distant method of two stages. The new version is written in
Rcpp and dramatically reduces computational time.
Version: 3.2.1
Depends: R (≥ 3.2.2)
Imports: Rcpp (≥ 0.12.7), expm
LinkingTo: Rcpp, RcppArmadillo
Published: 2020-09-10
DOI: 10.32614/CRAN.package.KoulMde
Author: Jiwoong Kim <jwboys26 at gmail.com>
Maintainer: Jiwoong Kim <jwboys26 at gmail.com>
License: GPL-2
NeedsCompilation: yes
CRAN checks: KoulMde results
Reference manual: KoulMde.pdf
Package source: KoulMde_3.2.1.tar.gz
Windows binaries: r-devel: KoulMde_3.2.1.zip, r-release: KoulMde_3.2.1.zip, r-oldrel: KoulMde_3.2.1.zip
macOS binaries: r-release (arm64): KoulMde_3.2.1.tgz, r-oldrel (arm64): KoulMde_3.2.1.tgz, r-release (x86_64): KoulMde_3.2.1.tgz, r-oldrel (x86_64): KoulMde_3.2.1.tgz
Old sources: KoulMde archive
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• Objectives are those things we want to minimize, or maximize.
• Expressed as a function of variables that provides a value
• Consider the example of building a fenced pasture. In this case when the area becomes too large, there is a reduced value. We want to maximize the value of V.
Figure 26.1 Example cost function for building a fence around a pasture
• The cost function can be written as..
Figure 26.2 A subroutine for cost function calculation
• Constraints are boundaries that cannot be crossed.
• Example of constraints, the pasture cannot be larger than one 1600m be 1600m beacuse of the constraints of an existing road system.
Figure 26.3 Example constraint functions for a pasture
• The cost function can be written as..
Figure 26.4 A subroutine for cost function calculation
• Slack variables allow constraints to be considered as part of the cost function. Helps with a system with many local minimum.
Figure 26.5 Example of slack variables for including constraints
• The cost function can be written as.. | {"url":"https://engineeronadisk.com/V2/book_modelling/engineeronadisk-246.html","timestamp":"2024-11-02T14:43:14Z","content_type":"text/html","content_length":"4291","record_id":"<urn:uuid:4be83f1c-937f-4e2e-be3e-435d9b655f4d>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00013.warc.gz"} |
Multiple Level Pie Chart Excel 2024 - Multiplication Chart Printable
Multiple Level Pie Chart Excel
Multiple Level Pie Chart Excel – You can create a multiplication graph in Excel using a design. You will find many examples of templates and figure out how to structure your multiplication graph
making use of them. Here are some tricks and tips to generate a multiplication graph. When you have a format, all you want do is duplicate the solution and mixture it within a new cellular. Then you
can utilize this solution to multiply a number of numbers by an additional established. Multiple Level Pie Chart Excel.
Multiplication table web template
You may want to learn how to write a simple formula if you are in the need to create a multiplication table. Initial, you should secure row one of several header column, then flourish the quantity on
row A by cellular B. A different way to create a multiplication table is to try using combined personal references. In such a case, you will enter $A2 into line A and B$1 into row B. The effect can
be a multiplication kitchen table by using a method that actually works for columns and rows.
If you are using an Excel program, you can use the multiplication table template to create your table. Just available the spreadsheet with the multiplication table change and template the title on
the student’s name. You can even alter the sheet to fit your person requires. It comes with an option to change the shade of the cellular material to improve the look of the multiplication kitchen
table, too. Then, it is possible to change the plethora of multiples to meet your requirements.
Building a multiplication graph in Shine
When you’re making use of multiplication kitchen table computer software, it is possible to produce a easy multiplication table in Stand out. Basically create a page with rows and columns numbered in
one to 30. In which the columns and rows intersect is the respond to. If a row has a digit of three, and a column has a digit of five, then the answer is three times five, for example. The same thing
goes for the opposite.
Very first, you are able to enter into the figures that you have to flourish. If you need to multiply two digits by three, you can type a formula for each number in cell A1, for example. To create
the amounts greater, pick the tissue at A1 and A8, and then go through the appropriate arrow to choose an array of cellular material. You can then sort the multiplication solution within the tissue
inside the other rows and columns.
Gallery of Multiple Level Pie Chart Excel
Create Multiple Pie Charts In Excel Using Worksheet Data And VBA
Pie Chart With Three Fields In Ms Excel Stack Overflow
HowTo Multilevel Pie In Excel YouTube
Leave a Comment | {"url":"https://www.multiplicationchartprintable.com/multiple-level-pie-chart-excel/","timestamp":"2024-11-12T06:38:13Z","content_type":"text/html","content_length":"52983","record_id":"<urn:uuid:24392a45-1a6e-4894-a75e-bef1312a8ce7>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00300.warc.gz"} |
Geometry and complexity theory
• Complexity theory addresses both practical and theoretical questions. Practical issues include developing new, efficient algorithms for computations that need to be done frequently (such as
multiplying matrices), as well as determining whether more efficient algorithms than the ones already known may exist. The theoretical aspects include questions almost philosophical in nature,
such as: Is there really a difference between intuition and systematic problem solving? (This question was asked by Godel and contributed to the development of the famous P versus NP conjecture.)
The project will use modern mathematical tools to address these questions, more specifically, algebraic geometry and representation theory. Although the research for this project is driven by
questions from computer science, these questions are of interest to mathematics in their own right and have the potential to guide future mathematical research in the way physics has done in
recent years.This project focuses on two central questions in complexity theory: the complexity of matrix multiplication and the Geometric Complexity Theory approach to Valiant''s conjectures.
Regarding matrix multiplication, linear algebra is central to all applications of mathematics, and matrix multiplication is the essential operation of linear algebra. In 1969 Strassen discovered
a new algorithm to multiply matrices significantly faster than the standard algorithm. This and subsequent work has led to the astounding conjecture that asymptotically, it is essentially almost
as easy to multiply matrices as it is to add them. It is a central question to determine just how efficiently one can multiply matrices, both practically and asymptotically. This project will
prove bounds for how efficiently matrices can be multiplied. Regarding Geometric Complexity Theory, a central question in theoretical computer science is whether brute force calculations can be
avoided in problems such as the traveling salesman problem. This is the essence of the P versus NP conjecture. The Geometric Complexity Theory (GCT) program addresses such questions using
geometric methods. GCT touches on central questions in algebraic geometry, differential geometry, representation theory and combinatorics. The goals of this project are (i) to better establish
the mathematical foundations of GCT by solving open problems in combinatorics and classical algebraic geometry and (ii) to solve complexity problems considered more tractable, such as determining
whether or not the determinant polynomial admits a small formula. | {"url":"https://vivo.library.tamu.edu/vivo/display/nc25bf573-e875-11ea-a81a-005056bb4fb7","timestamp":"2024-11-09T14:19:05Z","content_type":"text/html","content_length":"17657","record_id":"<urn:uuid:6029f9f6-7f03-4ab6-a8e4-1d3f6c2bee17>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00441.warc.gz"} |
Lens positive and negative judgment -factory custom- BORISUN
It is not rigorous to judge the positive and negative focal length of the lens simply by looking. If you have to “look”, judge by observing whether the thickness of the center of the lens is greater
than the thickness of the edge. The lens with the thickness of the center greater than the thickness of the edge is generally positive. Generally, the lenticular lens is positive, the lenticular lens
is negative, and the meniscus lens can be positive or negative (the above judgment is based on experience and is not very accurate).
The focal length of lens is determined by its thickness, refractive index and curvature, which is a complex formula [1]:
In Formula 1, R1 and R2 are the radius of curvature of the front and rear surfaces of the lens respectively, D is the thickness of the lens, and N is the refractive index of the lens material.
Based on Equation 1, the positive and negative focal lengths of various types of lenses are briefly discussed [2]:
1. Biconvex lens
For this type of lens, R1 > 0, R2 < 0. It can be seen from equation 1 that if the two surfaces R1 and R2 of the lens are kept unchanged, the focal length can be positive or negative with the
different thickness D.
2. Double concave lens
For this type of lens, R1 < 0, R2 > 0. It can be seen from equation 1 that the focal length is always negative.
3. Plane convex lens
It is assumed that the front surface of the lens R1 > 0, R2 = ∞. From equation 1, the lens focal length formula is simplified to
Therefore, the image square focal length of the plano convex lens is always positive, and its focal length is independent of the lens thickness.
4. Plane concave lens
Assuming that the front surface of the lens is concave, R1 < 0, R2 = ∞, its focal length is always negative, and its focal length is independent of the lens thickness.
5. Meniscus convex lens
For this type of lens, the symbols of the radius of curvature of the two surfaces are the same, but the absolute value of the radius of curvature of the convex surface is smaller. It can be seen from
equation 1 that the focal length is always positive.
6. Meniscus concave lens
For this type of lens, the two surface curvature radius symbols are the same, but the absolute value of convex curvature radius is larger. It is similar to biconvex lens, and the positive and
negative focal length changes with thickness. Because the radii on both sides of this lens are of the same number, the difference between the two radii is very small, and the effect of given positive
power can be obtained without a large thickness. Therefore, the meniscus concave lens plays an important role in correcting the bending of the image plane. | {"url":"https://www.borisuns.com/lens-positive-and-negative-judgment/","timestamp":"2024-11-11T21:12:57Z","content_type":"text/html","content_length":"296751","record_id":"<urn:uuid:6bcf9510-a63f-4f3c-a3a6-f1a3bd8e1759>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00376.warc.gz"} |
Scale Factor - Formula, Meaning, Examples (2024)
Scale factor is a number by which the size of any geometrical figure or shape can be changed with respect to its original size. It is used to draw the enlarged or reduced shape of any given figure
and to find the missing length, area, or volume of an enlarged or reduced figure. It should be noted that the scale factor helps in changing the size of the figure and not its shape.
1. What is a Scale Factor?
2. Scale Factor Formula
3. How to Find the Scale Factor?
4. FAQs on Scale Factor
What is a Scale Factor?
Scale factor is defined as the number or the conversion factor which is used to change the size of a figure without changing its shape. It is used to increase or decrease the size of an object. The
scale factor can be calculated if the dimensions of the original figure and the dimensions of the dilated (increased or decreased) figure are known. For example, a rectangle has a length of 5 units
and a width of 2 units. Now, if we increase the size of this rectangle by a scale factor of 2, the sides will become 10 units and 4 units, respectively. Hence, we can use the scale factor to get the
dimensions of the changed figures.
Observe the following figure which shows how the scale factor can change the original figure to its larger and smaller versions. In the following figure, the original rectangle has the dimensions
given as 3 units and 2 units. To create an enlarged figure, the dimensions are multiplied by the scale factor of 3 using the formula: Dimensions of the new shape = Dimensions of the original shape ×
Scale factor. This gives the new dimensions as 9 units and 6 units respectively. Similarly, to create a reduced figure, we multiply the original dimensions with the scale factor of 1/2. This gives
the reduced dimensions as 1.5 units and 1 unit.
Scale Factor Formula
The basic formula to find the scale factor of a figure is expressed as,
Scale factor = Dimensions of the new shape ÷ Dimensions of the original shape.
This formula can also be used to calculate the dimensions of the new figure or the original figure by simply substituting the values in the formula.
How to Find the Scale Factor?
The scale factor can be calculated when the new dimensions and the original dimensions are given. However, there are two terms that need to be understood when using the scale factor. When the size of
a figure is increased, we say that it has been scaled up and when it is decreased, we say that it has scaled down.
Scale Up
Scaling up means that a smaller figure is enlarged to a bigger one. In this case, the scale factor can be calculated by a formula, which is another version of the basic formula given in the previous
Scale factor = Larger figure dimensions ÷ Smaller figure dimensions
The scale factor for scaling up is always greater than 1. For example, if the dimension of the larger figure is 15 and that of the smaller one is 5, let us place this in the formula which makes it:
15 ÷ 5 = 3. Thus, we can see that the scale factor is greater than 1.
Scale Down
Scaling down means that a larger figure is reduced to a smaller one. Even in this case, the scale factor can be calculated by a formula, which is another version of the basic formula.
Scale factor = Smaller figure dimensions ÷ Larger figure dimensions
The scale factor for scaling down is always less than 1. For example, if the dimension of the smaller figure is 8 and that of the larger one is 24, let us place this in the formula which makes it: 8
÷ 24 = 1/3. Thus, we can see that the scale factor is less than 1.
Observe the following triangles which explain the concept of a scaled-up figure and a scaled-down figure.
Important Notes
The following points should be remembered while studying about the scale factor:
• The scale factor of a dilated figure is denoted by 'r' or 'k'.
• If the scale factor is more than 1 (k > 1), the image is enlarged.
• If the scale factor is less than 1 (0< k <1), the image is contracted.
• If the scale factor is 1 (k = 1), the image remains the same.
• The scale factor cannot be zero.
☛ Related Articles
• Dilation Geometry
• Scale Factor Worksheets 7th Grade
• Horizontal Scaling
• Vertical Scaling
FAQs on Scale Factor
What is a Scale Factor?
Scale factor is a number that is used to draw the enlarged or reduced shape of any given figure. It is a number by which the size of any geometrical figure or shape can be changed with respect to its
original size. It helps in changing the size of the figure but not its shape.
What Happens if the Scale Factor is Greater Than 1?
If the scale factor is more than 1 (k > 1), it means that the given figure needs to be enlarged.
What Does a Scale Factor of 0.5 Mean?
A scale factor of 0.5 means that the changed image will be scaled down. For example, the original figure of a square has one of its sides as 6 units. Now, let us use the scale factor of 0.5, to
change its size. We will use the formula: Dimensions of the new shape = Dimensions of the original shape × Scale factor. Substituting the values in the formula: the dimensions of the new square will
be = 6 × 0.5 = 3 units. This shows that a scale factor of 0.5 changed the figure to a smaller one.
How do you Find the Scale Factor?
The scale factor can be calculated when the new dimensions and the original dimensions are given. The basic formula to find the scale factor of a figure is: Scale factor = Dimension of the new shape
÷ Dimension of the original shape. For example, if the side length of a square is 6 units and if the size of the square has been increased such that the side length of the square becomes 18 units,
let us find the scale factor. We will use the formula, Scale factor = Dimension of the new shape ÷ Dimension of the original shape. After substituting the given values we get, Scale factor = 18 ÷ 6 =
3. Therefore, the scale factor that is used to increase the size of the square is 3.
What Happens if the Scale Factor is Less Than 1?
If the scale factor is less than 1 (0< k <1), then the new image that is formed will be contracted or scaled-down. In other words, the new figure will have smaller dimensions as compared to the
original figure after it is resized using the scale factor which is less than 1.
Where do we Use a Scale Factor?
Scale factor is a number by which the size of any geometrical figure or shape can be changed with respect to its original size. When things are too large, we use scale factors to calculate smaller,
proportional measurements. It is used to compare two similar geometric shapes and also in other fields like cooking, where the ingredients can be reduced or increased according to the given
situation. Scale factor can also be used to find any missing dimensions in similar figures.
What is the Formula For Scale Factor?
The basic formula that is used for calculating the scale factor is, Scale factor = Dimension of the new shape ÷ Dimension of the original shape. In case, if the original figure is scaled up, the
formula is written as, Scale factor = Larger figure dimensions ÷ Smaller figure dimensions. When the original figure is scaled down, the formula is expressed as, Scale factor = Smaller figure
dimensions ÷ Larger figure dimensions.
What Does a Negative Scale Factor Mean?
A negative scale factor makes the dilation rotate 180° and it creates an image on the other side of the centre of the enlargement.
What Scale Factor makes a Figure Smaller?
A scale factor which is less than 1 makes the original figure smaller. For example, let us use a scale factor of 1/3 to change the size of a figure with a given dimension of 36. We will place the
given values in the formula: Scale factor = Dimension of the new shape ÷ Dimension of the original shape. Substituting the values, we get, 1/3 = Dimension of the new shape ÷ 36. After solving this,
the dimension of the new shape is = 12. Since 12 is smaller than 36, it means the original figure has been reduced in size. Thus, it can be seen that the scale factor which is less than 1 makes a
figure smaller.
How is a Scale Different from a Scale Factor?
• Scale is a ratio that is used to define the relation of the actual figure or object with its model. It is commonly used in maps to represent the actual figures in smaller units. For example, a
scale of 1:3 means 1 on the map represents the size of 3 in the real world.
• Scale factor is a conversion factor - a number which is used to increase or decrease the size of a figure. For example, if a circle needs to be increased in size using a scale factor of 4, and
the circumference of the circle is 7 units. What will be the circumference of the new enlarged circle? We will use the formula, Scale factor = Dimension of the new shape ÷ Dimension of the
original shape. This can be written as, Dimension of the new shape = Scale factor × Dimension of the original shape. After substituting the given values, we will get, Dimension of the new shape =
7. After solving this, we get, Dimension of the new shape = 28 units. | {"url":"https://premiumh2o.biz/article/scale-factor-formula-meaning-examples","timestamp":"2024-11-14T07:47:10Z","content_type":"text/html","content_length":"74532","record_id":"<urn:uuid:742f18b7-0a92-4094-b146-9e5675d5d6df>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00202.warc.gz"} |
19 Year Cycle Lunar Standstill Upcoming
Back when Secret History was going through its first edits, Frank J (QFG
researcher) was quite fascinated by the discussion of the 19 year cycle.
He decided to do some research.
The graphs and things he included in his paper aren't included in the following, but they aren't necessary. What is important are his remarks as well as the upcoming event.
Those who have read "The Secret History of the World" are aware of the possible significance of this event.
19-year Lunar Cycles
By Frank J
I was inspired to put together this little paper by the discussion in
'Secret History' concerning the importance of the 19-year lunar cycle in the
cultures of some ancient civilizations, in particular the culture that
flourished around Stonehenge. The C's suggests that the 19-year cycle is a
window, gravitationally induced, for direct access by humans ("right
people, right place, right activity, right time") can directly access
higher dimensions without any help from other entities. by engaging in
group activities,
carried out in a geometric pattern, and
with certain amplification materials in particular geographic locations,
the participants having fused their magnetic centers,
And, of course, I am also interested in where we are now in that cycle.
The following explanations are cobbled together
from several scientific and archaeological sites on the Internet, and their
web addresses are provided at the end of this article.
Predictable motions of the earth's rotation axis on time scales less than
300 years are all referred to as nutation, a correction to the precession
cycle (26,000 years). The currently standard nutation theory is composed of
106 non-harmonically-related sine and cosine components, mainly due to
second-order torque effects from the sun and moon, plus 85 planetary
correction terms. The four dominant periods of nutation are 18.6 years
(precession period of the lunar orbit), 182.6 days (half a year), 13.7 days
(half a month), and 9.3 years (rotation period of the moon's perigee).
The primary nutation of 18.6 years drives two other observational cycles:
the Saros cycle (18 years and 11 days) and the Metonic Cycle (19 years)
Primary Nutation Cycle
The cause of this cycle is the precession of the lunar orbit about he Earth.
In addition, the Sun's gravitational pull leads to a precession of the
Moon's orbital axis, with a period of 18.6 years. This precession advances
the locations where the moon's orbit crosses the ecliptic (the nodes).
Eclipses will occur on the new or full moon nearest the time that the sun
passes one of the nodes
There are several observational consequences of this cycle, none of which
require a technical knowledge of the lunar precession.
Solar Eclipse Prediction
This cycle causes every eclipse of the sun to repeat itself at a different
place on the earth every 18.6 years. The effect of this "wobble" is that
eclipses seasons occur 365.24/18.6= 19.6 day earlier every year. Thus in
1997, the eclipse season of the fall will be centered about the node 19.6
days earlier than that of the previous year.
Lunar Standstills: Because of the 5.1 degree tilt of the moon's orbit with
respect to the ecliptic, the moon may be anywhere within 5.1 degrees above
or below the ecliptic. During major standstills the moon reaches a
declination of 23.5 plus 5.1 degrees or 28.6 degrees; major standstills
occur every 18.6 years. At minor standstill the greatest declination that
the moon reaches is 23.5 minus 5.1 degrees or 18.4 degrees.
This means that every 18.6 years, the rising or setting Moon reaches a
northern extreme in rising and setting azimuth at summer solstice, and a
southern extreme at winter solstice. These are called major standstills.
While such standstills can in principle be determined using horizon
observations, as with the summer solstice Sun the Moon's year-to-year
angular displacement along the horizon at summer solstice is very small near
standstill. It should be noted that 18.6 years is measured from the point of
view of the lunar orbit. Observationally, from the Earth's surface, the
length of time between two major standstills is not 18.6 years: it switches
back and forth between 18.5 years and 19 years, and 18.6 years is an
observational average. This may become clearer by looking closely at the
behavior of the moon at the time of her extreme positions. This chart shows,
for the current and three most recent nodal cycles, the maximum southern
azimuth position of the rising moon reached during each month for
approximately three years. The data shown are the maximum rising moons for
the periods Mar 1, 1949 - Mar 31, 1952; Sep 1, 1967 - Sep 30, 1970; Mar 1,
1986 - Mar 31, 1989; Sep 1, 2004 - Sep 30, 2007.
To understand what 'standstill" means observationally, it will be easier to
use the Sun as an example. The sun rises furthest to the north at the summer
solistice, around June 21 each year. Following the summer solistice it will
begin to rise a little further south each day, rising due east at the fall
equinox around September 21, and reaching its southernmost rising point at
the winter solistice around December 21. After the winter solstice the sun
will begin rising further north each day, rising due east at the spring
equinox around March 21, and finally reaching its northernmost rising point
again around June 21. The slow sweep of the sun's rising azimuth across the
eastern horizon takes a full year, and practically repeats itself exactly
from year to year. The rising point changes very little from day to day when
it's rising near either the northern or southern extremes of its motion.
This phenomenon is known as the "standstill." For several days around either
solstice the sun's rising azimuth will hardly change at all. In contrast,
when the rising point is between the extremes, say around the equinoxes, the
rising azimuth changes quite a bit from day to day. This phenomenon of
"standstills" near the extremes applies to periodic motion of many kinds,
including the motions of the moon.
The rising point of the moon changes from day to day in a very analogous
way, marking out a sweep from north to south and back again, except that it
takes only one month to accomplish one complete cycle. The actual period of
this cycle is the "draconitic month" of 27.21222 days, on the average.
Unlike the sun, however, the extremes of the northernmost and southernmost
rising azimuths will not remain the same for each cycle. After noting the
northernmost rising point for the moon during one month, one may very well
find it rises at a point even further north the next month. In fact, there
is an 18.61-year variation in the extremes of the moon's rising point.
A major standstill limit will happen at the moment the moon is near a
quarter moon and the lunar node is near the vernal (or autumnal) point. The
moon is at his highest point in its orbit and combining this with lunar
phase, the sun is near equinox.
Saros Cycle
The periodicity of solar eclipses depends upon two lunar orbital cycles
coinciding with the moon passing through a node. First, a new moon occurs on
average every 29.530588 days. The moon's average orbital period, perigee to
perigee, is 27.554548 days. These cycles repeat every 18 years, 11 1/3 days;
or 6585.3211 days; or 233 new moons, approximately 239 perigees and 242
nodes. Every eclipse in a Saros family shares the same 18 year, 11.33 day
223 lunar synodic months = 29.053059*223 = 6585.3216
242 lunar draconitic months = 27.21222*242 = 6585.3572
The Metonic Cycle
Another pattern evident in the table is that the solar calendar dates of
maximum moonrises often repeat, 19 years apart:
1 solar year = 365.2425 days
1 lunar synodic month (full moon to full moon) = 29.53059 days
19 years = 365.2425*19 = 6939.6075 days
235 lunar synodic months = 29.53059*235 = 6939.6887 days
This is a difference of only 0.0812 days, or about two hours. So, after
exactly nineteen solar years the sun will return to the same position
relative to the stars (by definition), and the moon will have very nearly
the same phase (just two hours difference). This fact was much appreciated
by the Greeks, as the dates of the new moon, full moon, etc., would repeat
every nineteen years.
For the purposes of maximum moonrises, the above coincidence alone would not
be enough to ensure that maximum moonrises will occur on the same date 19
years apart, it only guarantees the phase will be the same on the same
Another well-known aspect of the Metonic cycle is that since the sun, moon
and earth return to the same relative positions, the pattern of eclipses of
the moon and sun may repeat somewhat after 19 years elapses. The half-day
difference of the lunar draconitic cycle however is enough to throw the
eclipse repeatability out of kilter fairly rapidly, but three or four
eclipses may repeat, on the same dates 19 years apart, before this happens.
In any given year the solar calendar dates of Full Moon events will be
duplicated every 19 years. This creates a clear cycle connection that was
delineated by the mythic 19 Priestesses of Bridget. Each of these individual
"Priestesses" represented the "character" and experience of each of the 19
components of the Great Lunar Year. This relationship creates an excellent
basis for cyclic pattern divinations. A 19 year cycle upon which one can
hang other, shorter cycling patterns for delineation and understanding
So, I conjecture that the gravitational "window" in TIME that allows "the
right person, in the right group, at the right time and at the right place"
to access hyperdimensions is the standstill point, also called lunarstice.
When is the next one?
September 2, 2006
Sweet! Thanks for the head's up - you don't happen to know the time GMT do you?
interesting that the sun is on the verge of starting a new cycle as well
Based on my handy American Ephemeris for the 21st Century, it looks like it'll be 22:41 (few seconds either way) UT (GMT).
Interesting to note, but is there anything we should do or look for?
I mean, it seems you need to be in the right place, at the right time, and with a group of the right people in order to make use of this window. Did i miss something?
Laura said:
So, I conjecture that the gravitational "window" in TIME that allows "the
right person, in the right group, at the right time and at the right place"
to access hyperdimensions is the standstill point, also called lunarstice.
Hmmm... I suspect that the old places (Stonehedge, Coral Castle, Pyramids)
won't work since the magnetic lie lines have long shifted to other places or
more accurately, the land mass has itself shifted. Need any help to build
another stonehedge somewhere? :)
Perhaps that is why you moved to France since you knew where this
focal/converging/standstill point would all be? I guess it is all comming
together and the start of the "battle"?
What is to be expected, if my guesses are all off?
FOTCM Member
It would be cool if we could fugure out some action to be undertaken by a group to interact in some way with this hypothesised 'opening', but what if we should be more concerned with that which might
take the opportunity to come through it?
Indeed... and interacting with it could be just as dangerous if you don't know what you're doing!
Dagobah Resident
FOTCM Member
Is there any Aussies on this forum that
have read 'secret history'and
would like to meet on 2 .September?
We have some lovely hills around here
and we also poke through that 'electomagnetic blanket' aka you
get no mobil phone reception here .
And those of you in England maybe
stonehenge is out of bounds so to speak
but isnt that what the cropcircles are for?
Like the back-up system? RRR
I live in London, and im quite interested in this cycle thing. If anyone can give a student bum a lift to stonehenge or the likes, it would be most appreciated lol :)
It would be interesting to see if anything can be observed by any groups who do converge on this night. There are probaby enough of us in different countries to have a few groups gathering to look at
things from more than just one vantage point.
I think it is most important for everyone who wants to try something to have read Secret History so as to have all the clues I was able to gather about this event.
What strikes me as most significant is the "Dancing God" and music and the "celebratory" nature of the interaction.
So, ya'll do read up on it and then pool observations and suggestions.
I ordered Secret History 5 days ago, and if it takes 18 days like it said, that leaves me 2 days to cram up on knowledge before 2nd september ^^
FOTCM Member
Laura said:
I think it is most important for everyone who wants to try something to have read Secret History so as to have all the clues I was able to gather about this event.
What strikes me as most significant is the "Dancing God" and music and the "celebratory" nature of the interaction.
So, ya'll do read up on it and then pool observations and suggestions.
1) Lance
2) Chalice
3) Lyre
4) Dancing Shoes
5) Portable Stonehenge
6) Magnetic Center
Hmm... Some of those might be hard to acquire in the next month. :)
A Disturbance in the Force
Laura said:
What strikes me as most significant is the "Dancing God" and music and the "celebratory" nature of the interaction
Joe said:
It would be cool if we could fugure out some action to be undertaken by a group to interact in some way with this hypothesised 'opening', but what if we should be more concerned with that which
might take the opportunity to come through it?
Perhaps the Dancing God is a hyperdimensional being/beings who 'come through' -since there was a celebratory mood, it would indicate that it was not a negative experience for those humans involved,
but what if it wasn't considered negative because these humans didn't fully understand the nature of this Dancing God?
FOTCM Member
anart said:
Laura said:
What strikes me as most significant is the "Dancing God" and music and the "celebratory" nature of the interaction
Joe said:
It would be cool if we could fugure out some action to be undertaken by a group to interact in some way with this hypothesised 'opening', but what if we should be more concerned with that
which might take the opportunity to come through it?
Perhaps the Dancing God is a hyperdimensional being/beings who 'come through' -since there was a celebratory mood, it would indicate that it was not a negative experience for those humans
involved, but what if it wasn't considered negative because these humans didn't fully understand the nature of this Dancing God?
I am re-reading some stuff in SHoTW and Jessup indicates that maybe such nodal points are used as dimensional portals, which seems to imply a two-way street. So while on the one hand such an event as
the upcoming one could be highly beneficial in an STO way provided the opportunity is used properly, it might also be highly beneficial for 4D STS. I thought of looking back to around 1987 (about 18
to 19 years ago) to see how much UFO flap there was. There might be some interesting concentrations of events. | {"url":"https://cassiopaea.org/forum/threads/19-year-cycle-lunar-standstill-upcoming.2890/","timestamp":"2024-11-08T14:01:20Z","content_type":"text/html","content_length":"166073","record_id":"<urn:uuid:57acd1f0-6104-41df-90d4-7b13521c543e>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00714.warc.gz"} |
Weight Per Gallon Calculator - Savvy Calculator
Weight Per Gallon Calculator
About Weight Per Gallon Calculator (Formula)
The Weight Per Gallon Calculator is a valuable tool for various industries, including chemistry, manufacturing, and food production. Understanding the weight of a liquid per gallon is crucial for
accurate measurements, formulation, and transportation. This calculator allows users to quickly determine the weight of any liquid based on its total weight and volume. In this article, we will
explore the formula used for calculating weight per gallon, how to use the calculator effectively, and provide answers to common questions regarding this topic.
The formula for calculating Weight Per Gallon (WPG) is:
Weight Per Gallon (WPG) = Total Weight (W) / Total Volume (G)
In this formula:
• WPG represents the weight per gallon of the liquid.
• W is the total weight of the liquid in pounds or kilograms.
• G is the total volume of the liquid in gallons or liters.
How to Use
Using the Weight Per Gallon Calculator is a straightforward process. Follow these steps to obtain accurate results:
1. Collect Data: Gather the necessary information, including the total weight of the liquid and its total volume.
2. Input Values: Enter the values for total weight (W) and total volume (G) into the calculator. Make sure to use consistent units (e.g., pounds and gallons).
3. Calculate Weight Per Gallon: Click the calculate button to determine the weight per gallon of the liquid based on your inputs.
4. Review Results: Analyze the calculated weight per gallon to understand the properties of the liquid and make informed decisions.
Let’s consider an example to illustrate how to use the Weight Per Gallon Calculator:
• Total Weight (W): 50 lbs
• Total Volume (G): 10 gallons
1. Use the Formula: WPG = 50 lbs / 10 gallons
2. Calculate: WPG = 5 lbs/gallon
In this example, the weight per gallon of the liquid is 5 lbs.
1. What is weight per gallon?
Weight per gallon refers to the weight of a liquid for each gallon of volume, providing insight into its density.
2. Why is it important to know the weight per gallon?
Knowing the weight per gallon is essential for accurate measurements in shipping, formulation, and safety assessments.
3. Can I use the calculator for different liquids?
Yes, the calculator can be used for any liquid as long as you provide the total weight and volume.
4. What units can I use for weight and volume?
You can use pounds or kilograms for weight and gallons or liters for volume, but ensure consistency in your calculations.
5. How does temperature affect the weight per gallon?
Temperature can affect the density of liquids, which in turn impacts their weight per gallon.
6. Can the calculator help with chemical formulations?
Yes, understanding the weight per gallon can aid in precise chemical formulations and mixing.
7. What if I have a liquid that doesn’t conform to standard weights per gallon?
You can still use the calculator by entering the specific weight and volume values for that liquid.
8. How is weight per gallon related to density?
Weight per gallon is a measure of density; higher weight per gallon indicates a denser liquid.
9. Is the weight per gallon the same for all liquids?
No, different liquids have different weights per gallon due to variations in density.
10. What should I do if I only have the weight of a container?
Ensure to subtract the weight of the container to obtain the net weight of the liquid for accurate calculations.
11. Can I convert between different units using the calculator?
The calculator typically does not perform unit conversions, so ensure that you input values in the same units.
12. Is the Weight Per Gallon Calculator suitable for solids?
No, this calculator is specifically designed for liquids; solids have different volume and weight relationships.
13. What is the significance of knowing the weight per gallon in transportation?
It helps determine shipping costs, packaging requirements, and safety measures for transporting liquids.
14. Can the calculator be used for industrial applications?
Yes, it is widely used in industries such as food processing, pharmaceuticals, and chemical manufacturing.
15. What happens if I input incorrect values?
Inputting incorrect values will result in inaccurate calculations, so double-check your inputs for accuracy.
16. How can I verify the accuracy of my weight per gallon calculation?
You can compare your results with known values for specific liquids to verify accuracy.
17. Is it necessary to use a specific temperature for calculations?
While it’s not required, using the same temperature for measurements can improve accuracy, especially for temperature-sensitive liquids.
18. Can the calculator assist in recipe development?
Yes, knowing the weight per gallon helps in scaling recipes and ensuring accurate ingredient measurements.
19. What is the average weight per gallon for water?
The average weight per gallon of water is approximately 8.34 lbs at room temperature.
20. How does weight per gallon influence material selection for containers?
Understanding the weight per gallon of a liquid helps in selecting appropriate containers that can withstand the weight and pressure.
The Weight Per Gallon Calculator is an essential tool for anyone involved in the handling, formulation, or transportation of liquids. By accurately calculating the weight per gallon, users can make
informed decisions in various applications, from industrial processes to culinary arts. Mastering the use of this calculator enhances understanding of liquid properties, enabling better planning and
execution in both personal and professional endeavors.
Leave a Comment | {"url":"https://savvycalculator.com/weight-per-gallon-calculator","timestamp":"2024-11-08T11:56:37Z","content_type":"text/html","content_length":"147806","record_id":"<urn:uuid:7aa392df-b508-4dc1-b30a-3e974a98075e>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00876.warc.gz"} |
Aggregate Functions
Aggregate functions perform a computation against a set of values to generate a single result. For example, you could use an aggregate function to compute the average (mean) order over a period of
time. Aggregations can be applied as standard functions or used as part of a transformation step to reshape the data.
Output order of nested values from an aggregation function cannot be determined in advance. The work to generate output values is done in parallel, which results in different ordering of any nested
values for each execution run for each running environment.
Aggregate across an entire column:
Transformation Name New formula
Parameter: Formula type Single row formula
Parameter: Formula average(Scores)
Output: Generates a new column containing the average of all values in the Scores column.
Transformation Name Pivot columns
Parameter: Values average(Score)
Parameter: Max number of columns to create 1
Output: Generates a single-column table with a single value, which contains the average of all values in the Scores column. The limit defines the maximum number of columns that can be generated.
When aggregate functions are applied as part of a pivot transformation, they typically involve multiple parameters as part of an operation to reshape the dataset. See below.
Aggregate across groups of values within a column:
Aggregate functions can be used with the pivot transformation to change the structure of your data. Example:
Transformation Name Pivot columns
Parameter: Row labels StudentId
Parameter: Values average(Score)
Parameter: Max number of columns to create 1
In the above instance, the resulting dataset contains two columns:
• studentId - one row for each distinct student ID value
• average_Scores - average score by each student (studentId)
You cannot use aggregate functions inside of conditionals that evaluate to true or false.
A pivot transformation can include multiple aggregate functions and group columns from the pre-aggregate dataset. See Pivot Transform.
Null values are ignored as inputs to these functions.
These aggregate functions are available: | {"url":"https://help.alteryx.com/aws/en/trifacta-application/wrangle-language/aggregate-functions.html","timestamp":"2024-11-15T01:24:01Z","content_type":"application/xhtml+xml","content_length":"25464","record_id":"<urn:uuid:0379c3d8-4aef-40f3-ba94-a8654720ae6d>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00036.warc.gz"} |
Market Assumptions - Approximate Geometric mean
We created this feature after discussions with some of our subscribers. Please see the resources below for an explanation of the difference between an arithmetic average, on the one hand (i.e. the
simple mean value we all know, whereby the inputs are summed and then divided by the number of inputs), and the geometric mean, on the other hand (which takes account of compounding in a way that the
simple, arithmetic 'mean' value does not).
The following sites do a good job of explaining the difference:
Geometric Mean
Geometric Mean: Key Examples
After discussions with some other advisers, we came to a couple of conclusions:
1) For our standard projection, market assumptions using geometric averages would give a better result, as arithmetic averages skew the projection to the positive.
2) For our Monte Carlo simulation, arithmetic-based averages are most appropriate
Given that we believe most of the data provided to us by our subscribers are likely to be arithmetic assumptions, we decided that providing an option to approximate the geometric mean, given
arithmetic market assumptions, would allow us to provide better projections in our standard 'cash flow' calculation, while still allowing our monte carlo simulation to work as before.
The formula we use to estimate the 'true' geometric mean figure from an arithmetic mean figure is pretty standard, but an explanation is provided in the following paper written by William Bernstein:
The formula for calculating the approximate geometric mean (G) is: (from page 8 of the paper):
V = stdDev squared
R = arithmetic mean
G ~ R - ______
R - _____________
2 + 2 *R
When you provide updates for your market assumptions, we generally recommend you enable this feature as (in the absence of reasons to think otherwise) your assumptions are most likely to be
arithmetic mean values. | {"url":"https://support.planwithvoyant.com/hc/en-us/articles/202062668-Market-Assumptions-Approximate-Geometric-mean","timestamp":"2024-11-08T12:15:25Z","content_type":"text/html","content_length":"22293","record_id":"<urn:uuid:85332097-0844-4a04-8076-e45ebf17dbe5>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00243.warc.gz"} |
Project Euler - Problem 18 Solution
Project Euler – Problem 18 Solution
Yan Cui
I help clients go faster for less using serverless technologies.
This article is brought to you by
Step Functions, EventBridge, MSK, DynamoDB…stop hacking together AWS services and get back to building!
By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.
That is, 3 + 7 + 4 + 9 = 23.
Find the maximum total from top to bottom of the triangle below:
NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot
be solved by brute force, and requires a clever method! ;o)
With the warning of problem 67 in mind, I set out to tackle this problem with a somewhat efficient solution which will also work for problem 67. To start off, let’s look at the simple example, as you
traverse down the triangle from the top you have two options at every turn – go left, or go right. From the third row onwards, some of the paths will end up in the same place in the triangle:
From the above image you can see that the path 3-7-4 and 3-4-4 both exit the mini triangle at the same location but carrying two different totals – green = 14, red = 11. If you were to continue down
the triangle from this point on, regardless of whether you’re going to turn left or right in the next move, the max total will surely have to come from the path which has generated the highest total
so far.
If we apply the same logic to the entire triangle we can build up a triangle of totals for the raw triangle where every element represents the max total achievable by any path that has lead to it:
Let’s call this triangle of totals T, and the original triangle R, we have:
You can see that T(2,1) differed from T(2,0) and T(2,2) in that it required us to find the greater between T(1,0) and T(1,1). For simplicity sake, let’s say that for a given row n, in T (where n >=
2) we have:
The max total from top to bottom of the triangle R is equal to the max total in the last row of the triangle T. Seeing as we’re only interested in the max total rather than the triangle T itself, my
solution below only returns the last row of T which is all we need to solve the problem:
Whenever you’re ready, here are 3 ways I can help you:
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5 thoughts on “Project Euler – Problem 18 Solution”
1. why does it seem so complicated.
2. here’s my solution, basically I do the same thing but I start from the bottom of the triangle
open System.IO;
let getMaxSum (triangle: int list list) =
. let getResLine currLine prevLine =
. let rec loop resLine prevLine’ = function
. |[] -> resLine
. |hd::tl -> loop ((hd + (max (List.nth prevLine’ 0) (List.nth prevLine’ 1)))::resLine) (List.tail prevLine’) tl
. loop [] prevLine currLine |> List.rev
. let rec loop (prev:int list) (tri: int list list) =
. match tri with
. | [] -> prev.Head
. | hd::tl -> loop (getResLine hd prev) tl
. loop triangle.Head triangle.Tail
let triangle n =
. File.ReadAllLines(@”C:\euler\”+n+”.txt”)
. |> Array.map (fun s -> s.Split(‘ ‘) |> Array.map int32 |> Array.toList)
. |> Array.toList
. |> List.rev
let p18 = getMaxSum (triangle “18”)
let p67 = getMaxSum (triangle “67”)
3. I tried to put dots, in order to get indentation, but it didn’t really work, changing the font-family to Courier New for the comments will fix that
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PPT - Hypothesis Testing PowerPoint Presentation, free download - ID:6792717
1. Hypothesis Testing LIR 832 Lecture #3 January 30, 2007
2. Topics of the Day • A. Our Fundamental Problem Again: Learning About Populations from Samples • B. Basic Hypothesis Testing: One Tailed Tests Using a Z Statistic • C. Probability and Critical
Cutoff Approaches: Really the Same Thing • D. How do we do hypothesis tests on small samples (n = 30 or less). • E. How do we do hypothesis testing when we have information on population standard
deviation? On sample standard deviations? • F. How do we test a statement such as (two tailed test)? • G. How do we test for differences in means of two populations?
3. Hypothesis Testing • Fundamental Problem: We want to know about a population which is not observed and want to use a sample to learn about the population. Our problem is that sampling variability
makes sample an inexact estimator of the population. We need to come up with a method to learn about the population from samples which allows for sampling variability.
4. Hypothesis Testing: Example • We are considering implementing a training program which purports to improve quality and reduce the number of defects. Currently, 10 out of every 1000 parts produced
are not within spec. The standard deviation of defects is 8 parts (variance is 64). The typical employee produces 1,000 parts per day • The program costs $1,000 per employee and we have 10,000
production employees. We are unwilling to spend $10,000,000 for a pig in the poke. • Instead we decide to run a pilot on 100 employees to determine whether the program is effective for our
employees. The firm that does the training will do the program for free, so our only cost is lost production for the time during which employees are trained. Note that the 100 employees are a
pilot or, in our terminology, a sample.
5. Hypothesis Testing: Example • Employees are sent to the program and then given several days under instruction to apply what they have learned to their work. We run a one day test on the employees
and find that they average 8 parts per thousand. • Defects are down, but is this really an improvement or is it simply the result of sampling variation? Could we be reasonably certain if we
trained a second group of employees, or ran this same test next week, that defects would also be down?
6. Hypothesis Testing: Example • Abstractly, we are faced with the problem of distinguishing whether the program is effective or whether the improvement is reasonably explained by sampling variation
(aka luck).
7. Hypothesis Testing: Example • Let’s approach this as a statistical problem. We know that historically there have been 10 defects per 1000 with standard deviation of 8. So our question is, “How
likely is it that we have pulled a sample of 100 employees with a mean defect rate of 8 if, in fact, the training program did not work (in other words, that the population rate of defects remains
10 per 1000)?
8. Hypothesis Testing: Example • You can set this problem up as you have been already: • As the sample is larger than 30, we can use our z-table. The P(z<2.5) = 0.62%, very small. There is so small
a likelihood that the sample we have observed was drawn from a population with a mean of 10 that we reject the possibility that the training program was ineffective.
9. Hypothesis Testing: Example • What do I mean when I say that the probability is 0.62%? Two possible interpretations: • 1. Suppose we set up a population with a mean of 10 and a standard deviation
of 8 and draw samples of 100 from that population. Now imagine repeating this experiment 1000 times. We would expect that slightly over 6 of those samples would have a value of 8 or less. • 2.
Alternatively, if the population had a mean of 10, standard deviation of 8, we would expect that 0.62% of the time we would draw a sample values 8 or less by chance. • I find the first approach
makes it easier to understand what I mean when I say there is a 0.62% probability that an event occurred by chance, but you may prefer the second.
10. Hypothesis Testing: Example • Thinking about this graphically: • Two extreme possibilities: • 1. Training did little or nothing • 2. The training was highly effective (really useful in Thomas the
Tank Engine terminology)
11. Hypothesis Testing: Example • If the training did nothing, we would expect the distribution of defects post training to look a lot like the distribution of defects prior to training.
12. Hypothesis Testing: Example So we can use the old distribution as a standard against which to judge our sample results. If the sample looks a lot like the old distribution it would be reasonable
to believe that the training did not work. 1. For convenience, we add cut points at ± 2 standard deviations (1.6 = 2* .8)
13. Hypothesis Testing: Example • When we pull our sample, we calculate a mean and compare it to the old distribution. So, for example, our sample returns a value of 8. This lies to the left of the
low cut point and we conclude it is very unlikely that the training did nothing.
14. Hypothesis Testing: Example • Let’s go behind the graphs to the underlying, if unobserved population.
18. Hypothesis Testing: Formalizing the Steps • Step 1: State our beliefs about the world clearly (hypotheses). • Example: The consulting firms contention is that their training program reduces
defect rates. Another possibility is that the training program is ineffective and that any changes in defect rates are simply the result of sampling variability (randomness)
19. Hypothesis Testing: Formalizing the Steps • Step 2: Formalize this into the alternative and null hypothesis: • Alternative: HA μpost training < 10: the the training program is effective • Null:
HO μpost training ≥10: the training program has no effect on defect rates.
20. Hypothesis Testing: Formalizing the Steps • More about Step 2: • The two hypotheses are about the unobserved population. We will use samples to test these two hypotheses • Together, the null and
alternative cover all possible outcomes. • The null is about change being the result of sampling variability, not the result of systematic difference. So our null for the training program is that
it is not effective. Our alternative is that the program has a positive effect. • We always test to determine whether the null is reasonable (is sufficiently likely that we are unwilling to
conclude it is false).
21. Hypothesis Testing: Formalizing the Steps • Step 3: Set up the probability problem: • The sign of the test is taken from the alternative hypothesis. We ask, how likely are we to observe this
extreme an outcome if, indeed, the null is true.
22. Hypothesis Testing: Formalizing the Steps • Step 4: Solve the Probability Problem: • Note that our standard deviation is the root of the population variance divided by sample size. This is the
second part of the Central Limit Theorem and provides the predictive power to use of sample means in hypothesis testing. • The result of doing the test is a probability of the null being true
given the sample outcome. If that probability is sufficiently small, we conclude the null is not reasonable.
23. Hypothesis Testing: Formalizing the Steps • Step 5: Decide whether the probability of the null being true, given the sample outcome, is sufficiently low to allow you to reject (not believe) the
null. • The 0.62% likelihood is the probability that a population with a mean of 10 and a standard deviation of 8 would produce a sample of 100 observations with a mean of 8. It is very unlikely
and we may well decide the null isn’t realistic.
26. Hypothesis Testing: Example • A second example: We are considering the absence behavior of a new group of workers and wish to determine if it is different from the behavior of our typical company
employee. • The company has hired a large group of human resource managers over the last three years. They are well versed in corporate policy and, as they do many of the unpleasant tasks in the
firm, tend to have high stress levels. The Vice President of Human Resources has remarked that she is concerned about their stress levels and believes that our new HR managers they are taking
more days off than most employees to compensation for their problems. • We collect data on the 225 HR managers in the firm. We find they average 24 days of sick leave annually. The average
managers in the firm has taken 22 days of sick leave annually over the last ten years. The standard deviation for the pool of typical managers is 30 days. Is our VP correct that our HR managers
take more sick leave than other managers?
27. Hypothesis Testing: Example • Step 1: State our views of the world. • One view is that HR managers are taking more days of sick leave than has been typical of other managers. • Another is that
they are, as a group, no different than other managers.
28. Hypothesis Testing: Example • Step 2: Formalize the hypotheses we are testing. • Let μd be the population mean days off for the population of HR managers: • H0: Null: HR managers are not
different: • (NOTE: the null is about the population mean, not the sample) • μ ≤22 • HA: Alternative: HR managers take more sick time than other managers • (Similarly, the alternative is about
the population) • μ > 22
29. Hypothesis Testing: Example • Step 3: Set up the probability problem. • Note that our sample is 225, so we have central limit theorem results and can use a z distribution.
30. Hypothesis Testing: Example • Step 4: Solve the probability problem.
31. Hypothesis Testing: Example • Step 5: Determine whether the likelihood is sufficiently small to reject the null. • Our calculations indicate that there is a 15.87 percent likelihood that a
population with a mean of 22 would produce a sample of 225 individuals with a mean of 24. So, we would expect a sample of the type we got in about 1 in 6 experiments when HR managers were no
different than other managers. That type of outcome just isn’t too unlikely.
32. Hypothesis Testing: Example • There are, again, two ways of interpreting this result: • 1. (Repeat experiment approach) If we ran 100 experiments in which we pulled random samples of 225 from a
population in which the mean of 22 and a standard deviation of 30, between 15 and 16 of those samples would have a mean of 24 or greater. • 2. (Probability approach) Alternatively, 15.87% of the
time when we pull a sample of 225 from a population with a mean of 22 and standard deviation of 30, the mean will be 24 or greater.
35. Hypothesis Testing • Now consider what we are doing abstractly… • A little background. Employers are often very interested in employees views of their work and firm. There are many ways of
measuring employee attitudes, a common method used in survey research are Likert scaled questions. For example, we might ask a question such as: • How would you rate this job? Would you rate it
as • 1 - very bad 2 - bad 3 - good 4 - very good
36. Hypothesis Testing • A Somewhat Cranky Aside: • The response to this question is numeric, we get a number between 1 and 4, but it is ordinal data. We know the ordering of the response, a response
of very good is stronger than a response of good; similarly a response of bad is weaker than a response of good.; but there is no arithmetic relationship between the responses. Two responses of
“very bad” are not equal to one response of “bad”. Indeed, we cannot even be sure that the responses are equally spaced. The person who responds “bad” may be mildly unhappy relative to someone
who responds good; but the person who responds “very bad” could be somewhat more unhappy than the “bad” or they could just about ready to “go postal”. Still the data is ordered. • Despite this
limitation, it is common practice in behavior sciences to add the responses to form a mean response.
37. Hypothesis Testing • Back to our problem: • 1. We begin with a “theory” or observation about the population. These are typically a general statements such as: • a. Employees on evening or night
shifts are less satisfied with their work than other workers. • 2. Implicit in this statement is its opposite, the null hypothesis. This is what we are going to test • b. Null: Employees on
evening or night shifts are no less satisfied with their work than employees on the day shift.
38. Hypothesis Testing • 3. We turn these statements into operational hypotheses, statements which can be tested statistically. • c. We have been surveying employees for many years using Likert
scaled surveys. We know from these surveys that day shift employees rate the firm as good (3) on a four point Likert scale(very bad, bad, good, very good). Standard deviation is also 3. Our null
and alternative would be: • H0: Null: μevening or night ≥3 • HA: Alternative: μevening or night <3 • Note: The null and alternative exhaust all possibilities, no other outcome is possible.
39. Hypothesis Testing • 4. We pull a sample and test to determine the probability that the sample came from the population with the characteristics of the null. • Suppose we pull a sample of 36
night employees with a sample mean of 2.3 • We ask: • What is the probability of pulling this particular sample if the mean response of our night shift employees really is 3.0 or greater? • P
(x-bar < 2.3 if μnight really ≥3) or P(x-bar < 2.3 | μnight ≥3)? • Note: the sign in the statement is taken from the alternative hypothesis.
40. Hypothesis Testing • 5. We then calculate that probability:
41. Hypothesis Testing • 6. What does this 8.08% mean: • 1. Suppose we had a population with a mean value of 3 and standard deviation of 3 (this is exactly the population we build our null hypotheses
around). Then suppose we ran an experiment in which we drew 36 individuals at random from that population and repeated the experiment 100 times. We would expect that 8.0 of those samples would
have a mean of 2.3 or less! • 2. If our population has a mean and standard deviation of 3.0, we would expect that 8.08% of the time we would draw a sample of 36 individuals with a mean of 2.3 or
42. Hypothesis Testing • Criteria for rejecting or not rejecting the null: • We reject the null if it is sufficiently unlikely that the sample would be produced by the “null” population by chance. •
We do not reject the null if the probability that the observed sample was pulled from the null population is reasonably large. • So our logic is that we only believe our alternative if the
probability of the reverse (the null) is very low. • We have not established how low one has to go (how low the probability has to be) before one rejects the null. We are getting to this.
43. Hypothesis Testing: Example • We are told by a consultant that their program will, by making workers more aware of the effects of their absence, reduce the use of sick leave and personal leave.
We have no reason to disbelieve her, but given the price of the program $2,500 per employee, we want to run a pilot on 50 employees before taking it to the entire population of employees.
Reviewing our records, we find that, on average, employees take 10 days of sick leave and personal leave a year. The standard deviation in leave time is 8.5 days per year.
44. Hypothesis Testing: Example • Set up the null and alternative and test this problem. • 1. Let’s define μ as the number of absences due to sick leave or personal days. • 2. What is the null and
alternative? • 3. You run the pilot and find that the mean number of sick and personal days taken was 8.7. Set up the hypothesis test. • 4. What is the z score from this test? What is the
probability? • 5. Do you reject the null? Explain.
45. Graphical Example • We use an examination for testing new employees. We need to calibrate our criteria for passing this test to the performance of our current employees with whom we are very
satisfied). • The typical standard for hiring a person who has taken this test is a score of person hired after taking this test scores 500. We believe that our employees are superior and that,
if we gave this test to our current employees, the mean of the test would be more than 500. The company which designs the test cannot help us directly as, although they know the typical
performance of a test taker, they do not know the typical performance of our employees. They however, agree to administer the test to a sample of 24 of our employees. We are interested in using
this sample to test the hypothesis that the mean score for our employees would be more than 500. The standard deviation of the sample, after allowing for sample size, is 4. Once we know this
result, we can decide whether we can default to the typical performance standard or need to more closely determine our employees performance.
46. Graphical Example • Suppose we want to test our belief about the population and are willing to be wrong by chance about 5% of the time. We know that 5% of the area of a normal is 1.645 standard
deviations above the means, so we set up an upper cut point at +1.645 standard deviations (which is 506.58 = 1.645*4+ 500). What does this look like graphically? • What does this test look like?
47. Graphical Example • Note that the limits for accepting the null are set around the hypothesized population
50. Establishing Cut Points • Establishing the Appropriate Cut-offs for rejecting the null hypothesis. • To this point, we have produced p-values from our hypothesis tests and then tried to judge
whether a null was sufficiently unlikely that we were willing to reject it. • A more conventional approach is it use 10%, 5% and 1% as the standards | {"url":"https://fr.slideserve.com/amal-conway/hypothesis-testing-powerpoint-ppt-presentation","timestamp":"2024-11-03T16:47:50Z","content_type":"text/html","content_length":"109943","record_id":"<urn:uuid:3e749716-c8d4-4f4d-bcf8-1cd799f8896c>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00591.warc.gz"} |
Computational Fluid Dynamics (CFD)
Revision as of 20:32, 31 August 2023 by
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Computational Fluid Dynamics (CFD) is a branch of fluid mechanics that uses numerical analysis and data structures to solve and analyze fluid flow and heat transfer problems. It is often used in
concert with experimental and theoretical fluid dynamics to understand complex fluid systems. Like Finite Element Analysis (FEA), CFD is a subset of Computer-Aided Engineering (CAE) and plays a vital
role in various industries such as automotive, aerospace, and energy production.
The development of CFD dates back to the early days of computer science. Initial works were based on the method of characteristics and panel methods. The significant evolution in computational
capabilities since the 1980s has made CFD a mainstream tool for engineers. Today, it is employed in a variety of applications, from designing airplane wings to simulating the weather.
Basic Principles
CFD starts by discretizing a fluid domain into small volumes or elements. Equations of fluid dynamics—primarily the Navier-Stokes equations—are then solved numerically for each of these volumes. The
result is a detailed velocity and pressure field that represents the behavior of the fluid within the domain. Boundary conditions, turbulence models, and other physical phenomena can be included to
improve accuracy.
Types of CFD
• Laminar and Turbulent Flow: CFD can simulate both laminar flow, where fluid flows in parallel layers, and turbulent flow, characterized by chaotic behavior.
• Compressible and Incompressible Flow: Depending on the application, CFD can be adapted to model flow where density changes significantly (compressible) or remains relatively constant
• Steady-State and Transient Analysis: CFD can also simulate either steady-state conditions, where the flow field does not change over time, or transient conditions, which consider the
time-dependent nature of fluid flow.
• Multiphase Flows: It can simulate the interaction of multiple phases of matter, like the simultaneous flow of oil, water, and gas in pipelines.
Software Tools
• ANSYS Fluent
• OpenFOAM
• CFX
• STAR-CCM+
• COMSOL Multiphysics
• Aerospace: Aerodynamic design, heat transfer studies, and combustion simulation in engines.
• Automotive: Vehicle design for reduced drag, engine airflow, and cabin ventilation.
• Energy: Simulation of wind turbines, hydrodynamic studies of marine turbines, and combustion in power plants.
• Environmental Science: Weather simulation, pollutant dispersion, and flood modeling.
• Biomedical: Blood flow modeling, respiratory airflow.
Advantages and Disadvantages
• Can model highly complex fluid flow regimes, often impossible through analytical methods.
• Reduces the need for expensive and time-consuming experimental tests.
• Provides detailed insights into flow behavior, allowing for more optimized designs.
• Highly computationally intensive, requiring powerful hardware.
• Requires significant expertise for setting up simulations and interpreting results.
• Accuracy is highly dependent on the quality of the mathematical models and boundary conditions used.
Future Trends
The future of CFD lies in its integration with machine learning algorithms for improved modeling, cloud computing for resource optimization, and real-time simulations for control systems. Hybrid
models combining CFD with other simulation techniques like FEA are also becoming increasingly prevalent.
See Also | {"url":"https://cio-wiki.org/index.php?title=Computational_Fluid_Dynamics_(CFD)&oldid=17452","timestamp":"2024-11-13T02:45:44Z","content_type":"text/html","content_length":"24586","record_id":"<urn:uuid:4b8b22ac-0252-48ce-9626-b030d2c3e0d4>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00678.warc.gz"} |
Welcome to
VirtualTrebuchet 2.0
VirtualTrebuchet is a web based trebuchet simulator that will allow you to quickly evaluate different trebuchet configurations.
1. To begin, enter the parameters of your trebuchet in the input boxes.
2. Next, press the "Simulate" button located under the inputs.
3. Finally, watch your trebuchet go.
Any questions? Read the documentation or contact us. | {"url":"https://virtualtrebuchet.com/documentation/inputs/trebuchet-dimensions/lengthsling/","timestamp":"2024-11-12T08:50:53Z","content_type":"text/html","content_length":"22217","record_id":"<urn:uuid:7c7fda74-de5c-493f-96ea-c9a6587a3c09>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00247.warc.gz"} |
Compound Interest Calculator - CalculatorBox
Compound Interest Calculator
Figures rounded to max decimal places.
Our compound interest calculator can calculate the value of an investment if given the appropriate values for the principal, interest rate in percentage, frequency of compounding per year, and the
compounding period you want to calculate the value of your investment for.
Using our interest calculator
In order to use our calculator, you simply input the values you have available, and the calculator will calculate the resulting amount for you! Here is a quick explanation of what each variable you
need to input means.
Input Explanation
Principal This is the sum you are investing and hoping to increase the value of. You may hear some synonyms used, such as sum invested, initial investment, or simply investment.
Interest rate The interest rate is the percentage value that your investment will be increased by during each compounding period. This value can sometimes also be expressed in an alternative to the
percentage, such as a fraction or a decimal number.
Compounding This is the frequency at which your investment will be compounded. You can choose a variety of ways to express the frequency, but the default is usually the number of times an
frequency investment is compounded per year. So, for example, if your investment is to be compounded every month, the frequency would be 12, because there are 12 months in a year.
Time This refers to the duration of the investment. Usually, it is expressed as the number of full compounding periods repeated. The default is usually the number of years you plan to keep
the investment with the financial institution.
What is compound interest?
The main difference between simple and compound interest is that simple interest calculates the sum of interest paid once at the beginning and then keeps adding the same value each interest period.
Compound interest calculates a new value of the interest for each period, hence compounding the value with previous increases. This means, that the percentage stays the same, but the value the
percentage is calculated out of keeps increasing each compounding period.
How is compound interest calculated?
The calculation of compounded interest is done using a formula which is a simplified version of continuously calculating a simple interest rate towards a new principle.
A simple example would work like this: “We invested 100$ at a 10% monthly compounded interest for 2 months”.
Hence this problem calls for first calculating 10% of 100, which is 100×1.1 = 110$. For the second month, 10% will be calculated again, but this time from the new value of 110$. Hence, there are
110$x1.1 = 121$ after 2 months.
Using the rules of exponents, we can write this as an expression of 110×1.1×1.1 = 110×1.1^2. Here we see that the interest rate is put to the power equal to the number of compounding periods.
The full formula for the compound interest rate is hence
The key to this formula is:
** (Number of times the interest is compounded during the investment period. Keep in mind, that in most cases this number is 1 and we simply use the investment period as the baseline)
t - time ~of ~the ~investment
How to calculate monthly compound interest
Referring to our example, the numbers would turn out as follows:
P = 100 as this is the invested sum.
r = 10% = 0.1 in decimal form, as this is the interest rate.
n = 1 as we have a monthly compounding period and we compound once a month.
t = 2 as we invested for 2 months
A=100(1+0.1/1)^{1x2} \\= 100*1.1^2 = 121
The benefits of compound interest
The most important benefit of compound investing is that the money is increasing in value at an exponential rate. Compared to simple interest, which increases at a linear rate, compound interest has
a perspective of making a lot more money on the interest with the same principal over time, as each compound period is working with an increased number.
The initial increases are only slightly higher than simple interest, but after a reasonable period of time, the interest starts increasing at a very high rate. The example below shows two graphs. The
red line is a simple monthly interest of 10%l. The blue line is a compound interest of the same principal and time period, but with a percentage interest rate of only 2%. Watch how over time the
compound interest surpasses the simple interest, despite having a 5 times lower rate. This is why compound interest is the best wealth-creation tool in the long run.
What will $10,000 be worth in 20 years?
Let us compare the difference between investing 10,000 $ at a 2% yearly interest rate with simple and compounded interest.
Simple interest
We simply calculate 2% of 10,000$ and add that sum to the 10,000$ exactly 20 times. Or we can apply the formula
We get
A= 10,000*(1+0.02*20) \\= 10,000*1.4 = 14,000\$.
Here we can see that the simple interest made us 4,000$ in profit over the course of 20 years.
Compound interest
We directly apply the formula and get
A=10,000*(1+0.02/1)^{1*20} \\= 10,000*1.0220 \\= 10,000*1.486 = 14,860\$
when rounded to the whole $. As we can see, with all conditions the same, the compound interest made 860$ more over the same period of time as the simple interest, out of the same principal and with
the same interest rate.
Compounding with additional deposits
When considering compounded interest as a wealth creation tool, it is important to realize that for most people it will be the case of regular deposits and contributions. In other words, most people
will not have a large sum of money available the day they decide to start investing and accumulating wealth. However, it makes only sense to start saving what we can at an earlier time, so that sum
is already producing some interest.
Compounding with additional deposits is simply a way to start building your wealth by regularly depositing money into a savings account or other investment. It is also wise, if possible, to deposit
further investments before compound periods, in order to maximize your returns.
Where to invest for compound interest
Source Explanation
Checking Account Compound interest is usually a standard in the financial sector for checking accounts.
Savings Accounts Most saving accounts come with compounded interest as a default.
Funds (ETFs) Various funds promise compound interest if investing for longer periods of time.
Retirement funds Retirement funds work on a compound interest basis in most countries, which only makes sense considering that they are the funds that need to accumulate a lot of wealth over a long
period of time, in order to ensure retirement.
Cryptocurrencies Cryptocurrencies and digital financial platforms offer a wide variety of investment tools that work on the basis of compound interest, with the most used being staking.
Mutual Funds Active investment into bonds and stocks can also compound profits if sales are done during peaks and purchases during slumps, even if for the same assets. Additionally, stocks that
payout dividends that can be re-invested also offer the opportunity for the investment itself to create more liquidity that can be used as additional deposits.
Before you go…
Do not forget to check out our articles on Compounding and how it affects annual growth rate of savings! | {"url":"https://calculatorbox.com/calculator/compound-calculator/","timestamp":"2024-11-10T16:09:50Z","content_type":"text/html","content_length":"187902","record_id":"<urn:uuid:7bd0d3ed-c205-427c-b330-f49117d0bad3>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00417.warc.gz"} |
Pseudorandom Number Generators from Elliptic Curves
We give a survey of several recently suggested constructions of generating sequences of pseudorandom points on elliptic curves. Such constructions are of interest for both classical and elliptic
curve cryptography and are also of intrinsic mathematical interest. We also explain some main underlying ideas behind the proofs, pose several open questions and outline several directions for
further research.
Original language English
Title of host publication Recent trends in cryptography
Editors Ignacio Luengo
Place of Publication USA
Publisher American Mathematical Society
Pages 121-141
Number of pages 21
ISBN (Print) 9780821839843
Publication status Published - 2009
Event Summer School on Recent Trends in Cryptography - Santander, Spain
Duration: 11 Jul 2005 → 15 Jul 2005
Publication series
Name Contemporary Mathematics
Publisher American Mathematical Society
Volume 477
ISSN (Print) 0271-4132
Conference Summer School on Recent Trends in Cryptography
Country/Territory Spain
City Santander
Period 11/07/05 → 15/07/05
Dive into the research topics of 'Pseudorandom Number Generators from Elliptic Curves'. Together they form a unique fingerprint. | {"url":"https://researchers.mq.edu.au/en/publications/pseudorandom-number-generators-from-elliptic-curves","timestamp":"2024-11-10T06:19:13Z","content_type":"text/html","content_length":"42965","record_id":"<urn:uuid:39f65b1c-5951-425c-aa83-2e3cf102c486>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00371.warc.gz"} |
Convert 165 CM to Inches | Visual Length Comparison
Visualize Centimeters
1 centimeter is equal to 0.39 inches, so if you want to change centimeters into inches, you just need to multiply the centimeter value by 0.39.
How to convert 1 centimeters to inches?
1 cm * 1 in = 0.39370078740157477 in
2.54 cm
Convert 1 cm to common lengths
How many inches are 1 centimeters? 0.3937 in
How many feet are 1 centimeters? 0.0328 ft
How many yards are 1 centimeters? 0.0109 yd
How many milimeters are 1 centimeters? 10.0000 mm
How many meters are 1 centimeters? 0.0100 m
How many kilometers are 1 centimeters? 0.00001 km
FAQs about CM and Inch
How many centimeters is an inch?
An inch is equivalent to 2.54 centimeters. This conversion factor is widely used for converting between the imperial system (inches) and the metric system (centimeters). So, if you need to convert
inches to centimeters or vice versa, simply multiply or divide by 2.54, respectively.
How long is 1 cm on your finger?
The average pinky finger is about 1 centimeter wide, but the length of a finger depends on the individual and gender.
How many centimeters are in a meter?
There are 100 centimeters in a meter. This is a standard conversion in the metric system where 1 meter equals 100 centimeters.
How many millimeters in a centimeter?
There are 10 millimeters in a centimeter. This is a basic unit of measurement in the metric system, where 1 centimeter equals 10 millimeters.
160 centimeters is approximately 5 feet 3 inches. 165 centimeters is approximately 5 feet 5 inches. 170 centimeters is approximately 5 feet 7 inches. 175 centimeters is approximately 5 feet 9 inches.
180 centimeters is approximately 5 feet 11 inches. 185 centimeters is approximately 6 feet 1 inch. 190 centimeters is approximately 6 feet 3 inches.
Common food items that can be used to show size in cm include: a pea (1 cm), a peanut (2 cm), a grape (3 cm), a walnut (4 cm), a lime (5 cm ), an egg (6 cm), a peach (7 cm), and a grapefruit (10 cm
User Rating
out of 53,590 ratings | {"url":"https://cm-to-inch.com/en/165_cm_to_inches","timestamp":"2024-11-10T12:21:14Z","content_type":"text/html","content_length":"58436","record_id":"<urn:uuid:c523ad9a-c75b-4369-b7c9-6c5bec0e09c6>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00393.warc.gz"} |
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• Ji, Kathy Q.; Zhang, Helen W. J.; Zhao, Alice X. H.,Ranks of overpartitions modulo 6 and 10,JOURNAL OF NUMBER THEORY,2018
• Wang, Feng-Yu,Estimates for invariant probability measures of degenerate SPDEs with singular and path-dependent drifts,PROBABILITY THEORY AND RELATED FIELDS,2018
• Shao, Jinghai,Invariant measures and Euler-Maruyama's approximations of state-dependent regime-switching diffusions,SIAM JOURNAL ON CONTROL AND OPTIMIZATION,2018
• Huang, Harry H. Y.; Wang, Larry X. W.,The corners of core partitions ,SIAM JOURNAL ON DISCRETE MATHEMATICS,2018
• Ma, Jie; Ning, Bo,Coloring graphs with two odd cycle lengths,SIAM JOURNAL ON DISCRETE MATHEMATICS,2018
• Chang, Huibin; Lou, Yifei; Duan, Yuping; Marchesini, Stefano,Total variation-based phase retrieval for Poisson noise removal,SIAM JOURNAL ON IMAGINS SCIENCES,2018
• Nie, Jiawang; Yang, Zi; Zhang, Xinzhen,A complete semidefinite algorithm for detecting copositive matrices and tensors,SIAM JOURNAL ON OPTIMIZATION,2018
• Wang, Feng-Yu,Distribution dependent SDEs for Landau type equations,STOCHASTIC PROCESSES AND THEIR APPLICATIONS,2018
• Chen, William Y. C.;Fu, Amy M.,Context-free grammars for permutations and increasing trees,ADVANCES IN APPLIED MATHEMATICS,2017
• Wu, Jiahong; Wu, Yifei,Global small solutions to the compressible 2D magnetohydrodynamic system without magnetic diffusion,ADVANCES IN MATHEMATICS,2017
• Wang, Feng-Yu,Integrability conditions for SDEs and semilinear SPDEs,ANNALS OF PROBABILITY,2017
• Jiang, Renjin; Kauranen, Aapo,Korn's inequality and John domains,CALCULUS OF VARIATIONS AND PARTIAL DIFFERENTIAL EQUATIONS,2017
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• Shen, Ruipeng,Scattering of solutions to the defocusing energy subcritical semi-linear wave equation in 3D,COMMUNICATIONS IN PARTIAL DIFFERENTIAL EQUATIONS,2017
• Fang, Teng; Fang, Xin Gui; Xia, Binzhou; Zhou, Sanming,Vertex-imprimitive symmetric graphs with exactly one edge between any two distinct blocks,JOURNAL OF COMBINATORIAL THEORY SERIES A,2017
• Chen, Zhen-Qing; Hu, Eryan; Xie, Longjie; Zhang, Xicheng,Heat kernels for non-symmetric diffusion operators with jumps,JOURNAL OF DIFFERENTIAL EQUATIONS,2017
• Lin, Tai-Chia; Wang, Xiaoming; Wang, Zhi-Qiang,Orbital stability and energy estimate of ground states of saturable nonlinear Schrödinger equations with intensity functions in R2,JOURNAL OF
DIFFERENTIAL EQUATIONS,2017
• Ning, Cui; Ohta, Masahito; Wu, Yifei,Instability of solitary wave solutions for derivative nonlinear Schrödinger equation in endpoint case,JOURNAL OF DIFFERENTIAL EQUATIONS,2017
• Wang, Feng-Yu,Hypercontractivity and applications for stochastic Hamiltonian systems,JOURNAL OF FUNCTIONAL ANALYSIS,2017
• Chen, Shaowei; Liu, Zhaoli; Wang, Zhi-Qiang,A variant of Clark's theorem and its applications for nonsmooth functionals without the Palais-Smale condition,SIAM JOURNAL ON MATHEMATICAL
• ,Global existence and finite time blow-up of solutions of a Gierer-Meinhardt system, J. Differential Equations,2017
• Aksoy, Sinan; Chung, Fan; Peng, Xing,Extreme values of the stationary distribution of random walks on directed graphs,ADVANCES IN APPLIED MATHEMATICS,2016
• Wu, Jie; Zhang, Liqun,The Landis-Oleinik conjecture in the exterior domain,ADVANCES IN MATHEMATICS,2016
• Peng, Shuangjie; Peng, Yan-fang; Wang, Zhi-Qiang,On elliptic systems with Sobolev critical growth,CALCULUS OF VARIATIONS AND PARTIAL DIFFERENTIAL EQUATIONS,2016
• Ding, Yanheng; Li, Jiongyue; Xu, Tian,Bifurcation on compact spin manifold,CALCULUS OF VARIATIONS AND PARTIAL DIFFERENTIAL EQUATIONS,2016
• Tian, Wenyi; Yuan, Xiaoming, Linearized primal-dual methods for linear inverse problems with total variation regularization and finite element discretization,INVERSE PROBLEMS,2016
• Byeon, Jaeyoung; Sato, Yohei; Wang, Zhi-Qiang,Pattern formation via mixed attractive and repulsive interactions for nonlinear Schrödinger systems,JOURNAL DE MATHEMATIQUES PURES ET APPLIQUEES,2016
• Liu, Jiaquan; Liu, Xiangqing; Wang, Zhi-Qiang,Sign-changing solutions for coupled nonlinear Schrödinger equations with critical growth,JOURNAL OF DIFFERENTIAL EQUATIONS,2016
• Shen, Ruipeng,A semi-linear energy critical wave equation with an application,JOURNAL OF DIFFERENTIAL EQUATIONS,2016
• Hou, Qing-Hu; Jin, Hai-Tao; Mu, Yan-Ping; Zhang, Li,Congruences on the number of restricted m-ary partitions,JOURNAL OF NUMBER THEORY,2016
• Chen, William Y. C.; Wang, Larry X. W.; Xie, Gary Y. B.,Finite differences of the logarithm of the partition function,MATHEMATICS OF COMPUTATION,2016
• Chen, William Y. C.; Guo, Alan J. X.; Guo, Peter L.; Huang, Harry H. Y.; Liu, Thomas Y. H. ,s-inversion sequences and P-partitions of type B,SIAM JOURNAL ON DISCRETE MATHEMATICS,2016
• Chung, Fan; Peng, Xing,Decomposition of random graphs into complete bipartite graphs,SIAM JOURNAL ON DISCRETE MATHEMATICS,2016
• Guo, Bao-Zhu; Yu, Huaiqiang,Optimal state estimation for non-time invertible evolutionary systems,SIAM JOURNAL ON CONTROL AND OPTIMIZATION,2016
• J Nie, X Zhang,Positive maps and separable matrices, SIAM Journal on Optimization,2016
• Chen Yan-Hong,Liu Chungen and Zheng Youquan,Existence results for the fractional Nirenberg problem,Journal of Functional Analysis,2016
• Mengwei Xu, Jane J. Ye and Liwei Zhang,Smoothing SQP methods for solving degenerate nonsmooth constrained optimization problems with applications to bilevel programs, SIAM Journal on | {"url":"https://math.tju.edu.cn/kxyj/fblw.htm","timestamp":"2024-11-14T20:05:08Z","content_type":"text/html","content_length":"45517","record_id":"<urn:uuid:ca62f1f6-a3e9-448a-aaf4-1ac1d7e7fc8e>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00048.warc.gz"} |
What is the space complexity of a Segment Tree for an array of size n - ITEagers
Data Structure - Question Details
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[Solved] Consider a post office with two employees | SolutionInn
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Consider a post office with two employees and a single waiting line. On average, the inter-arrival time is 6 minutes. Standard deviation of inter-arrival time
Consider a post office with two employees and a single waiting line. On average, the inter-arrival time is 6 minutes. Standard deviation of inter-arrival time is 3.94 minutes. Average service time is
5 minutes. Standard deviation of service time is 2.83 minutes.
What is the average number of customers in the system?
What is the average number of customers waiting?
What is the average time a customer waits for service?
What is the average time a customer spends in the system?
What is the system utilization
There are 3 Steps involved in it
Step: 1
Number of servers m 2 6 minutes 4 minutes Interarrival time t...
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Force Conversion Calculator
Force Conversion - History and Science Behind Common Force Units
Force is a central concept in physics, as it is what causes an object to accelerate. There are various units used to measure force, with the most common being the Newton (N), Kilogram-force (kgf),
and Pound-force (lbf). These units have rich historical backgrounds, rooted in both scientific discoveries and practical applications. Let’s explore their history, the scientists associated with
them, and how they can be converted to one another.
Newton (N)
The Newton (N) is the SI unit of force and is named after Sir Isaac Newton, one of the most influential scientists in history. In 1687, Newton formulated the three laws of motion, which laid the
foundation for classical mechanics. His second law of motion, $F = ma$, defines force as the product of an object’s mass and acceleration.
• Definition:
$1\text{\hspace{0.17em}}\text{N}=1\text{\hspace{0.17em}}\text{kg}\cdot {\text{m/s}}^{2}$
A force of 1 Newton is the force needed to accelerate a 1-kilogram mass by 1 meter per second squared.
Newton’s groundbreaking work on gravity and motion revolutionized our understanding of the physical world, and the Newton as a unit of force is a tribute to his immense contribution.
Kilogram-force (kgf)
The kilogram-force (kgf) is a unit of force based on the gravitational force exerted by the Earth on a mass of 1 kilogram. It was widely used in various engineering fields before the adoption of the
SI system. One kilogram-force is equal to the force exerted by a mass of 1 kilogram under standard gravity ($9.80665 \, \text{m/s}^2$).
• Definition:
The use of kilogram-force has diminished over time as the SI system (which uses Newtons) became more standardized. However, it is still commonly used in some regions, particularly in the mechanical
and civil engineering fields.
Pound-force (lbf)
The pound-force (lbf) is a unit of force in the imperial system, commonly used in the United States. It is defined as the force exerted by gravity on a pound of mass. Just as kilogram-force relates
to the gravitational pull on 1 kg, pound-force corresponds to the force of gravity on a mass of 1 pound.
• Definition:
The unit arose from the widespread use of pounds in both commerce and engineering in the English-speaking world. Despite the growing prevalence of SI units globally, lbf remains widely used in
engineering and industry, especially in the US aerospace and automotive industries.
Force Unit Conversion Equations
To convert between Newtons (N), Kilogram-force (kgf), and Pound-force (lbf), use the following equations:
1. Newton to Kilogram-force:
$\text{kgf} = \frac{\text{N}}{9.80665}$
2. Newton to Pound-force:
$\text{lbf} = \frac{\text{N}}{4.44822}$
3. Kilogram-force to Newton:
$\text{N} = \text{kgf} \times 9.80665$
4. Pound-force to Newton:
$\text{N} = \text{lbf} \times 4.44822$
5. Kilogram-force to Pound-force:
$\text{lbf} = \frac{\text{kgf} \times 9.80665}{4.44822}$
6. Pound-force to Kilogram-force:
$\text{kgf} = \frac{\text{lbf} \times 4.44822}{9.80665}$
Converting 10 Newtons to Kilogram-force and Pound-force
Let’s convert 10 Newtons to Kilogram-force and Pound-force:
• Convert 10 N to Kilogram-force:
$\text{kgf} = \frac{10}{9.80665} \approx 1.0197 \, \text{kgf}$
• Convert 10 N to Pound-force:
$\text{lbf} = \frac{10}{4.44822} \approx 2.2481 \, \text{lbf}$
Thus, 10 Newtons is approximately 1.02 kgf and 2.25 lbf.
Fig. Screen Shot from CHEMIX School - Force Conversion Calculator
How to Use the Force Conversion Calculator
The calculator you have created consists of 7 editable text fields, each corresponding to a specific force unit. Here's a guide on how to use it correctly:
1. Input a Value:
Insert a numerical value in one of the text fields. Leave one field blank where you want the result to appear.
2. Ensure Valid Input:
Check that the input values are valid (positive numerical values). Invalid values (e.g., letters or negative numbers) may result in errors.
3. Press Enter to Calculate:
After entering a valid value, click inside the empty field where the result should appear and press Enter. The calculator will then compute the conversion and display the result.
4. Auto-Clearing Feature:
When you click on any of the text fields, the existing content will be cleared. You can then input a new value and repeat the process to calculate the corresponding forces in the other units.
By understanding the science and history behind these force units, you gain a deeper appreciation for how we measure the forces around us. Whether you're calculating the force exerted by an engine,
determining load capacities in engineering projects, or even understanding the forces at play in everyday life, these units and conversions play a crucial role in providing accurate and useful
Related topics:
Pressure Conversion Calculator
Energy Conversion Calculator
Length Conversion Calculator
Mass Conversion Calculator
Power Conversion Calculator
Temperature Conversion Calculator
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Adding To Ten Worksheets
Adding To Ten Worksheets - Ad includes lesson plans, printables, quiz games, practice problems & more. Memorizing all of the complements of 10 is a very helpful mental math skill in our base 10
number system. Get deals and low prices on addition worksheets on amazon Worksheet #5 worksheet #6 3 more similar: Web adding to 10 worksheets | free download: Web addition facts with 10's. Web
adding multiples of 10 and comparing. Web these math worksheets should be practiced regularly and are free to download in pdf formats. Web build addition skills with sums up to 10. Worksheet #3
worksheet #4 64 questions:
Adding whole tens Addition Year 1 (aged 56) by
Web these math worksheets should be practiced regularly and are free to download in pdf formats. Worksheet #5 worksheet #6 3 more similar: This bundled set of worksheets is a fun and engaging way for
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Adding Tens
Students fill in the missing addends so the sum is 10. Web with this worksheet, students can practice making 10 by cutting and gluing addends into the appropriate blank space in the math problem.
Memorizing all of the complements of 10 is a very helpful mental math skill in our base 10 number system. The first parameter will be determined.
Make a Ten Strategy for Addition.pdf First grade math, Kindergarten math, Teaching math
Web breadcrumbs worksheets math drills addition adding ten to numbers under 100 adding ten to numbers under 100 add ten more worksheets students add ten to random. Web fun resources to help students
practice the make a ten strategy of addition!sheets included:4 sheets: Web with this worksheet, students can practice making 10 by cutting and gluing addends into the appropriate.
Free Making 10 Worksheet Math addition worksheets, Math for kids, Kindergarten reading activities
Web build addition skills with sums up to 10. Solve for the missing addend or sum (practice numbers that make. Addition interactive worksheet number sentence: Now it's time to put those skills to the
test! Worksheet #3 worksheet #4 64 questions:
Adding to Ten with Fluency Worksheet Have Fun Teaching
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add. Addition interactive worksheet number sentence: Web breadcrumbs worksheets math drills addition adding ten to numbers under 100 adding ten to numbers under.
️Addition Making 10 Worksheets Free Download Gambr.co
Web addition facts with 10's. Ad includes lesson plans, printables, quiz games, practice problems & more. Web build addition skills with sums up to 10. The first parameter will be determined by which
numbers you want to add. Web with this worksheet, students can practice making 10 by cutting and gluing addends into the appropriate blank space in the math.
Adding Ten Worksheet Have Fun Teaching
Ad includes lesson plans, printables, quiz games, practice problems & more. Worksheet #5 worksheet #6 3 more similar: Memorizing all of the complements of 10 is a very helpful mental math skill in
our base 10 number system. Students fill in the missing addends so the sum is 10. Web adding multiples of 10 and comparing.
Making Tens Worksheet 2nd Grade Unique Miss Giraffe S Class Making A 10 to Add in 2020 Math
Web addition facts with 10's. Web fun resources to help students practice the make a ten strategy of addition!sheets included:4 sheets: Web 1 interactive story download all ( 5) grade your students
have been plugging away at their math skills all year. Ad includes lesson plans, printables, quiz games, practice problems & more. Web breadcrumbs worksheets math drills addition adding.
Third Grade Addition Worksheets
Web adding numbers ending with zero or adding number which are multiples of 10 (ten). Ad includes lesson plans, printables, quiz games, practice problems & more. Students add ten to numbers between 1
and 12. Web build addition skills with sums up to 10. Web adding multiples of 10 and comparing.
Addition To 10 Worksheets Free
Web 1 interactive story download all ( 5) grade your students have been plugging away at their math skills all year. This worksheet will help kids to practice counting ability. Web fun resources to
help students practice the make a ten strategy of addition!sheets included:4 sheets: Ad discover a wide selection of entertaining and informational books for children. The first.
Web adding to make 10. Students fill in the missing addends so the sum is 10. Ad includes lesson plans, printables, quiz games, practice problems & more. Compare addition problems with multiples of
10. This bundled set of worksheets is a fun and engaging way for children to practice strategies and build fluency with addition and subtraction facts within 10. Web addition facts with 10's.
Addition interactive worksheet number sentence: 34 + 30 < 43 + 20. Get deals and low prices on addition worksheets on amazon Web 1 interactive story download all ( 5) grade your students have been
plugging away at their math skills all year. Web adding multiples of 10 and comparing. Worksheet #1 worksheet #2 24 questions: Learn to add multiples of ten with these free math worksheets. The first
parameter will be determined by which numbers you want to add. Now it's time to put those skills to the test! Web build addition skills with sums up to 10. Ad discover a wide selection of
entertaining and informational books for children. Solve for the missing addend or sum (practice numbers that make. Worksheet #3 worksheet #4 64 questions: Grade = preschool 1st grade kindergarten
subject = counting worksheet
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Preprint 2019-09
Raimund Bürger, Stefan Diehl, María Carmen Martí:
A system of conservation laws with discontinuous flux modelling flotation with sedimentation
The continuous unit operation of flotation is extensively used in mineral processing, wastewater treatment, and other applications for selectively separating hydrophobic particles (or droplets) from
hydrophilic ones, where both are suspended in a viscous fluid. Within a flotation column, the hydrophobic particles are attached to gas bubbles that are injected and float as aggregates forming a
foam or froth at the top that is skimmed. The hydrophilic particles sediment and are discharged at the bottom. The hydrodynamics of a flotation column is described in simplified form by studying
three phases, namely the fluid, the aggregates, and solid particles, in one space dimension. The relative movements between the phases are given by constitutive drift flux functions. The resulting
model is a system of two scalar conservation laws with a multiply discontinuous flux for the aggregates and solids volume fractions as functions of height and time. The model is of triangular nature
since one equation can be solved independently of the other. Based on the theory of conservation laws with discontinuous flux, steady-state solutions that satisfy all jump and entropy conditions are
constructed. For the existence of the industrially relevant steady states, conditions on feed flows and concentrations are established and mapped as “operating charts”. A numerical method that
exploits the triangular structure, is formulated on a pair of staggered grids, and is employed for the simulation of the fill-up and transitions between steady states of the flotation column.
This preprint gave rise to the following definitive publication(s):
Raimund BüRGER, Stefan DIEHL, María Carmen MARTí: A system of conservation laws with discontinuous flux modelling flotation with sedimentation. IMA Journal of Applied Mathematics, vol. 84, 5, pp.
930-973, (2019). | {"url":"https://ci2ma.udec.cl/publicaciones/prepublicaciones/prepublicacionen.php?id=359","timestamp":"2024-11-03T02:53:25Z","content_type":"text/html","content_length":"9664","record_id":"<urn:uuid:44cf4b18-a8f6-4b0e-b516-b7dc65c398bb>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00339.warc.gz"} |
Create a chi squared distribution in Excel
In my last blog post I discussed how to create a gamma distribution in Excel, and that post can be read here:- https://tracyrenee61.medium.com/create-a-gamma-distribution-in-excel-3e05a2408544
In this post I intend to discuss how to create a chi squared distribution in Excel.
The chi squared distribution is one of the most important continuous probability distributions with many uses in statistical theory and inference. Chi squared is not only a distribution,but a
statistical test as well. In 1900, English mathematician Karl Pearson introduced the distribution in a seminal paper. Pearson showed that the chi squared distribution arose from such a multivariate
normal approximation to the multinomial distribution, taking careful account of the statistical statistical dependence (negative correlations) between the numbers of observations in different
The chi squared distribution is a continuous probability distribution that is used in many hypothesis tests including the chi squared goodness of fit test and the chi squared test of independence.
The main purpose of the chi squared distribution is to perform hypothesis testing and not for real world situations.
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Fluid Structure Interaction of Buoyant Bodies with Free Surface Flows: Computational Modelling and Experimental Validation
Department of Economy, Engineering, Society, and Business Organisation, University of Tuscia, 01100 Viterbo, Italy
Department of Enterprise Engineering “Mario Lucertini”, University of Rome “Tor Vergata”, 001133 Roma, Italy
Johna A. Paulson School of Engineering and Applied Sciences, Harvard University, 33 Oxford St., Cambrdige, MA 02138, USA
UNESCO Chair in Water Resources Management and Culture, University for Foreigners of Perugia, 06125 Perugia, Italy
Department of Engineering, University of Naples “Parthenope”, 80143 Napoli, Italy
Author to whom correspondence should be addressed.
Submission received: 12 April 2019 / Revised: 10 May 2019 / Accepted: 14 May 2019 / Published: 20 May 2019
In this paper we present a computational model for the fluid structure interaction of a buoyant rigid body immersed in a free surface flow. The presence of a free surface and its interaction with
buoyant bodies make the problem very challenging. In fact, with light (compared to the fluid) or very flexible structures, fluid forces generate large displacements or accelerations of the solid and
this enhances the artificial added mass effect. Such a problem is relevant in particular in naval and ocean engineering and for wave energy harvesting, where a correct prediction of the hydrodynamic
loading exerted by the fluid on buoyant structures is crucial. To this aim, we develop and validate a tightly coupled algorithm that is able to deal with large structural displacement and impulsive
acceleration typical, for instance, of water entry problems. The free surface flow is modeled through the volume of fluid model, the finite volume method is utilized is to discretize the flow and
solid motion is described by the Newton-Euler equations. Fluid structure interaction is modeled through a Dirichlet-Newmann partitioned approach and tight coupling is achieved by utilizing a
fixed-point iterative procedure. As most experimental data available in literature are limited to the first instants after the water impact, for larger hydrodynamic forces, we specifically designed a
set of dedicated experiments on the water impact of a buoyant cylinder, to validate the proposed methodology in a more general framework. Finally, to demonstrate that the proposed numerical model
could be used for a wide range of engineering problems related to FSI in multiphase flows, we tested the proposed numerical model for the simulation of a floating body.
1. Introduction
Fluid structure interaction (FSI) studies the physical phenomena related to the motion of a solid body, possibly compliant, within an encompassing fluid [
]. The development of reliable models for FSI is relevant for several engineering fields [
], including naval or marine engineering [
], bio-engineering [
], fluid machinery [
], energy harvesting [
], environmental engineering [
] and micro-mechanical systems and sensing devices [
] and a large scientific literature deals with the numerical evaluation of the fluid forces caused by the prescribed motion of a solid structure [
Recently, due to the increased availability of computational power, significant research efforts head towards the development of numerical methods that simultaneously model the fluid flow and the
solid motion and deformation [
]. Such methodologies require the modeling of two non-overlapping domains, the fluid and the solid, separated by a common interface. Each field of application has different modeling requirements. For
instance, aero-elasticity couples a stiff structure with a light and possibly compressible fluid [
], while in bio-engineering a very flexible structure and an incompressible fluid with comparable densities are coupled [
]. In naval or marine engineering structures interact with a free surface flows and are subject to a slamming pressure impulsive load [
When one of the two domains is predominant, as for example when a very stiff structure immersed in a light fluid, weakly or loosely coupled partitioned approaches are usually employed [
]. In this case, the two domains are solved sequentially through separate models that exchange informations through the fluid-structure interface. The dominant one (e.g., the solid) is modeled first,
determining the boundary conditions for the other one whose reaction is lagged in time. In the conventional serial staggered approach (CSS) [
], at each time-step, the solid motion is predicted using the fluid flow from the previous time-step, then such a motion is utilized to determine the fluid flow and, finally, the solid motion is
corrected utilizing the actual fluid flow [
]. Such a strategy is called Dirichlet-Neumann partitioning [
]. The flow field is the Dirichlet partition subject to the prescribed interface displacement and the solid is the Neumann partition subject to the interface forces [
However, loosely coupled partitioned algorithms do not guarantee convergence [
]. In particular, the artificial added mass effect is a well known issue for these approaches [
], especially at certain fluid to structure density ratios and certain geometries. This effect becomes relevant when the fluid forces generate large displacements or accelerations, in the solid, that
is for light (compared to the fluid) or very flexible structures [
]. In such cases, strong or tight coupling, is needed [
] and can be achieved through both partitioned and monolithic algorithms.
In the realm of Dirichlet-Neumann partitioned methodologies, tight coupling is achieved through iterative algorithms that solve each domain multiple times until convergence within the same time-step.
Different approaches such as fixed-point (also known as Block-Gauss-Seidel) [
] and interface Newton-Krylov [
] iterations are possible. Fixed-point iteration is the most straightforward, yet very effective, coupling strategy [
]. It allows to reuse existing solvers for the solid and fluid domains, reducing the modeling effort. Dynamic under-relaxation is often employed to improve the convergence speed of fixed-point
iterations [
Monolithic algorithms obtain the strong coupling by solving simultaneously the non-linear system of equations that represents the fluid flow and solid displacement [
]. These methodologies are reliable and robust but partitioned approaches are far more popular for FSI problems [
]. In fact, monolithic solvers are more computationally intensive compared to partitioned ones and lack modularity hindering the code reuse that, on the contrary, is a characteristic of partitioned
The fluid-structure interaction modelling becomes even more challenging in the presence of free surface flows, which is a common occurrence in several relevant engineering problems, in ocean
engineering, aerospace and energy harvesting applications [
]. In particular, water entry problems [
], which occur when a solid body impacts the water, represent a challenging issue involving fluid-structure interaction in the presence of a free surface. A correct prediction of the impulsive loads
originating from such an impact, as well as of the responses of the fluid and the structure, are crucial in the design of marine vessels and aerospace structures, such that several analytical,
experimental and numerical studies have been published so far. In energy harvesting, when the fluid being treated as a source of exploitable energy rather than a mere source of mechanical damping, a
proper simulation of fluid-structure interaction is crucial and in some cases involves solid bodies interacting with a free surface (i.e., wave energy harvesting). In such a case, the interaction
between the solid and the water may be responsible for large impulsive loadings, with the humbling presence of the free-surface, which represents a jump of density of approximately three orders of
magnitude and whose dynamics is very complex to be predicted. Therefore, the development of reliable numerical methods and experimental techniques to characterize such impulsive loading is crucial
for the design of several kind of engineering devices, such as marine vessels and hydraulic structures. Several works are available in literature dealing with the simulation of fluid-structure
interaction in the presence of a free surface [
]. Different numerical models, alternative to Navier-Stokes, may be also employed to simulate the fluid flow with a free surface, such as Smoothed Particles Hydrodynamics (SPH) [
], Lattice Boltzmann Method (LBM) [
] and constrained interpolation profile (CIP) method [
]. However, only few reliable numerical methods with strong coupling modelling techniques have been proposed in the scientific literature so far and only part of them refer to buoyant rigid bodies
(e.g., References [
]), a problem relevant to the design of marine vessels [
] and aerospace structures [
] that are often designed to resist to large displacements and impulsive accelerations.
In this paper, we propose a computational model for the FSI of buoyant rigid bodies in the presence of free surface flows and we validate it with literature and custom designed experimental data for
different fluid structure interaction problems. We adopt tightly coupled algorithm through a Dirichlet-Neumann partitioned approach with fixed-point iterations. The incompressible free surface flow
is modelled through the volume of fluid approach (VOF), the partial differential flow equations are discretised thorugh the finite volume method and solid displacement is described by the
Newton-Euler equations. The proposed algorithm is implemented within the foam-extend-3.2 open source framework [
], allowing a substantial code reutilization. Specifically, the flow solution is based on the interDyMFoam solver that utilizes the Arbitrary Lagrangian Eulerian method (ALE) and the PIMPLE algorithm
] to solve the multiphase flow equations in a deformable domain, while we implement the solution algorithm for the Newton-Euler equations and the coupling procedure.
The rest of the paper is organized as follows. In
Section 2
we describe the proposed methodology. In particular, in
Section 2.1
Section 2.2
we discuss on the fluid and the structure modeling, respectively, while in
Section 2.3
we describe the proposed coupling algorithm. The methodology is experimentally validated in
Section 3
, by comparing numerical results to literature ones and ad-hoc experiments. Conclusions are drawn in
Section 4
2. Methodology
The problem in study is schematically represented in
Figure 1
. Solving an FSI problem requires the physical modeling of two non-overlapping domains, the fluid,
and the solid,
, ones, separated by an interface
Herein we consider two immiscible fluid phases, namely phase 1 and phase 2 and we assume that the flow is incompressible. The solid domain is rigid and free to move subject to fluid, gravity or other
external forces.
We consider an inertial reference frame
$O x y z$
and a body fixed frame
$G x ¯ y ¯ z ¯$
, which defines the orientation of
with respect to
$O x y z$
Figure 1
is the center of mass of
and the superimposed bar (
$· ¯$
) identifies the vectors expressed in body fixed coordinates.
We solve the FSI problem through a partitioned Dirichlet-Neumann approach with fixed-point iteration [
]. Thereafter separate models are utilized for
Section 2.1
) and
, (see
Section 2.2
2.1. Fluid Flow
The incompressible flow of two immiscible fluids is described by the following continuity and momentum conservations:
$D ( ρ u ) D t = − ∇ p + ∇ · μ ∇ u + ρ g ,$
is the fluid velocity,
is the pressure,
is the dynamic viscosity,
is the density,
is the gravitational acceleration and
$D ( · ) / D t$
denotes the material derivative. Despite the flow is incompressible,
vary in space and time according to the fluid phase.
The force exerted by the fluid on the solid body is,
$F = ∫ Σ − p I + μ ∇ u − ∇ u T n Σ ,$
is the identity matrix,
is the body surface and
is the unit normal to the solid body. The moment of
with respect to
$M = ∫ Σ d × − p I + μ ∇ u − ∇ u T n Σ ,$
$d = r P − r G$
the distance of each surface element from the center of gravity of the body.
The flow is numerically approximated utilizing the interDyMFoam algorithm included within the foam-extend-3.2 software package [
]. It implements the volume of fluid (VOF) [
] and models the incompressible flow of two immiscible fluids through a single set of the Navier-Stokes and continuity equations (Equation (1)). The interface between the two phases is tracked by
solving Equation (
) [
$D α D t + ∇ · u r α 1 − α = 0 ,$
is the water volume fraction and
$u r$
is the relative velocity of the two phases. The second term of Equation (
) is introduced to contrast the numerical diffusion of the fluid interface [
Equations (1) and (
) are discretized through a generalized unstructured finite volume approach with a collocated variable arrangement [
]. The pressure velocity coupling and the non linearity of Equation (1) are solved using the PIMPLE algorithm [
]. The discrete version of Equation (
) is solved through the Multidimensional Universal Limiter with Explicit Solution (MULES) procedure [
Turbulent flows can be modeled with different levels of approximations including Reynolds averaged Navier Stokes (RANS), large eddy simulation (LES) and direct numerical simulation (DNS) [
]. The most common turbulence models for RANS and LES are already included in the foam-extend-3.2 environment.
As the motion of the solid body distorts the fluid domain, the computational nodes are re-allocated after each time-step to avoid depletion of the mesh quality. Several procedures for the mesh
deformation are implemented within the foam-extend library [
]. The numerical solution of Equations (1) and (
) accounts for the mesh deformation through the arbitrary Lagrangian Eulerian (ALE) procedure [
2.2. Rigid Body Dynamics in Space
The position (
$r p$
) of the generic point
$O x y z$
can be expressed as
$r G$
is the position vector of
is the rotation matrix that defines the orientation of
with respect to
$O x y$
. The motion in space of an unconstrained body follows the Newton-Euler equations [
$m r ¨ G = F + m g + F ext$
$J ¯ ω ˙ ¯ = M ¯ + M ¯ ext − ω ¯ × J ¯ ω ¯ ,$
the mass,
$F ext$
the external force,
$J ¯$
the inertia matrix,
$M ¯ ext$
the external moment, and
$ω ¯$
the angular velocity of
. The system of Equation (6) can be regarded as a system of differential equations in
$ω ¯$
. Since
$ω ¯$
cannot be directly integrated, we introduce the four component vector
$p = e 0 , e 1 , e 2 , e 3$
containing the Euler parameters to obtain the orientation of the body as follows [
$A = 2 e 0 2 − 1 I + 2 e e T + e 0 e ˜ ,$
$e T = [ e 1 , e 2 , e 3 ]$
$e ˜$
is the skew matrix associated to
. The following equation provides the relation between
$ω ¯$
$G = [ − e , e ˜ + e 0 I ]$
The system of ordinary differential equations from (
) to (
) is numerically integrated through the Euler explicit method.
Equations from (
) to (
) describe the 6 degrees of freedom (6-DOF) motion of an unconstrained body. Reduced order models (i.e., 1-DOF, 2-DOF or 3-DOF) are obtained by assuming zero the proper components of
$r ¨ G$
$ω ˙ ¯$
. For instance, 3-DOF motion in the
$x y$
-plane is obtained by setting
$ω ˙ x = ω ˙ y = 0$
$r ¨ z = 0$
. Similarly, 2-DOF motion in the
$x y$
-plane is obtained by setting
$ω ˙ = 0$
$r ¨ z = 0$
. Finally, 1-DOF, for instance along the
-direction, is achieved by setting
$ω ˙ = 0$
$r ¨ z = r ¨ x = 0$
2.3. Coupling Algorithm
The FSI problem is solved through a partitioned (or segregated) approach, which means that the mathematical models for the fluid flow (i.e., Equations (1) and (
)) and for the solid motion (i.e., Equations (
)) are integrated separately [
]. To accomplish the interaction between the fluid and the structure,
to the solid and
communicates the displacement to
. A strong or tight, coupling is required to ensure convergence when the density of the solid is comparable or lower, than the fluid density [
]. Being the algorithm segregated, such a coupling is accomplished through the iterative procedure reported in
Figure 2
At the beginning of each time step, $τ$, the time step duration $t ( τ )$ is calculated according to the Courant condition, assuming that the fluid flow determines the most stringent stability
constraint. Then, the interface fields are initialized. In particular, the residuals of the FSI sub-cycle and the interface displacement are set to zero (i.e., $R 0 = 0$, $Δ 0 = 0$) and the
relaxation factor $ϕ 1$ is initialized $0.25$, with the superscripts referring to the FSI subcycle.
After initialization, the FSI sub-cycle, identified in the followings by the superscript
, begins with the calculation of the interface forces. Fluid force and moment are calculated through Equations (
) and (
) and then utilized to solve the system of Equations (
). The displacement of the points on the fluid-solid interface is then estimated as:
$ξ i n = r i n ( τ ) − r i ( τ − 1 ) .$
The algorithm proceeds with the calculation of the interface displacement. The displacement of the generic node
with respect to the previous time-step is estimated as:
$Δ i n = Δ i n − 1 + ϕ n R i n − 1 .$
Under-relaxation is necessary to guarantee the convergence of the segregated algorithm, as widely recognized in the literature [
]. Having determined the displacement of
, the computational mesh for the fluid domain can be adjusted to meet the new boundary configuration. For each FSI sub-cycle, the flow equations have to be integrated within the same time-step. The
ALE procedure calculates the convective fluxes accounting for the displacement of the faces of each computational cell within a time-step. A correct estimation of such fluxes requires that each
computational node is displaced from its position at time-step
$( τ − 1 )$
to the final position at sub-cycle
of time-step
. Within the foam-extend-3.2 framework, this is accomplished by first moving back the mesh to its configuration at time-step
and then smoothing the nodes to their final position. The procedure described in
Section 2.1
is then utilized to solve the flow equations. At any step, mass is updated guaranteeing continuity.
The residuals are defined as:
$R i n = ξ i n − Δ i n .$
The convergence of the FSI sub-cycle is assessed by comparing the magnitude of the residuals to a user-defined threshold
$R max$
$∑ i − 1 N Σ R i n R i 1 < R max ,$
$N Σ$
is the number of computational nodes in the fluid solid interface. If the inequality reported in Equation (
) is not verified, the relaxation factor is estimated before cycling the FSI sub-cycle starting from the flow solution (see
Figure 2
), according to the Aitken’s procedure [
$ϕ n + 1 = max − ϕ n ∑ i = 1 N Σ R i n − 1 · R i n − R i n − 1 ∑ i = 1 N Σ R i n − R i n − 1 2 , ϕ max if n ≥ 3 ϕ 1 if n < 3 ,$
$ϕ max$
is a limiting value that can be varied for each application.
3. Validation
The methodology described in
Section 2
is validated by comparison against experimental data. Both independent literature data [
] and custom designed experiments are considered in this validation.
Despite the proposed methodology is able to simulate 3D hydrodynamics and 6-DOF motions, for validation we concentrate on 2D hydrodynamics and 1-DOF structure motion to facilitate the comparison
between numerical results and experimental data.
We assume that the densities and the viscosities of water and air are $ρ water = 1000$ kg/m$3$, $ρ air = 1$ kg/m$3$, $μ water = 10 − 6$ Pa·s and $μ air = 1 . 48 × 10 − 5$ Pa·s. For all the cases we
considered $R max = 10 − 3$ and $ϕ max = 1$.
3.1. Water Entry of a Non-Buoyant Wedge
Zhao et al. [
] experimentally studied the water impact of a rigid wedge characterized by a total mass of 241 kg, a deadrise angle of
$30 ∘$
, a width of
m and a breadth of 1 m. The specimen is bounded to move in the vertical direction and impacts the water at a velocity of
Following the indications in Reference [
], we assume that the flow is two dimensional (2D), being the breadth of the specimen larger than its width. Thereafter the fluid domain is discretized through a 2D structured mesh with about 5200
computational cells. Moreover, according to References [
], we assumed that turbulence modeling is not required for impulsive impact phenomena.
Temporal discretization, is performed through the first order accurate Euler implicit scheme [
]. The Gauss upwind [
] scheme is utilized to discretize the convective terms in Equation (1) and the Gauss interface compression scheme is employed for the convective term in Equation (
) [
]. Other spatial discretizations are performed through the Gauss linear procedure.
Figure 3
we compare the numerical and experimental results for the vertical velocity of the wedge. We note that numerical results compare very well with the experiments. Specifically, the average relative
error is
$ε v = 1 N sample ∑ j = 1 N sample v CFD j − v EXP j max ( v EXP ) − min ( v EXP ) = 3.2 % ,$
$N sample$
the number of the considered experimental samples,
$v CFD$
the numerical approximation vertical component of the velocity of the specimen center and
$v EXP$
its experimental evaluation.
3.2. Water Entry of a Buoyant Wedge
The test performed in the previous section is not particularly challenging, as the specimen is relatively heavy and not buoyant, being the maximum mass of the displaced water 144 kg much lower than
the mass of the specimen (241 kg). In this section, the proposed numerical model is verified against the water impact of buoyant bodies performed by Panciroli et al. [
]. Therein, the water impact of several buoyant wedges with different average deadrise angles, curvatures and impact velocities are reported. All the wedges have a square base of width
m. Here, we focus on the specimens with a flat wetted surface, a constant deadrise angle of
$35 ∘$
and a mass of 0.412 kg. The impact velocity varies between
m/s and
m/s as a function of the drop height
and is reported in
Figure 4
together with its theoretical value, for all the considered experiments.
Analogously to the numerical test described in
Section 3.1
, we consider a 2D flow [
] and we utilize a laminar modeling approach [
]. A hexa-dominant unstructured mesh, with about 15,000 computational cells is used to discretize the fluid domain. We utilized the same numerical setting described in
Section 3.1
except for the discretization of the convective term of Equation (
) that is now performed through the linear-upwind procedure.
Numerical results are compared to experimental measurement from Reference [
] in terms of vertical displacement of the center of mass. We denote with
$δ EXP$
$δ CFD$
the experimental and numerical displacement, respectively.
Figure 5
Table 1
show the good agreement between the proposed numerical model and the experimental data for all the considered cases. The accuracy of CFD is quantitatively assessed by the average relative error,
evaluated as
$ε δ = 1 N sample ∑ j = 1 N sample δ CFD j − δ EXP j max ( δ EXP ) − min ( δ EXP ) ,$
and reported in
Table 1
. Notably,
$ε δ$
is always below
$4 %$
Figure 5
shows that CFD slightly underestimates the wedge penetration. This behavior can be explained by the 2D approximation that overestimates the fluid forces on the specimen. In fact, border effects,
which are not considered in the 2D simulation, are expected to decrease the pressure close to the wedge borders, thus decreasing the fluid reaction [
Figure 6
depicts the free surface evolution for
$h = 0.5$
m, qualitatively demonstrating the capability of the proposed methodology to capture the most relevant features of hull water entry. In fact, bulk flow and pile-up can be clearly identified in
Figure 6
3.3. Water Entry and Exit of a Buoyant Cylinder
The test cases proposed in
Section 3.1
Section 3.2
focus on the water entry problem. Experimental data are, in fact, limited to the first instants after the water impact, when hydrodynamic forces are larger. To validate the proposed methodology in a
more general framework, we specifically designed a set of dedicated experiments on the water entry and exit of a buoyant cylinder [
3.3.1. Experimental Setup
The experimental setup consists of a water tank made of stainless steel and Lexan^®, with an internal volume of $1.5 m × 1.85 m × 0.7 m$. The tank is filled with water up to a level of $0.4 m$. Two
transparent sides grant optical access to the tank. The specimen, a hollow PVC cylinder filled with polystyrene with a diameter $d = 0.16$ m and a breadth $l = 0.295$ m, is rigidly connected to a
wood sledge that is allowed to move along two vertical aluminum rails. We consider four drop heights of $0.25$ m, $0.50$ m, $0.75$ m and $1.00$ m and two specimen masses, $m = m 0$ and $m = m 0 + 1$
kg, being $m 0 = 2.214$ kg.
The specimen position over time is measured through two Spectrasymbol thinpot linear potentiometers embedded in one of the rails. A wiper with tip diameter of $2.7$mm is connected to the sledge and
actuates the position sensor.
The optical acquisition of the experiments is performed via a Phantom^® Miro 110 monochromatic high speed camera operating at a frame-rate of 2500 Hz for $m = m 0$ and of 2700 Hz for $m = m 0 + 1$
A capacitive accelerometer Adafruit ADXL335-5V is rigidly connected to the sledge and measures the specimen acceleration before water impact. Acceleration data are acquired only before the water
impact because the measurement range of the accelerometer is ±3 g, while acceleration ensuing from the water impact is expected to be much larger.
The analogue signals from the accelerometer and the displacement sensor are synchronously acquired by a National Instrument NI USB-6009 acquisition board at a sampling rate of
kHz at a resolution of 14 bits. The acquisition system is controlled through an ad-hoc realized LabVIEW [
] dashboard.
Each drop test is repeated three times to assess the data fidelity against random and systematic errors. We average the data from the three repetitions and apply a smoothing average filtering to
obtain the effective position and acceleration as functions of time. We set $t = 0$ when the specimen impacts the water free surface.
The impact velocity is of paramount importance in water entry problems, as discussed in Reference [
] and we obtain a robust estimation of it averaging the values yielded by three procedures:
Numerical differentiation of the displacement measurement through central difference approximation;
Numerical integration of the measured acceleration through the trapezoid rule;
Least square fitting of the specimen displacement to a second order function and algebraic differentiation.
The obtained impact velocity is reported in
Figure 7
, as a function of the specimen mass and drop height. For more details on the experimental setup, the reader can refer to References [
3.3.2. Numerical Setup
We consider here a 2D flow [
] and we utilize a laminar modeling approach [
]. We adopt the same numerical setup described in
Section 3.2
. In this case the hexa-dominant computational mesh is characterized by about 19,000 computational cells.
3.3.3. Results and Dicussion
Numerical and experimental evaluations of the cylinder displacement are reported in
Figure 8
for all the combinations of
. We observe a general good agreement between CFD and experiments. A visual estimation of the cylinder position through the high speed camera is also reported in
Figure 8
as a further confirmation of the accuracy of the proposed methodology. The reliability of the present simulations is quantitatively assessed by the percent relative errors estimated through Equation
) and reported in
Table 2
. Remarkably,
$ε δ < 11 %$
for all the considered tests.
We note that the differences between numerical results and experimental data may be generally related to an overestimation of the fluid forces by the numerical model. In fact, numerical simulations
underpredict the maximum penetration of the specimen and larger fluid forces also cause a higher rebound after motion inversion (see
Figure 8
). As already highlighted in
Section 3.1
Section 3.2
, the 2D approximation may explain such an underprediction. Moreover, we discarded any friction dissipation on the experimental setup after water impact. Such a dissipation also concur in reducing
the specimen momentum, in particular after the motion inversion, thus smearing the amplitude of the specimen oscillations.
Figure 9
represents the free surface dynamics ensuing from the water impact, by comparing the high speed camera images to the CFD evaluation of the water volume fraction for different time instants. The
proposed numerical model correctly captures the most relevant features of the hydrodynamics related to the water impact of the cylinder, such as the pile-up and water jet formation (
Figure 9
b), the air cavity formation (
Figure 9
c) and closure ((
Figure 9
d,e) in the proximity of the maximum penetration. Slight differences between CFD and experiments are detected for the free surface pattern in
Figure 9
b,c, before the closure of the air cavity. Images from the high speed camera show a larger cavity compared to CFD. Moreover, the cavity closure is slightly anticipated in the CFD with respect to the
experiments, as evidenced in
Figure 9
c, where the experimental air cavity borders points away from the cylinder, while the proposed numerical model predicts that the air cavity borders are directed inward. In this respect, we comment
that such a discrepancy may be related to a poor prediction of the air flow. As a matter of facts, we focused on the hydrodynamics ensuing from water impact, that largely influence the cylinder
motion, rather than on the modeling of the air flow. In particular, initial conditions are not realistic for the air flow, as the fluid domain is initialized with a constant null velocity and the
cylinder is impulsively started at the impact onset. On the contrary, in the experiments the air flow at the impact onset is already completely developed, as a result of the specimen free fall path
in air. We also note that the lack of turbulence modeling may also hinder the correct evaluation of the air flow behind the cylinder.
Figure 10
shows the convergence history for
$h = 0.5$
m and
$m = m 0$
. For sake of clarity we reported only 25 equally spaced time-steps from
$t = 0$
$t = 2$
s. Convergence is relatively fast being reached with a number of FSI subcycles comprised between 4 and 7. We also comment that
is equal to
for the first 3 iterations (see Equation (
)) and is then increased up to
thanks to the Aitken’s procedure. Similar convergence histories are obtained also for the other cases.
3.4. Roll Motion of a Rectangular Structure
The previous tests are all focused on hull slamming problems. However, the proposed numerical model could be used for many other engineering problems related to FSI in multiphase flows and, among
them, for the simulation of floating structures and their response to wave actions, which have attracted much interest in the hydraulic and ocean scientific community in the last decades [
In this section, we test the proposed numerical model for the simulation of a floating body. Jung et al. [
] experimentally studied the roll motion of a rectangular acrylic structure
m large,
m high and
m wide (normal to the motion plane). The structure is hinged on a water tank 35 m long,
m deep and
m wide and is free to roll about its center of gravity. Translations, as well as yaw and pitch, are not permitted. The water depth is
m. At equilibrium the width of the rectangle is horizontal and the draft of the body is
m. The moment of inertia with respect to an axis normal to the motion plane through
$J = 0.236$
The free roll decay test was performed by releasing the structure from an initial roll angle of 15 degrees in calm water and measuring the roll angle variation as a function of time with a rotary
sensor [
We utilize the numerical setup described in
Section 3.2
and the same mesh structure. In this case the domain discretization results in about 7000 computational cells.
As evidenced in Reference [
], contrarily to the hull slamming problems described in the previous sections, the turbulent eddy dissipation is expected to significantly impact the amplitude of the roll motion. Thereafter, a
proper turbulence treatment is necessary to capture the hydrodynamics induced by the solid body motion and, in turn, the rotation amplitude. To this aim, we adopt a RANS modeling approach with a
$k − ε$
standard closure [
]. Standard wall functions [
] are utilized for the solution of the turbulent kinetic energy and turbulent kinetic energy dissipation velocity equations at solid boundaries.
The reliability of the proposed methodology is assessed in
Figure 11
that reports the rotation angle (
) as a function of time. The agreement between CFD and experimental results from Reference [
] is remarkable. Specifically, the average relative error is
$ε β = 1 N sample ∑ j = 1 N sample β CFD j − β EXP j max ( β EXP ) − min ( β EXP ) = 3.8 % .$
To demonstrate the potentiality of the proposed methodology in terms of flow analysis we represent the flow field generated by the solid body motion in
Figure 12
. Therein, the velocity vectors are represented only within water to allow the visualization of the vortex structures ensuing from the rotation of the rectangle. We comment that, is this case, the
air motion is expected to have a minor impact on the structure dynamics, with respect to water. As expected, the fluid velocity is maximized at the onset of the motion, when the angular velocity of
the solid structure is larger. Two counter-rotating vortexes are generated at the corners of the rotating box. Such vortexes subtract mechanical energy from the structure contributing to the
reduction of the oscillation amplitude. This mechanical energy is dissipated through a turbulent dissipation mechanism [
4. Conclusions
In this paper we develop a computational methodology to model the FSI of rigid bodies in free surface flows. We adopt a tightly coupled partitioned approach to cope with structures whose density is
comparable or lower to the fluid one.
The methodology is implemented within the foam-extend-3.2 open-source framework that allows a significant model re-utilization. The solution of the flow equation relies on the interDyMFoam solver
that implements the VOF model and approximates Equations (1) and (
) through a generalized finite volume method. The 6-DOF solid body motion is described through the Newton-Euler equations that are numerically integrated utilizing the Euler explicit method.
Quaternions or Euler parameters, are employed to parametrize the solid body orientation with respect to an inertial reference frame. Simplified solid body motions are obtained by nullifying the
proper acceleration components in the Newton-Euler equations. Within the Dirichlet-Newmann approach, the tight coupling is obtained through fixed point iterations, whose convergence is boosted
utilizing the Aitken’s dynamic under-relaxation.
We validate the proposed methodology by comparing numerical results to experiments in four different simplified 2D geometries (i.e., 2D flow field and 1-DOF solid body motion), yet computationally
demanding test cases. Specifically, we select three different water impact problems and the free decay of the roll motion of a buoyant rectangle.
Both literature and custom designed experiments are considered. We first studied the water impact of elongated wedges varying the deadrise angle, the mass and the impact velocity. Comparison with
experiments from the literature is very satisfactory with an average relative error below 4%. Nevertheless, literature data were limited to a relatively short time after the water impact. Thereafter,
we performed a dedicated set of experiments to assess the capabilities of the proposed methodology also for the water exit. By contrasting CFD with those experiments, we note that the average
relative error is around or below 10%. Moreover, the method is capable of correctly describing the initial water impact, the motion inversion and the successive oscillations about its equilibrium
position of a buoyant cylinder. Finally, since rotations were not included in the previous tests, we simulated the natural decay of the roll motion of a buoyant rectangle to assess the capability of
the proposed methodology to simulate floating bodies. Also in this case the agreement between numerical predictions and experimental data is very good, being the average relative error below 4%.
Summarizing, the proposed numerical model performs remarkably well for all the tests cases. Even in the worst case, the discrepancies fall within the confidence interval expected for the adopted
numerical models. The simplifying hypotheses in the computation, such as 2D flow, laminar modeling or the eddy viscosity model for the turbulent closure may explain the differences between CFD and
We comment that, despite we considered only 1-DOF cases for validation, due to the availability of experimental results, the combinations of the four proposed test cases includes both pure
translations and rotations. Thus, the proposed methodology can be considered reliable also in the general 6-DOF case.
Besides demonstrating the reliability of the proposed methodology, the test cases described throughout the paper also evidence its relevance for engineering applications. In fact, such problems are
important in several engineering fields such as naval, civil and hydraulic engineering. Moreover, FSI combined with multiphase flow modeling can be useful for other applications such as
bio-engineering or energy harvesting.
Future work will be devoted to implementing kinematic constraints that limit the solid body motion and to modeling compliant bodies.
Author Contributions
Conceptualization, A.L.F. and G.F.; methodology, A.L.F. and G.F.; formal analysis and investigation, A.L.F. and G.F.; original CFD coding, A.L.F. and G.F.; coding implementation, A.A.; writing
original draft preparation, A.L.F.; writing review and editing, A.A. and C.B.; supervision, S.U., C.B. and E.J.
This research has been supported by the Italian Ministry Program PRIN, grant n. 20154EHYW9 “Combined numerical and experimental methodology for fluid structure interaction in free surface flows under
impulsive loading”, with Chiara Biscarini as the Principal Investigator.
The authors would like to thank two Anonymous Reviewers are also acknowledged for constructive comments to this paper.
Conflicts of Interest
The authors declare no conflict of interest. The funders had no role in the design of the study; in the collection, analyses, or interpretation of data; in the writing of the manuscript, or in the
decision to publish the results.
Figure 3.
Comparison between numerical and experimental results for the specimen velocity during the water impact of a rigid wedge. Experimental data are retrieved from Reference [
Figure 4.
Impact velocity of the specimen as a function of the drop height. Experimental data retrieved from Reference [
Figure 5.
Displacement of the specimen center of mass. Comparison between CFD and experiments [
] for different
: (
$h = 0.25$
m; (
$h = 0.50$
m; (
$h = 0.75$
m; (
$h = 1.00$
Figure 6. Evolution of the free surface during water impact for $h = 0.5$ m. CFD results for: (a) $t = 0$ ms; (b) $t = 6$ ms; (c) $t = 12$ ms; (d) $t = 18$ ms; (e) $t = 24$ ms.
Figure 7. Impact velocity of the specimen as a function of the drop height and of the mass of the impacting cylinder.
Figure 8. Displacement of the specimen center of mass. Comparison between CFD and experiments for different h and m: (a) $h = 0.25$ m and $m = m 0$; (b) $h = 0.5$ m and $m = m 0$; (c) $h = 0.75$ m
and $m = m 0$; (d) $h = 1.00$ m and $m = m 0$; (e) $h = 0.25$ m and $m = m 0 + 1$ kg; (f) $h = 0.5$ m and $m = m 0 + 1$ kg; (g) $h = 0.75$ m and $m = m 0 + 1$ kg; (h) $h = 1.00$ m and $m = m 0 + 1$
Figure 9. Evolution of the free surface for $m = m 0$ and $h = 0.5$ m. Comparison between CFD and experiments for different times after water impact: (a) $t = 0$ S; (b) $t = 0.08$ s; (c) $t = 0.16$
s; (d) $t = 0.24$ s; (e) $t = 0.32$ (s).
Figure 10. Convergence history for $h = 0.5$ m and $m = m 0$. Note that only 25 time-steps are represented for clarity. Each mark represents a FSI subcycle.
Figure 11.
Rotation of the specimen about its center of mass. Comparison between CFD and experiments from Reference [
Figure 12. Representation of the flow field in the vicinity of the oscillating box for different times. For clarity, vectors are represented only in water. (a) $t = 0.08$ s, (b) $t = 0.8$ s, (c) $t =
0.16$ s, (d) $t = 0.24$ s and (e) $t = 0.32$ s.
h [m] $0.25$ $0.50$ $0.75$ $1.00$
$ε δ$ [%] $2.81$ $2.24$ $2.68$ $3.47$
Table 2. Percent error on the displacement of the center of mass of the buoyant cylinder as a function of h and m.
$m = m 0$ $m = m 0$ + 1 kg
$h = 0.25$ m $9.93 %$ $7.33 %$
$h = 0.50$ m $9.82 %$ $6.32 %$
$h = 0.75$ m $10.0 %$ $9.03 %$
$h = 1.00$ m $8.93 %$ $10.8 %$
© 2019 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http:/
Share and Cite
MDPI and ACS Style
Facci, A.L.; Falcucci, G.; Agresta, A.; Biscarini, C.; Jannelli, E.; Ubertini, S. Fluid Structure Interaction of Buoyant Bodies with Free Surface Flows: Computational Modelling and Experimental
Validation. Water 2019, 11, 1048. https://doi.org/10.3390/w11051048
AMA Style
Facci AL, Falcucci G, Agresta A, Biscarini C, Jannelli E, Ubertini S. Fluid Structure Interaction of Buoyant Bodies with Free Surface Flows: Computational Modelling and Experimental Validation. Water
. 2019; 11(5):1048. https://doi.org/10.3390/w11051048
Chicago/Turabian Style
Facci, Andrea Luigi, Giacomo Falcucci, Antonio Agresta, Chiara Biscarini, Elio Jannelli, and Stefano Ubertini. 2019. "Fluid Structure Interaction of Buoyant Bodies with Free Surface Flows:
Computational Modelling and Experimental Validation" Water 11, no. 5: 1048. https://doi.org/10.3390/w11051048
Note that from the first issue of 2016, this journal uses article numbers instead of page numbers. See further details
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Houdini & Autoencoders: Escaping the shackles of high-dimensional space
From autoencoders to stable diffusion
Over the next few editions, I'm going to be writing about generative AI techniques. But before we can jump headfirst into Stable Diffusion and the fancy images it produces, we need to look back at
some of the fundamentals. That's why I'm covering the humble autoencoder this week. Although it isn't a generative model in the strict sense of the term, learning how it works will help us understand
more complicated networks in the coming weeks. Let's dive into the magical world of autoencoders.
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This Week on Gradient Ascent:
• Understanding autoencoders - The magical basics 🔢
The Prestige of Autoencoders:
"Every great magic trick consists of three parts or acts. The first part is called "The Pledge". The magician shows you something ordinary: a deck of cards, a bird or a man. He shows you this object.
Perhaps he asks you to inspect it to see if it is indeed real, unaltered, normal. But of course... it probably isn't. The second act is called "The Turn". The magician takes the ordinary something
and makes it do something extraordinary. Now you're looking for the secret... but you won't find it, because of course you're not really looking. You don't really want to know. You want to be fooled.
But you wouldn't clap yet. Because making something disappear isn't enough; you have to bring it back. That's why every magic trick has a third act, the hardest part, the part we call "The
- Christopher Priest, The Prestige
I'm a huge Christopher Nolan fan. If you are, too, then you'll probably remember these lines from one of the most underrated movies he's made, "The Prestige". In a Wolverine vs. Batman showdown, Hugh
Jackman and Christian Bale feud with each other over a single magic trick. The movie opens with this quote and introduces us to how a magician tricks us into believing that an illusion is real.
Guess what? We have a magic trick or two in the machine learning world too. Let's take a look at the humble autoencoder and how it performs an incredible trick - making data disappear! Let's figure
out how it does this.
The Pledge: Introducing the Autoencoder
The pledge, or the first part of this magic trick, simply shows what an autoencoder looks like. As you might expect, it looks deceptively simple. It consists of two components, an encoder, and a
It's an unassuming neural network and, in our example below, consists entirely of linear layers. The input to our autoencoder will be images. Given this input, the autoencoder will make it disappear
and magically bring it back.
How does it do so? It learns to represent data in an unsupervised1 manner.
Here's what our autoencoder looks like in code. Note that this architecture is called a stacked (or deep) autoencoder since it consists of several layers stacked together. Also notice that the size
of the input gets progressively smaller as it moves through the encoder.
The Turn: Encoding and Compression
The Turn is where the magician takes the ordinary object and makes it do something extraordinary. Our humble autoencoder will now make the input data disappear.
Given an image, the encoder transforms it into a lower-dimensional representation called the latent space. By compressing the input, the autoencoder has to learn the most important features of the
input data and can't simply copy over the input.
How does the compression happen? The encoder uses a series of layers that have lower dimensionality than the input. With each successive layer, the input becomes increasingly sparse. Compressing
input data is easy, right? Well, how do you recover the original data from the compressed representation?
The Prestige: Decoding and Reconstruction
In the final part of the trick, our autoencoder brings back the original data. The decoder reconstructs the original data from the compressed latent representation. The decoder is typically a mirror
image of the encoder and progressively increases the size of the data flowing through it.
Although the reconstruction may not be identical to the original, it will still be very close.
Here's how our autoencoder reconstructed some pictures of clothing items from the Fashion MNIST dataset.
Pretty cool, huh? The real magic of the autoencoder is that it learns how to do this without explicitly being told so.
Behind the Curtain
So far, we've taken for granted that the encoder can compress the input in any way it likes, and the decoder can wave a magic wand and bring the original input back. But, like any magic trick,
there's some sleight of hand involved. Let's unpack that next.
Squiggly Deconstruction
Mathematically, here's how the encoder and decoder are represented.
What are X, Z, and X_hat? They are our input, compressed representation, and reconstruction, respectively. W and b are the weights and biases of the encoder, while W’ and b’ are the weights of the
decoder. f and g are activation functions that introduce non-linear magic into the picture. This allows our network to learn complex representations.
In order for the decoder and encoder to work harmoniously, we have to ensure that X_hat is as close to X as possible. That's where the reconstruction loss comes in. It gives us an idea of how close
the reconstructed output is to the input. In this case, we use the mean squared loss (MSE). This computes the difference between the reconstruction and the input and squares it.
Over the course of training, the autoencoder adjusts its weights2 so that this loss becomes smaller and smaller.
Curtain Call: Applications
Autoencoders are pretty nifty and can be used for:
• Dimensionality reduction: If you haven't already guessed, autoencoders learn compact representations of data, making them great for reducing the dimensionality of data.
• Anomaly Detection: Since autoencoders learn the important features from the input, they can spot abnormalities in data that deviate significantly from normal observations. These weird samples
will have a higher reconstruction loss, and thus we can spot them!
• Denoising: Autoencoders can be used to remove noise from input data. For example, I added some noise to the images and passed them through the autoencoder. It removed the noise reasonably well.
In the next edition, we'll look at more complex models that overcome the limitations of the simple autoencoder we saw this week. The biggest of these limitations is that a standard autoencoder cannot
produce novel samples since it only learns a direct mapping of the input data. How do we solve that? Stay tuned!
P.S.: The code for this week’s article can be found here.
Some would say self-supervised.
Both the encoder and decoder’s weights
I always enjoy the way you simplify and make the challenging more understandable. This is a great metaphor, well deployed.
Expand full comment
nice metaphor!
Expand full comment
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Tree Distance
Submit solution
Points: 12 (partial)
Time limit: 2.0s
Memory limit: 1G
You are given a tree by your parents for Christmas. They want you to find the length of the path between some nodes. A tree is a graph, such that all nodes are connected, and there is only one simple
path between any nodes.
Your parents will give you questions, of the form . For each question, print the length of the path between node and node .
The length of a path is defined as the number of edges between nodes.
Input Specification
The first line will contain integers , the number of nodes in the tree, and the number of questions, respectively.
The next lines will each contain integers, , which means that nodes and node are connected.
The next lines will each contain integers, .
Output Specification
For each question, print the length of the path between node and node on its own line.
Subtask 1 [30%]
Subtask 2 [70%]
No further constraints.
Sample Input
Sample Output
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How the math editor works
A Math Editor is a digital tool that facilitates the creation, editing, and formatting of mathematical expressions and formulas. Trelson now offers a math editor that can be made available to
students in the ‘Writing Space’ module, as well as in the ‘Form’ module through the question type 'Text with special characters'.
Math editor in the writing space
Note that students do not automatically have access to the math editor in the Writing Space module. As a teacher, you need to actively choose to enable support for math in the writing space. You can
also create a “Custom template” in the writing space and use the math editor to write prepared questions or exercises for students to work on in the locked mode when the task starts.
How to enable the math editor in the writing space:
1. Modules: Enter the task and click on the "Modules" tab.
2. The pen: Click on the icon that looks like a pen next to the writing space module.
3. Settings: Click on the gear icon to bring up the settings for the writing space.
4. Enable math support: Check the box that says "Enable math support".
5. Save: Click "Save". Now all students will have access to the math editor in the writing space when they open the task.
Math editor in form questions
Below are some examples of how you and your students can use the math editor in form questions.
Include a mathematical formula in the question title
As a teacher, you can use the math editor to include mathematical expressions in question titles and answer choices. When you see the "Square root" icon, you can use the math editor. Teachers can,
for example, use it in the titles of form questions. Students see the same icon when they have access to the math editor.
Let the student answer the question with mathematical expressions
Choose the question type "Text with special characters" if your students are to perform a mathematical calculation as an answer to the question. Students will then get the "Square root" icon which
they click on to open the math editor.
Present several formulas for the student to choose as an answer to a question
When you create a form question and choose the question type "Multiple choice" or "Checkboxes" you will see a "Square root" icon to the right of each answer choice. This means you can enter a formula
as an answer choice. Click on the icon to open the math editor. When you have finished your formula, click "Insert". Note that you still need to name each answer choice in words, as shown in the
image below.
Not just math – also chemical formulas
The math editor can also be used to write chemical formulas and expressions. Students can press the icon that looks like a “C” to access this feature.
Important to note
• Automatic grading: In the "Form" module, Trelson supports automatic grading where you as a teacher can specify the correct answer in advance. However, if a student answers a question with the
question type "Text with special characters", you cannot use automatic grading. If you create equations as answer choices, you can specify which option is correct and thus enable automatic
• Not a calculator function: The math editor is not a calculator but is only used to visualize mathematical or chemical expressions. If students need access to a calculator, we recommend using
GeoGebra as a resource, or providing students with a physical calculator.
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According to legend, the famous Achilles, hero of the Trojan War, was the fastest runner in all of Greece. One day an lazy old tortoise challenged Achilles to a race, claiming that he could win as
long as Achilles gave him a small head start. Achilles laughed and told the tortoise he would surely lose, on which the tortoise replied: “On the contrary, I will win, and I can prove it to you by a
simple argument.” The argument given by the tortoise is the famous Zeno’s paradox, a piece from a series of paradoxes aimed to prove the irrationality of motion. For years mathematics has been the
gateway for paradoxes that keep even the brightest puzzled for days. In this article you can find a series of mind-boggling mathematical questions, facts and phenomena that show mathematics is about
more than just numbers.
Zeno’s Paradox
The tortoise’s argument is structured as follows: if the tortoise is granted a hundred meters head start and we assume that Achilles is ten times faster than him, we can say that when Achilles
reaches the starting point of the tortoise, the tortoise will have run ten meters. Hence, Achilles will have to catch up and by the time he reaches this next point, the tortoise will have moved
again. Hence in each moment Achilles must be catching up the distance between him and the tortoise who is adding new distance every time. As a result, Achilles will never reach the tortoise no matter
how much faster he is.
Benford’s law
Everyone must have heard at some point in their life about the normal distribution and how it reoccurs in almost all phenomena. There is however another law of numbers that shows random data might
not be that random at all. If you take a list of numbers representing absolutely anything, it turns out about thirty percent of all of these numbers will start with the digit one, a bit less with the
digit two, continuing until the smallest percent of numbers will begin with the digit nine. This phenomenon is a simplification of Benford’s law which states that the leading digits of the numbers
found in real-world data sets are not uniformly distributed as one might expect, but occur in a decreasing order. Again, random patterns turn out to be less random than we thought.
Russell’s Paradox
At the turn of the 20th century, mathematician Bertrand Russell was interested in a new branch of mathematics: Set Theory. Until then, it was believed that anything could be made into a set (a
collection of objects). However in 1901, Russell made a mind blowing discovery when he realized that not everything could be turned into a set. Take for example the set that contains all sets that do
not contain themselves. What about this set? Russell’s set certainly does not contain itself, hence it should be included. However now Russell’s set does contain itself, and hence a logical step
would be to exclude it again, after which is should be included again… and so on. This contradiction is now known as Russell’s Paradox and made a huge impact on Set Theory as we know it today.
1 = 0.999…
A final mind-boggling concept of mathematics is the idea that the repeating decimal 0.999… equals one as proven below to be true.
The reason why for many this might seem illogical or deceiving is because of the fact that for our minds the concept of infinity is hard to grasp. We still imagine 0.999… to end somewhere and hence
be just a tiny bit smaller than one. While in the end, 1 and 0.999… are just the same number expressed in different ways. Even though most people abandon their interest in mathematics the moment
letters and numbers get mixed up, it keeps producing mind-blowing results.
This article is written by Fenna Beentjes
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Quantity of Cement, Sand & Water required for Plastering Estimate
How to Calculate Quantity of Cement, Sand & Water required for Plastering Estimate
In this Article today we will talk about the Quantity of Cement Sand and Water in Plastering | Calculation of Materials for Plastering | How much Cement and Sand Required for Plastering | Plastering
Estimate | How to Find Plastering Quantity | Quantity of Materials for Plaster | Requirements of Good Plaster | Strength of Plastering | Cement Mortar Calculation for Plastering
Quantity of Cement, Sand & Water required for Plastering:
Plastering refers to protecting a wall or ceiling by laying a plaster (Cement plaster). Plastering is done to remove surface imperfections caused by brickwork and to keep the surface smooth for
There are many different types of plastering materials out of them, cement plaster is extensively used. Some other types of plastering materials include lime plaster, clay plaster, etc. Calculation
of Materials for Plastering
How to Select the Right Plastering Material for your House?
In any type of plastering two major factors are considered they are Surface Protection and the cost of material. If the quality of plastering is increased and taken higher, then the cost is affected.
If price is considered and Quality of plaster are taken lesser importance then the surface protection is compromised. Calculation of Materials for Plastering
1. Plastering material should be cheap and economical.
2. It should be durable enough to sustain any climatic changes in the entire life span of structure.
3. Plastering material should have excellent workability which can be applied during any weather conditions. Calculation of Materials for Plastering
Let us calculate the quantity raw materials (Cement, Sand & water) required for cement plaster. Though the same process is applicable for any other types of plastering materials. Cement Mortar
Calculation for Plastering
1. It should adhere to the background and should remain adhered during all climatic changes. Requirements of Good Plaster
2. It should be cheap and economical.
3. It should be hard and durable.
4. It should be possible to apply it during all weather conditions.
5. It should effectively check the entry or penetration of moisture from the surfaces.
6. It should possess good workability. Plastering Estimate
Architects refer the following concrete code table in drawing. Knowing this table will help you in deciding the proportion of cement, sand and gravel in Plastering. Plastering Estimate
Plaster type Description
12mm 12 MM thick cement plaster is done where the plain surface of brick masonry is plastered.
15mm 15 MM cement plaster is required on the rough side of 9” and 4.5” wall.
18mm 18 MM thick cement plaster with neat cement slurry is required for making ‘Dado’ with the cement concrete flooring.
20mm 20 MM thick cement plaster is done in two coats in some cases on rough side of wall or according to the design requirement.
Calculating the quantities of Cement & Sand required for plastering Work:
How to Calculate Quantity of Cement Sand and Water in Plastering
General points to be remembered in Plastering work calculation Plastering Estimate
• The ratios mentioned in plastering are volumetric ratios of Cement & Sand (Ex. Cement:Sand = 1:5, 1 part of cement and 5 parts of sand in a mortar).
• The overall thickness of plastering should be minimum 20mm including two coats.
• Cement has a dry density volume of 1440Kg/m^3
• Each bag of cement weighs = 50 Kgs or 110 lbs
• The Volume of each cement bag = 50Kgs/1440 = 0.0348 m^3.
• The dry density of sand = 1600Kgs/m^3
• The plastering is done in two layers (two coats): The first coat of plastering is laid with the thickness of minimum 12mm (usually ranges between 12-15mm) and this coat is called as a Rough coat
or Primary coat.
• The second coat should be laid with the thickness of 8mm and this is called as a Finish coat or Secondary coat of plastering. Plastering Estimate
Total Plastering = First coat + Second Coat How to Find Plastering Quantity
Different ratios of Cement mortar used for plastering are tabulated below:
Mix Ratio Areas of usage
1:6 & 1:5 This ratio is usually used for Internal plastering of bricks
1:4 Used for Ceiling and external walls
1:3 As its a rich mortar mix and it is used where external walls are prone to severe climatic conditions. It is also used for repair works.
How to Find Plastering Quantity Cement Mortar Calculation for Plastering
Calculate Quantity of Cement and Sand in Plaster:
Steps involved in calculation of plastering quantities:
1. Find the total area of wall to be plastered in Sqm (m^2).
2. Consider the ratio and thickness of plastering
3. Calculate the Total Volume of Plastering
4. Find out the Volume of Cement and Sand individually for both the coats
5. Calculate the total volume of cement & Sand required for plastering
Now coming to the calculation part, We are considering the below values for calculation purpose: How to Find Plastering Quantity
1. Wall width and height is 10m and 10m.
2. Ratio of First coat of plastering (Cement:Sand = 1:5) with the thickness of 12mm.
3. the ratio of secondary coat of plastering (Cement:Sand = 1:3) with the thickness of 8mm. How to Find Plastering Quantity
Step 1: Calculate the Area of Plastering
Area = width x height = 10 x 10 = 100m^2 Quantity of Materials for Plaster
Step 2: Find the Wet Volume of Plastering
Wet Volume of First Coat = Area of Plastering x Thickness of Plastering
= 100m^2 x 12mm (Convert mm to m)
= 100×0.012 = 1.2m^3
Hence, Wet Volume of First Coat of Plastering = 1.2m^3
The Wet volume of Second Coat = Area of Plastering x Thickness of Plastering
= 100m^2 x 8mm (Convert mm to m)
= 100×0.008 = 0.8m^3
Therefore Wet Volume of Second coat of plastering = 0.8m^3
Step 3: Find the Dry Volume of Plastering
Dry Volume of First Coat = Wet Volume of First Coat x 1.33
= 1.2 m^2 x 1.33
= 1.596 m^3
Hence, Dry Volume of First Coat of Plastering = 1.596 m^3
The Dry volume of Second Coat = Wet Volume of Second Coat x 1.33
= 0.8 m^3 x 1.33
= 1.064 m^3
Therefore Dry Volume of Second coat of plastering = 1.064 m^3
Step 4: Finding the individual quantities of Cement and Sand.
First coat ratio = 1:5 (1 part of Cement and 5 parts of Sand)
Total parts = 1+5 = 6 ^Quantity of Materials for Plaster
Quantity of Cement required for First coat
= (Total Volume of first coat plastering x No. of Parts of cement) ÷ Total Parts Cement Mortar Calculation for Plastering
= 1.596 x 1/6 = 0.266 m^3
Quantity of Sand required for First coat =
(Total Volume of first coat plastering x No. of parts of sand) ÷ Total Parts
= 1.596 x 5/ 6 = 1.330 m^3
Similarly for Second coat,
Second coat ratio = 1:3 (1 part of cement and 3 parts of sand)
Total parts = 1+3 =4 ^Quantity of Materials for Plaster
Quantity of Cement required for Second coat
= (Total Volume of second coat plastering x No. of Parts of cement) ÷ Total Parts
= 1.064 x 1/4 = 0.266 m^3
Quantity of Sand required for Second coat
= (Total Volume of second coat plastering x No. of parts of sand)÷ Total Parts
= 1.064 x 3/4 = 0.798 m^3 Quantity of Materials for Plaster
Step 5: Finding the quantity of Cement Bags:
Total Quantity of Cement = Quantity of Cement for First Coat + Quantity of Cement for Second Coat
Total Quantity of Cement = 0.266 + 0.266 = 0.532 m^3
As mentioned above volume of 1 bag of cement (50kgs) = 0.0348m^3
For 0.532 m^3 = 0.532 x 50 / 0.0348 = 764.36 Kgs = 15.28 bags
Similarly, Total Quantity of Sand = Quantity of Sand for First Coat + Quantity of Sand for Second Coat
Total Quantity of Sand = 1.330 + 0.798 = 2.128 m^3
Similarly, for Sand 1m^3 = 1600Kgs. 2.128 m^3 = 2.128 x 1600 = 3404.8 Kgs = 3.4 tonnes
Step 6: Finding the quantity of water required for plastering:
Amount of water to be added in mix depends upon the moisture content present in cement, sand & atmosphere.
Quantity of water = 20% of total dry material (Cement + Sand)
= 20% of (764.36 + 3404.8) = 0.2 x 4169.16 = 833.832 = 834 litres. Requirements of Good Plaster
Final Result:
The Quantity of cement required = 764.36 Kgs
Calculated Quantity of Sand (Fine aggregate) required = 3404.8 Kgs
Quantity of Water required = 834 litres.
How many bags of cement are required for 1 square meter of plastering?
If the above mentioned values are considered for this then (Rough estimate)
From above, 100m^2= 574 Kgs of cement 1m^2 = 574/100 = 5.7Kgs Requirements of Good Plaster
Quantity of Cement, Sand & Water required for Plastering. For 100m^2 of Wall, if first & second coat of cement mortar ratio 1:5 & 1:3 laid then
The Quantity of cement required = 764.36 Kgs
Calculated Quantity of Sand (Fine aggregate) required = 3404.8 Kgs
Quantity of Water required = 834 litres.
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University of Edinburgh University of Edinburgh
An asymptotically optimal mixing-time bound on the bases-exchange Markov chain for a matroid
Probability Seminar
21st April 2023, 3:30 pm – 4:30 pm
Fry Building, 2.04
We discuss our results on the problem of sampling bases of a matroid. In particular, we will discuss a modified log-Sobolev inequality for r-homogeneous strongly log-concave distributions. As a
consequence, we obtain an asymptotically optimal mixing time bound for the bases-exchange chain and a concentration result for such distributions. I will also mention some continuing results by other
authors, which extended this work
Joint work with Heng Guo and Giorgos Mousa. | {"url":"https://www.bristolmathsresearch.org/seminar/mary-cryan/","timestamp":"2024-11-08T06:05:40Z","content_type":"text/html","content_length":"54113","record_id":"<urn:uuid:922411c6-100d-4cc4-9669-0e0d94cb596d>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00221.warc.gz"} |
Do it in Java 8: The State Monad
1. Do it in Java 8: The State Monad
Do it in Java 8: The State Monad
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In a previous article (Do it in Java 8: Automatic memoization), I wrote about memoization and said that memoization is about handling state between function calls, although the value returned by the
function does not change from one call to another as long as the argument is the same. I showed how this could be done automatically. There are however other cases when handling state between
function calls is necessary but cannot be done this simple way. One is handling state between recursive calls. In such a situation, the function is called only once from the user point of view, but
it is in fact called several times since it calls itself recursively. Most of these functions will not benefit internally from memoization. For example, the factorial function may be implemented
recursively, and for one call to f(n), there will be n calls to the function, but not twice with the same value. So this function would not benefit from internal memoization. On the contrary, the
Fibonacci function, if implemented according to its recursive definition, will call itself recursively a huge number of times will the same arguments.
Here is the standard definition of the Fibonacci function:
f(n) = f(n – 1) + f(n – 2)
This definition has a major problem: calculating f(n) implies evaluating n² times the function with different values. The result is that, in practice, it is impossible to use this definition for n >
50 because the time of execution increases exponentially. But since we are calculating more than n values, it is obvious that some values are calculated several times. So this definition would be a
good candidate for internal memoization.
Warning: Java is not a true recursive language, so using any recursive method will eventually blow the stack, unless we use TCO (Tail Call Optimization) as described in my previous article (Do it in
Java 8: recursive and corecursive Fibonacci). However, TCO can't be applied to the original Fibonacci definition because it is not tail recursive. So we will write an example working for n limited to
a few thousands.
The naive implementation
Here is how we might implement the original definition:
static BigInteger fib(BigInteger n) {
return n.equals(BigInteger.ZERO) || n.equals(BigInteger.ONE)
? n
: fib(n.subtract(BigInteger.ONE)).add(fib(n.subtract(BigInteger.ONE.add(BigInteger.ONE))));
Do not try this implementation for values greater that 50. On my machine, fib(50) takes 44 minutes to return!
Basic memoized version
To avoid computing several times the same values, we can use a map were computed values are stored. This way, for each value, we first look into the map to see if it has already been computed. If it
is present, we just retrieve it. Otherwise, we compute it, store it into the map and return it.
For this, we will use a special class called Memo. This is a normal HashMap that mimic the interface of functional map, which means that insertion of a new (key, value) pair returns a map, and
looking up a key returns an Optional:
public class Memo extends HashMap<BigInteger, BigInteger> {
public Optional<BigInteger> retrieve(BigInteger key) {
return Optional.ofNullable(super.get(key));
public Memo addEntry(BigInteger key, BigInteger value) {
super.put(key, value);
return this;
Note that this is not a true functional (immutable and persistent) map, but this is not a problem since it will not be shared.
We will also need a Tuple class that we define as:
public class Tuple<A, B> {
public final A _1;
public final B _2;
public Tuple(A a, B b) {
_1 = a;
_2 = b;
With this class, we can write the following basic implementation:
static BigInteger fibMemo1(BigInteger n) {
return fibMemo(n, new Memo().addEntry(BigInteger.ZERO, BigInteger.ZERO)
.addEntry(BigInteger.ONE, BigInteger.ONE))._1;
static Tuple<BigInteger, Memo> fibMemo(BigInteger n, Memo memo) {
return memo.retrieve(n).map(x -> new Tuple<>(x, memo)).orElseGet(() -> {
BigInteger x = fibMemo(n.subtract(BigInteger.ONE), memo)._1
.add(fibMemo(n.subtract(BigInteger.ONE).subtract(BigInteger.ONE), memo)._1);
return new Tuple<>(x, memo.addEntry(n, x));
This implementation works fine, provided values of n are not too big. For f(50), it returns in less that 1 millisecond, which should be compared to the 44 minutes of the naive version. Remark that we
do not have to test for terminal values (f(0) and f(1). These values are simply inserted into the map at start.
So, is there something we should make better? The main problem is that we have to handle the passing of the Memo map by hand. The signature of the fibMemo method is no longer fibMemo(BigInteger n),
but fibMemo(BigInteger n, Memo memo). Could we simplify this? We might think about using automatic memoization as described in my previous article (Do it in Java 8: Automatic memoization ). However,
this will not work:
static Function<BigInteger, BigInteger> fib = new Function<BigInteger, BigInteger>() {
public BigInteger apply(BigInteger n) {
return n.equals(BigInteger.ZERO) || n.equals(BigInteger.ONE)
? n
: this.apply(n.subtract(BigInteger.ONE)).add(this.apply(n.subtract(BigInteger.ONE.add(BigInteger.ONE))));
static Function<BigInteger, BigInteger> fibm = Memoizer.memoize(fib);
Beside the fact that we could not use lambdas because reference to this is not allowed, which may be worked around by using the original anonymous class syntax, the recursive call is made to the non
memoized function, so it does not make things better.
Using the State monad
In this example, each computed value is strictly evaluated by the fibMemo method, and this is what makes the memo parameter necessary. Instead of a method returning a value, what we would need is a
method returning a function that could be evaluated latter. This function would take a Memo as parameter, and this Memo instance would be necessary only at evaluation time. This is what the State
monad will do.
Java 8 does not provide the state monad, so we have to create it, but it is very simple. However, we first need an implementation of a list that is more functional that what Java offers. In a real
case, we would use a true immutable and persistent List. The one I have written is about 1 000 lines, so I can't show it here. Instead, we will use a dummy functional list, backed by a
java.util.ArrayList. Although this is less elegant, it does the same job:
public class List<T> {
private java.util.List<T> list = new ArrayList<>();
public static <T> List<T> empty() {
return new List<T>();
public static <T> List<T> apply(T... ta) {
List<T> result = new List<>();
for (T t : ta)
return result;
public List<T> cons(T t) {
List<T> result = new List<>();
return result;
public <U> U foldRight(U seed, Function<T, Function<U, U>> f) {
U result = seed;
for (int i = list.size() - 1; i >= 0; i--) {
result = f.apply(list.get(i)).apply(result);
return result;
public <U> List<U> map(Function<T, U> f) {
List<U> result = new List<>();
for (T t : list) {
return result;
public List<T> filter(Function<T, Boolean> f) {
List<T> result = new List<>();
for (T t : list) {
if (f.apply(t)) {
return result;
public Optional<T> findFirst() {
return list.size() == 0
? Optional.empty()
: Optional.of(list.get(0));
public String toString() {
StringBuilder s = new StringBuilder("[");
for (T t : list) {
s.append(t).append(", ");
return s.append("NIL]").toString();
The implementation is not functional, but the interface is! And although there are lots of missing capabilities, we have all we need.
Now, we can write the state monad. It is often called simply State but I prefer to call it StateMonad in order to avoid confusion between the state and the monad:
public class StateMonad<S, A> {
public final Function<S, StateTuple<A, S>> runState;
public StateMonad(Function<S, StateTuple<A, S>> runState) {
this.runState = runState;
public static <S, A> StateMonad<S, A> unit(A a) {
return new StateMonad<>(s -> new StateTuple<>(a, s));
public static <S> StateMonad<S, S> get() {
return new StateMonad<>(s -> new StateTuple<>(s, s));
public static <S, A> StateMonad<S, A> getState(Function<S, A> f) {
return new StateMonad<>(s -> new StateTuple<>(f.apply(s), s));
public static <S> StateMonad<S, Nothing> transition(Function<S, S> f) {
return new StateMonad<>(s -> new StateTuple<>(Nothing.instance, f.apply(s)));
public static <S, A> StateMonad<S, A> transition(Function<S, S> f, A value) {
return new StateMonad<>(s -> new StateTuple<>(value, f.apply(s)));
public static <S, A> StateMonad<S, List<A>> compose(List<StateMonad<S, A>> fs) {
return fs.foldRight(StateMonad.unit(List.<A>empty()), f -> acc -> f.map2(acc, a -> b -> b.cons(a)));
public <B> StateMonad<S, B> flatMap(Function<A, StateMonad<S, B>> f) {
return new StateMonad<>(s -> {
StateTuple<A, S> temp = runState.apply(s);
return f.apply(temp.value).runState.apply(temp.state);
public <B> StateMonad<S, B> map(Function<A, B> f) {
return flatMap(a -> StateMonad.unit(f.apply(a)));
public <B, C> StateMonad<S, C> map2(StateMonad<S, B> sb, Function<A, Function<B, C>> f) {
return flatMap(a -> sb.map(b -> f.apply(a).apply(b)));
public A eval(S s) {
return runState.apply(s).value;
This class is parameterized by two types: the value type A and the state type S. In our case, A will be BigInteger and S will be Memo.
This class holds a function from a state to a tuple (value, state). This function is hold in the runState field. This is similar to the value hold in the Optional monad.
To make it a monad, this class needs a unit method and a flatMap method. The unit method takes a value as parameter and returns a StateMonad. It could be implemented as a constructor. Here, it is a
factory method.
The flatMap method takes a function from A (a value) to StateMonad<S, B> and return a new StateMonad<S, B>. (In our case, A is the same as B.) The new type contains the new value and the new state
that result from the application of the function.
All other methods are convenience methods:
• map allows to bind a function from A to B instead of a function from A to StateMonad<S, B>. It is implemented in terms of flatMap and unit.
• eval allows easy retrieval of the value hold by the StateMonad.
• getState allows creating a StateMonad from a function S -> A.
• transition takes a function from state to state and a value and returns a new StateMonad holding the value and the state resulting from the application of the function. In other words, it allows
changing the state without changing the value.
There is also another transition method taking only a function and returning a StateMonad<S, Nothing>. Nothing is a special class:
public final class Nothing {
public static final Nothing instance = new Nothing();
private Nothing() {}
This class could be replaced by Void, to mean that we do not care about the type. However, Void is not supposed to be instantiated, and the only reference of type Void is normally null. The problem
is that null does not carry its type. We could instantiate a Void instance through introspection:
Constructor<Void> constructor;
constructor = Void.class.getDeclaredConstructor();
Void nothing = constructor.newInstance();
but this is really ugly, so we create a Nothing type with a single instance of it. This does the trick, although to be complete, Nothing should be able to replace any type (like null), which does not
seem to be possible in Java.
Using the StateMonad class, we can rewrite our program:
static BigInteger fibMemo2(BigInteger n) {
return fibMemo(n).eval(new Memo().addEntry(BigInteger.ZERO, BigInteger.ZERO).addEntry(BigInteger.ONE, BigInteger.ONE));
static StateMonad<Memo, BigInteger> fibMemo(BigInteger n) {
return StateMonad.getState((Memo m) -> m.retrieve(n))
.flatMap(u -> u.map(StateMonad::<Memo, BigInteger> unit).orElse(fibMemo(n.subtract(BigInteger.ONE))
.flatMap(x -> fibMemo(n.subtract(BigInteger.ONE).subtract(BigInteger.ONE))
.flatMap(z -> StateMonad.transition((Memo m) -> m.addEntry(n, z), z)))));
Now, the fibMemo method only takes a BigInteger as its parameter and returns a StateMonad, which means that when this method returns, nothing has been evaluated yet. The Memo doesn't even exist!
To get the result, we may call the eval method, passing it the Memo instance.
If you find this code difficult to understand, here is an exploded commented version using longer identifiers:
static StateMonad<Memo, BigInteger> fibMemo(BigInteger n) {
* Create a function of type Memo -> Optional<BigInteger> with a closure
* over the n parameter.
Function<Memo, Optional<BigInteger>> retrieveValueFromMapIfPresent = (Memo memoizationMap) -> memoizationMap.retrieve(n);
* Create a state from this function.
StateMonad<Memo, Optional<BigInteger>> initialState = StateMonad.getState(retrieveValueFromMapIfPresent);
* Create a function for converting the value (BigInteger) into a State
* Monad instance. This function will be bound to the Optional resulting
* from the lookup into the map to give the result if the value was found.
Function<BigInteger, StateMonad<Memo, BigInteger>> createStateFromValue = StateMonad::<Memo, BigInteger> unit;
* The value computation proper. This can't be easily decomposed because it
* make heavy use of closures. It first calls recursively fibMemo(n - 1),
* producing a StateMonad<Memo, BigInteger>. It then flatMaps it to a new
* recursive call to fibMemo(n - 2) (actually fibMemo(n - 1 - 1)) and get a
* new StateMonad<Memo, BigInteger> which is mapped to BigInteger addition
* with the preceding value (x). Then it flatMaps it again with the function
* y -> StateMonad.transition((Memo m) -> m.addEntry(n, z), z) which adds
* the two values and returns a new StateMonad with the computed value added
* to the map.
StateMonad<Memo, BigInteger> computedValue = fibMemo(n.subtract(BigInteger.ONE))
.flatMap(x -> fibMemo(n.subtract(BigInteger.ONE).subtract(BigInteger.ONE))
.flatMap(z -> StateMonad.transition((Memo m) -> m.addEntry(n, z), z)));
* Create a function taking an Optional<BigInteger> as its parameter and
* returning a state. This is the main function that returns the value in
* the Optional if it is present and compute it and put it into the map
* before returning it otherwise.
Function<Optional<BigInteger>, StateMonad<Memo, BigInteger>> computeFiboValueIfAbsentFromMap = u -> u.map(createStateFromValue).orElse(computedValue);
* Bind the computeFiboValueIfAbsentFromMap function to the initial State
* and return the result.
return initialState.flatMap(computeFiboValueIfAbsentFromMap);
The most important part is the following:
StateMonad<Memo, BigInteger> computedValue = fibMemo_(n.subtract(BigInteger.ONE))
.flatMap(x -> fibMemo_(n.subtract(BigInteger.ONE).subtract(BigInteger.ONE))
.flatMap(z -> StateMonad.transition((Memo m) -> m.addEntry(n, z), z)));
This kind of code is essential to functional programming, although it is sometimes replaced in other languages with “for comprehensions”. As Java 8 does not have for comprehensions we have to use
this form.
At this point, we have seen that using the state monad allows abstracting the handling of state. This technique can be used every time you have to handle state.
More uses of the state monad
The state monad may be used for many other cases were state must be maintained in a functional way. Most programs based upon maintaining state use a concept known as a State Machine. A state machine
is defined by an initial state and a series of inputs. Each input submitted to the state machine will produce a new state by applying one of several possible transitions based upon a list of
conditions concerning both the input and the actual state.
If we take the example of a bank account, the initial state would be the initial balance of the account. Possible transition would be deposit(amount) and withdraw(amount). The conditions would be
true for deposit and balance >= amount for withdraw.
Given the state monad that we have implemented above, we could write a parameterized state machine:
public class StateMachine<I, S> {
Function<I, StateMonad<S, Nothing>> function;
public StateMachine(List<Tuple<Condition<I, S>, Transition<I, S>>> transitions) {
function = i -> StateMonad.transition(m ->
Optional.of(new StateTuple<>(i, m)).flatMap((StateTuple<I, S> t) ->
transitions.filter((Tuple<Condition<I, S>, Transition<I, S>> x) ->
x._1.test(t)).findFirst().map((Tuple<Condition<I, S>, Transition<I, S>> y) ->
public StateMonad<S, S> process(List<I> inputs) {
List<StateMonad<S, Nothing>> a = inputs.map(function);
StateMonad<S, List<Nothing>> b = StateMonad.compose(a);
return b.flatMap(x -> StateMonad.get());
This machine uses a bunch of helper classes. First, the inputs are represented by an interface:
public interface Input {
boolean isDeposit();
boolean isWithdraw();
int getAmount();
There are two instances of inputs:
public class Deposit implements Input {
private final int amount;
public Deposit(int amount) {
this.amount = amount;
public boolean isDeposit() {
return true;
public boolean isWithdraw() {
return false;
public int getAmount() {
return this.amount;
public class Withdraw implements Input {
private final int amount;
public Withdraw(int amount) {
this.amount = amount;
public boolean isDeposit() {
return false;
public boolean isWithdraw() {
return true;
public int getAmount() {
return this.amount;
Then come two functional interfaces for conditions and transitions:
public interface Condition<I, S> extends Predicate<StateTuple<I, S>> {}
public interface Transition<I, S> extends Function<StateTuple<I, S>, S> {}
These act as type aliases in order to simplify the code. We could have used the predicate and the function directly. In the same manner, we use a StateTuple class instead of a normal tuple:
public class StateTuple<A, S> {
public final A value;
public final S state;
public StateTuple(A a, S s) {
value = Objects.requireNonNull(a);
state = Objects.requireNonNull(s);
This is exactly the same as an ordinary tuple with named members instead of numbered ones. Numbered members allows using the same class everywhere, but a specific class like this one make the code
easier to read as we will see.
The last utility class is Outcome, which represent the result returned by the state machine:
public class Outcome {
public final Integer account;
public final List<Either<Exception, Integer>> operations;
public Outcome(Integer account, List<Either<Exception, Integer>> operations) {
this.account = account;
this.operations = operations;
public String toString() {
return "(" + account.toString() + "," + operations.toString() + ")";
This again could be replaced with a Tuple<Integer, List<Either<Exception, Integer>>>, but using named parameters make the code easier to read. (In some functional languages, we could use type aliases
for this.)
Here, we use an Either class, which is another kind of monad that Java does not offer. I will not show the complete class, but only the parts that are useful for this example:
public interface Either<A, B> {
boolean isLeft();
boolean isRight();
A getLeft();
B getRight();
static <A, B> Either<A, B> right(B value) {
return new Right<>(value);
static <A, B> Either<A, B> left(A value) {
return new Left<>(value);
public class Left<A, B> implements Either<A, B> {
private final A left;
private Left(A left) {
this.left = left;
public boolean isLeft() {
return true;
public boolean isRight() {
return false;
public A getLeft() {
return this.left;
public B getRight() {
throw new IllegalStateException("getRight() called on Left value");
public String toString() {
return left.toString();
public class Right<A, B> implements Either<A, B> {
private final B right;
private Right(B right) {
this.right = right;
public boolean isLeft() {
return false;
public boolean isRight() {
return true;
public A getLeft() {
throw new IllegalStateException("getLeft() called on Right value");
public B getRight() {
return this.right;
public String toString() {
return right.toString();
This implementation is missing a flatMap method, but we will not need it. The Either class is somewhat like the Optional Java class in that it may be used to represent the result of an evaluation
that may return a value or something else like an exception, an error message or whatever. What is important is that it can hold one of two things of different types.
We now have all we need to use our state machine:
public class Account {
public static StateMachine<Input, Outcome> createMachine() {
Condition<Input, Outcome> predicate1 = t -> t.value.isDeposit();
Transition<Input, Outcome> transition1 = t -> new Outcome(t.state.account + t.value.getAmount(), t.state.operations.cons(Either.right(t.value.getAmount())));
Condition<Input, Outcome> predicate2 = t -> t.value.isWithdraw() && t.state.account >= t.value.getAmount();
Transition<Input, Outcome> transition2 = t -> new Outcome(t.state.account - t.value.getAmount(), t.state.operations.cons(Either.right(- t.value.getAmount())));
Condition<Input, Outcome> predicate3 = t -> true;
Transition<Input, Outcome> transition3 = t -> new Outcome(t.state.account, t.state.operations.cons(Either.left(new IllegalStateException(String.format("Can't withdraw %s because balance is only %s", t.value.getAmount(), t.state.account)))));
List<Tuple<Condition<Input, Outcome>, Transition<Input, Outcome>>> transitions = List.apply(
new Tuple<>(predicate1, transition1),
new Tuple<>(predicate2, transition2),
new Tuple<>(predicate3, transition3));
return new StateMachine<>(transitions);
This could not be simpler. We just define each possible condition and the corresponding transition, and then build a list of tuples (Condition, Transition) that is used to instantiate the state
machine. There are however to rules that must be enforced:
• Conditions must be put in the right order, with the more specific first and the more general last.
• We must be careful to be sure to match all possible cases. Otherwise, we will get an exception.
At this stage, nothing has been evaluated. We did not even use the initial state!
To run the state machine, we must create a list of inputs and feed it in the machine, for example:
List<Input> inputs = List.apply(
new Deposit(100),
new Withdraw(50),
new Withdraw(150),
new Deposit(200),
new Withdraw(150));
StateMonad<Outcome, Outcome> = Account.createMachine().process(inputs);
Again, nothing has been evaluated yet. To get the result, we just evaluate the result, using an initial state:
Outcome outcome = state.eval(new Outcome(0, List.empty()))
If we run the program with the list above, and call toString() on the resulting outcome (we can't do more useful things since the Either class is so minimal!) we get the following result:
(100,[-150, 200, java.lang.IllegalStateException: Can't withdraw 150 because balance is only 50, -50, 100, NIL])
This is a tuple of the resulting balance for the account (100) and the list of operations that have been carried on. We can see that successful operations are represented by a signed integer, and
failed operations are represented by an error message.
This of course is a very minimal example, and as usual, one may think it would be much easier to do it the imperative way. However, think of a more complex example, like a text parser. All there is
to do to adapt the state machine is to define the state representation (the Outcome class), define the possible inputs and create the list of (Condition,Transition). Going the functional way does not
make the whole thing simpler. However, it allows abstracting the implementation of the state machine from the requirements. The only thing we have to do to create a new state machine is to write the
new requirements!
Machine Monad (functional programming) Java (programming language) Implementation
Opinions expressed by DZone contributors are their own. | {"url":"https://dzone.com/articles/do-it-in-java-8-state-monad","timestamp":"2024-11-09T12:42:08Z","content_type":"text/html","content_length":"153416","record_id":"<urn:uuid:22532a35-b161-4bfd-aaf9-f26524a308cc>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00361.warc.gz"} |
Solve the Number Bonds Worksheet - 15 Worksheets.com
Solve the Number Bonds
Worksheet Description
This worksheet presents a series of number bond exercises, each with the number 15 at the center, linked to one known number and one unknown number. Students are tasked with determining the missing
number in each bond, using their knowledge of addition and subtraction. The worksheet comprises 12 individual problems, all of which follow the same format to allow for repetition and practice. The
aim is for students to fill in the empty circle with the correct number that, when added to the given number, equals 15.
The worksheet is designed to teach students about number bonds, an essential concept in early math education that focuses on the relationships between numbers. It reinforces the idea that two smaller
numbers can be added to make a larger number, specifically the sum of 15 in this case. This practice helps students to become fluent in addition and subtraction facts, an important skill for more
advanced math. Moreover, it encourages mental math capabilities, as students work out the solutions without the use of calculators or number lines. | {"url":"https://15worksheets.com/worksheet/decomposing-numbers-3/","timestamp":"2024-11-05T14:09:37Z","content_type":"text/html","content_length":"108939","record_id":"<urn:uuid:70622d80-2e54-4329-8b59-beb5fb3fc6cd>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00434.warc.gz"} |
ideal capacitor element energy storage formula - Suppliers/Manufacturers
This physics video tutorial explains how to calculate the energy stored in a capacitor using three different formulas. It also explains how to calculate the... AP Physics 2: Algebra-Based
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About ideal capacitor element energy storage formula - Suppliers/Manufacturers
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محتويات ذات صلة | {"url":"https://kennel-blackzorro.nl/Mon-27-May-2024-27442.html","timestamp":"2024-11-10T01:42:19Z","content_type":"text/html","content_length":"51163","record_id":"<urn:uuid:afed90f2-778c-436e-b3f3-55314483dcfc>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00244.warc.gz"} |
Leagues (statute) to Poles Converter
Enter Leagues (statute)
β Switch toPoles to Leagues (statute) Converter
How to use this Leagues (statute) to Poles Converter π €
Follow these steps to convert given length from the units of Leagues (statute) to the units of Poles.
1. Enter the input Leagues (statute) value in the text field.
2. The calculator converts the given Leagues (statute) into Poles in realtime β using the conversion formula, and displays under the Poles label. You do not need to click any button. If the input
changes, Poles value is re-calculated, just like that.
3. You may copy the resulting Poles value using the Copy button.
4. To view a detailed step by step calculation of the conversion, click on the View Calculation button.
5. You can also reset the input by clicking on button present below the input field.
What is the Formula to convert Leagues (statute) to Poles?
The formula to convert given length from Leagues (statute) to Poles is:
Length[(Poles)] = Length[(Leagues (statute))] / 0.0010416645004435415
Substitute the given value of length in leagues (statute), i.e., Length[(Leagues (statute))] in the above formula and simplify the right-hand side value. The resulting value is the length in poles,
i.e., Length[(Poles)].
Calculation will be done after you enter a valid input.
Consider that an ancient road stretches for 25 statute leagues.
Convert this distance from statute leagues to Poles.
The length in leagues (statute) is:
Length[(Leagues (statute))] = 25
The formula to convert length from leagues (statute) to poles is:
Length[(Poles)] = Length[(Leagues (statute))] / 0.0010416645004435415
Substitute given weight Length[(Leagues (statute))] = 25 in the above formula.
Length[(Poles)] = 25 / 0.0010416645004435415
Length[(Poles)] = 24000.0499
Final Answer:
Therefore, 25 st.league is equal to 24000.0499 pole.
The length is 24000.0499 pole, in poles.
Consider that a historical expedition covered 50 statute leagues.
Convert this distance from statute leagues to Poles.
The length in leagues (statute) is:
Length[(Leagues (statute))] = 50
The formula to convert length from leagues (statute) to poles is:
Length[(Poles)] = Length[(Leagues (statute))] / 0.0010416645004435415
Substitute given weight Length[(Leagues (statute))] = 50 in the above formula.
Length[(Poles)] = 50 / 0.0010416645004435415
Length[(Poles)] = 48000.0998
Final Answer:
Therefore, 50 st.league is equal to 48000.0998 pole.
The length is 48000.0998 pole, in poles.
Leagues (statute) to Poles Conversion Table
The following table gives some of the most used conversions from Leagues (statute) to Poles.
Leagues (statute) (st.league) Poles (pole)
0 st.league 0 pole
1 st.league 960.002 pole
2 st.league 1920.004 pole
3 st.league 2880.006 pole
4 st.league 3840.008 pole
5 st.league 4800.01 pole
6 st.league 5760.012 pole
7 st.league 6720.014 pole
8 st.league 7680.016 pole
9 st.league 8640.018 pole
10 st.league 9600.02 pole
20 st.league 19200.0399 pole
50 st.league 48000.0998 pole
100 st.league 96000.1996 pole
1000 st.league 960001.9964 pole
10000 st.league 9600019.964 pole
100000 st.league 96000199.6395 pole
Leagues (statute)
A league (statute) is a unit of length used to measure distances. One statute league is equivalent to 3 miles or approximately 4.828 kilometers.
The statute league is defined as three miles, and it was historically used in various English-speaking countries for measuring distances, especially in land navigation and mapping.
Statute leagues are less commonly used today but may still appear in historical documents, literature, and some regional contexts. They provide a way to express distances in a scale larger than miles
but smaller than other large units like leagues nautical.
A pole is a unit of length used primarily in land measurement and surveying. One pole is equivalent to 16.5 feet or approximately 5.0292 meters.
The pole is defined as 16.5 feet, which is the same length as a rod or a perch, and is used in various practical applications such as land measurement and construction.
Poles are used in land surveying, property measurement, and agricultural contexts. The unit provides a practical measurement for shorter distances and has historical significance in land measurement
Frequently Asked Questions (FAQs)
1. What is the formula for converting Leagues (statute) to Poles in Length?
The formula to convert Leagues (statute) to Poles in Length is:
Leagues (statute) / 0.0010416645004435415
2. Is this tool free or paid?
This Length conversion tool, which converts Leagues (statute) to Poles, is completely free to use.
3. How do I convert Length from Leagues (statute) to Poles?
To convert Length from Leagues (statute) to Poles, you can use the following formula:
Leagues (statute) / 0.0010416645004435415
For example, if you have a value in Leagues (statute), you substitute that value in place of Leagues (statute) in the above formula, and solve the mathematical expression to get the equivalent value
in Poles. | {"url":"https://convertonline.org/unit/?convert=leagues_statute-poles","timestamp":"2024-11-10T16:14:58Z","content_type":"text/html","content_length":"91368","record_id":"<urn:uuid:6e14f721-d222-4221-a675-91040109c9c5>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00824.warc.gz"} |
ict 9 and 10 in science question
ICT – 7 (marks )-class 9
Group 'A' (VSQ) (1×1=1)
1. Which of the following is an example of hacking?
a) Cyber law
b) Cyber Crime
c) Cyber Security
d) Cyberbullying
Group 'B' (SQ) (1×2=2)
1. What is cyber bullying? Write 1 effective methods to prevent your friend from being bullied online.
Group 'D' (LQ) (1×4=4)
1. What is Bloom’s Taxonomy? Make a clear pyramid showing Bloom’s Taxonomy.
ICT – 7 (marks )-class 10
Group 'A' (VSQ) (1×1=1)
1. What is the short cut key to shut down the computer?
a) CTRL+S
b) CTRL+O
c) ALT+F4
d) ALT+F2
Group 'B' (SQ) (1×2=2)
1. What is Info graphics? Give some examples.
Group 'D' (LQ) (1×4=4)
1. What is telecommunication? How do you keep your friends and family safe online? Write in points. | {"url":"https://studylib.net/doc/26282710/ict-9-and-10-in-science-question","timestamp":"2024-11-09T16:37:39Z","content_type":"text/html","content_length":"50238","record_id":"<urn:uuid:a4e58f26-12be-4fef-8785-f93a621906d8>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00324.warc.gz"} |
Lesson 4: How to Estimate Animal Size and Numbers at a Distance
Lesson 4: How to Estimate Animal Size and Numbers at a Distance
Teach students about size-distance relationships and making group estimates.
Nature observation is an inexact science. Unpredictable events that can occur, such as a flutter of butterflies passing overhead or a herd of deer crossing your path, make it difficult to collect
perfect data. Therefore, nature researchers often use estimations.
In this lesson, you will teach students how to make good estimations of animal size and numbers. This lesson brings in math and art concepts to help students understand spatial relationships and
approximations. Start with size-distance relationships, and end with the grid system.
Lesson Objectives and Materials
Estimating Size
Teach students how to estimate size at a distance. This lesson is broken into three parts. The first part introduces techniques for measuring size; the last parts expand on the concept through an
outdoor activity and perspective drawing. Be prepared to take digital pictures throughout the lesson.
Here are the steps to follow:
Part 1: Estimating Size at a Distance
Part 2: Practice Outdoors
Pre-Lesson Preparation: Prepare three wooden stakes by attaching to each an identical, life-size picture of an animal. The great horned owl works well.
1. Take your students outside, and place the stakes at different distances -- approximately ten, fifty, and one hundred feet. Then ask students to
2. Take a digital picture of the stakes for later use.
3. Gather the stakes to show the students that the photos are the same size.
4. Ask students to discuss their observations.
Part 3: Practice with Perspective Drawing
1. Explain the basic concepts of perspective drawing, describing how one can represent objects at a distance on paper by making objects appear smaller and closer together as they near the vanishing
point. Explain the basic meaning of "vanishing point" and the "horizon line," using pictures and examples for emphasis.
2. Upload to a computer the photos you took previously of the students and stakes. Open each image separately in a program that allows you to edit the image using lines (such as Adobe PhotoShop).
Credit: Damien Scogin
Click here to enlarge perspective drawing.
3. Demonstrate how to draw using one-point linear perspective on the computer.
4. Ask students to practice perspective drawing in their field journals or on handouts.
Practical and Assessment
Test your students' understanding of making estimates and drawing using perspective. Show younger students different pairs of animal pictures in different sizes and ask them to speculate which one of
the two animals in the pair is farther away. Ask older students to draw their nature animals using perspective -- creating the horizon line, the vanishing point, and three different sketches of their
animals. (See image below.)
How'd your students do? Here are some ways to assess your students' comprehension of the material, reflective of grade level. Assess students by point scale or qualitatively.
Meets standard: Student was able to meet the above standards with only one error in size-distance relationship interpretation.
Below standard: Student made more than one mistake in size-distance relationship interpretation.
Estimating Group Numbers
Teach students how to estimate the number of animals in a group. This lesson is split into two parts. The first part gives a general introduction to group estimates, using an ant mound as example;
the second part teaches the grid system.
Here are the steps to follow:
Part 1: Introduction to Group Estimates
Pre-Lesson Preparation: Make copies of a picture of an ant mound. Also, create a large class chart that has at least four columns. Mount the chart on the wall for recording data throughout the
Refer to the NM data-collection form, and introduce the sections that relate to number and estimates. Explain that scientists need good estimates to determine the population of different species
across the country.
1. Show a picture of an ant mound and ask
2. Ask each student or pair of students to mark their estimates on the class chart.
3. Record on the class chart the different methods students come up with for determining the number of ants.
4. Pass out pictures of an ant mound. Ask students to choose a strategy for estimating the number of ants in the picture. Some students might count by ones; some students might count groups of ants.
5. Record each student's or pair of students' estimates in another column of the class chart. Ask students what they notice about the two groups of estimates. The second group of numbers should be
more similar than the first estimates.
Part 2: The Grid System
1. Still using the ant-mound example, ask students to speculate how to determine the number of ants without counting them or without a picture (for example, if they encountered an ant mound
2. Overlay a grid onto the picture of the ant mound. Ask students how they would use the grid to determine the number of ants.
3. Explain how to use the grid system, which is the method scientists use when estimating large numbers of animals in nature: Count the number of animals in one box and multiply by the total number
of boxes.
4. Practice the grid system using other examples, such as pictures of birds in a tree, blood cells, or gum balls. (See example below.)
5. Challenge students to imagine the grid in their heads without looking at it. Explain that scientists use an imaginary grid to recognize the number of animals traveling together in nature.
6. Ask students to practice using the imaginary grid on different objects, such as cookies on a cookie sheet or pens on a desk.
7. Take the lesson outside and practice using the imaginary grid for trees in the park, birds in a flock, plants in a flower bed, and rocks in a pile.
Practical and Assessment
Test your students' ability to estimate the number of animals in groups. Show different groups of objects to students. Then ask them to guess the number of objects from the picture or diagram.
How'd your students do? Here are some ways to assess your students' comprehension of the material, reflective of grade level.
Teacher Tips
Keep previous lessons fresh in students' minds. Ask students to elaborate on the characteristics and behavior of the animals you use as examples.
We use these terms throughout this and other NatureMapping lessons. | {"url":"https://www.edutopia.org/naturemapping-lesson-distance-size-estimates","timestamp":"2024-11-15T00:45:26Z","content_type":"text/html","content_length":"118984","record_id":"<urn:uuid:e022d7ab-0735-43ad-aaac-b40b83fb9c8b>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00251.warc.gz"} |
Decimeters to Leagues Converter
β Switch toLeagues to Decimeters Converter
How to use this Decimeters to Leagues Converter π €
Follow these steps to convert given length from the units of Decimeters to the units of Leagues.
1. Enter the input Decimeters value in the text field.
2. The calculator converts the given Decimeters into Leagues in realtime β using the conversion formula, and displays under the Leagues label. You do not need to click any button. If the input
changes, Leagues value is re-calculated, just like that.
3. You may copy the resulting Leagues value using the Copy button.
4. To view a detailed step by step calculation of the conversion, click on the View Calculation button.
5. You can also reset the input by clicking on button present below the input field.
What is the Formula to convert Decimeters to Leagues?
The formula to convert given length from Decimeters to Leagues is:
Length[(Leagues)] = Length[(Decimeters)] / 48280.32716680902
Substitute the given value of length in decimeters, i.e., Length[(Decimeters)] in the above formula and simplify the right-hand side value. The resulting value is the length in leagues, i.e., Length
Calculation will be done after you enter a valid input.
Consider that a luxury watch has a case diameter of 5 decimeters.
Convert this diameter from decimeters to Leagues.
The length in decimeters is:
Length[(Decimeters)] = 5
The formula to convert length from decimeters to leagues is:
Length[(Leagues)] = Length[(Decimeters)] / 48280.32716680902
Substitute given weight Length[(Decimeters)] = 5 in the above formula.
Length[(Leagues)] = 5 / 48280.32716680902
Length[(Leagues)] = 0.00010356185
Final Answer:
Therefore, 5 dm is equal to 0.00010356185 lea.
The length is 0.00010356185 lea, in leagues.
Consider that a premium sound system's speaker has a woofer diameter of 3 decimeters.
Convert this diameter from decimeters to Leagues.
The length in decimeters is:
Length[(Decimeters)] = 3
The formula to convert length from decimeters to leagues is:
Length[(Leagues)] = Length[(Decimeters)] / 48280.32716680902
Substitute given weight Length[(Decimeters)] = 3 in the above formula.
Length[(Leagues)] = 3 / 48280.32716680902
Length[(Leagues)] = 0.00006213711
Final Answer:
Therefore, 3 dm is equal to 0.00006213711 lea.
The length is 0.00006213711 lea, in leagues.
Decimeters to Leagues Conversion Table
The following table gives some of the most used conversions from Decimeters to Leagues.
Decimeters (dm) Leagues (lea)
0 dm 0 lea
1 dm 0.00002071237 lea
2 dm 0.00004142474 lea
3 dm 0.00006213711 lea
4 dm 0.00008284948 lea
5 dm 0.00010356185 lea
6 dm 0.00012427422 lea
7 dm 0.00014498659 lea
8 dm 0.00016569896 lea
9 dm 0.00018641133 lea
10 dm 0.0002071237 lea
20 dm 0.0004142474 lea
50 dm 0.0010356185 lea
100 dm 0.002071237 lea
1000 dm 0.02071237 lea
10000 dm 0.2071 lea
100000 dm 2.0712 lea
A decimeter (dm) is a unit of length in the International System of Units (SI). One decimeter is equivalent to 0.1 meters or approximately 3.937 inches.
The decimeter is defined as one-tenth of a meter, making it a convenient measurement for intermediate lengths.
Decimeters are used worldwide to measure length and distance in various fields, including science, engineering, and everyday life. They provide a useful scale for measurements that are larger than
centimeters but smaller than meters, and are commonly used in educational settings and certain industries.
A league is a unit of length that was traditionally used in Europe and Latin America. One league is typically defined as three miles or approximately 4.83 kilometers.
Historically, the league varied in length from one region to another. It was originally based on the distance a person could walk in an hour.
Today, the league is mostly obsolete and is no longer used in modern measurements. It remains as a reference in literature and historical texts.
Frequently Asked Questions (FAQs)
1. What is the formula for converting Decimeters to Leagues in Length?
The formula to convert Decimeters to Leagues in Length is:
Decimeters / 48280.32716680902
2. Is this tool free or paid?
This Length conversion tool, which converts Decimeters to Leagues, is completely free to use.
3. How do I convert Length from Decimeters to Leagues?
To convert Length from Decimeters to Leagues, you can use the following formula:
Decimeters / 48280.32716680902
For example, if you have a value in Decimeters, you substitute that value in place of Decimeters in the above formula, and solve the mathematical expression to get the equivalent value in Leagues. | {"url":"https://convertonline.org/unit/?convert=decimeters-leagues","timestamp":"2024-11-02T18:04:42Z","content_type":"text/html","content_length":"90963","record_id":"<urn:uuid:7205b9bc-e2fa-4abb-b17f-1824430708d9>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00769.warc.gz"} |
Logic number puzzle: Kendoku (KenKen, CalcuDoku).
│Download Kendoku desktop application (Windows 64bit) │Download Kendoku source code (Lazarus/Free Pascal)│
Description: Puzzle game with numbers to be placed onto a grid. The rules of the game are as follows:
1. In each field of the NxN grid, write numbers from 1 to N.
2. In each row and each column (but not in each area), each number must be written exactly once.
3. With the numbers of each area, the given result must be obtained by applying the given arithmetic operation.
4. If no operation is given, any of addition, subtraction, multiplication or division may be used.
The game is a typical logic game: In order to find the correct solution of the puzzle, you have to explore the arithmetic operations and area results given and the numbers already written, and by
logical thinking (concluding from what you know, that a given field must be a given number, or on the contrary, can't be a given number), determine, which fields have to contain which numbers.
The game is played exclusively with the mouse: Click the field, where you want to write a number, then click one of the number buttons to write the corresponding number or Clear to remove the number
from the actually selected field.
Note: Kendoku may be considered as an extended combination of Sumdoku (area sums) and Factors (area products).
To do: The fourth rule isn't implemented in the actual version of the puzzle (there is an arithmetic operation given for each area).
Free Pascal features: Changing the color of shapes and the caption of static-texts during runtime. Using buttons to insert characters into a given control's caption. Two-dimensional arrays, arrays of
records (classic Pascal).
If you like this application, please, support me and this website by signing my guestbook. | {"url":"https://www.streetinfo.lu/computing/lazarus/doc/Kendoku.html","timestamp":"2024-11-06T21:16:22Z","content_type":"text/html","content_length":"9845","record_id":"<urn:uuid:fa56adab-60c1-494d-b726-bf154a9b4a2e>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00465.warc.gz"} |
Back EMF in a Simple DC Motor - HSC Physics
Back EMF in a Simple DC Motor
This is part of the HSC Physics syllabus under the topic Applications of the Motor Effect.
HSC Physics Syllabus
• investigate the operation of a simple DC motor to analyse:
- the functions of its components
- production of a torque `\tau = nIAB_(_|_) sin \theta`
- effects of back emf (ACSPH108)
What is Back EMF?
This video investigates the operation of a simple DC motor to analyse the effects of back emf.
How Does a Simple DC Motor Produces Back EMF?
As the armature rotates within the external magnetic field, the amount of magnetic flux passing through the coil changes. This is because whilst the field strength is constant, the angle between the
surface’s normal (area vector) and the flux lines changes. This causes the coil to experience an induced emf according to Faraday's Law.
Lenz's law states that the current associated with this induced emf will produce a magnetic field that opposes the external magnetic field. Thus, the induced emf’s direction is always opposite to the
battery's supply voltage This is why this emf is called back emf.
As such, the net voltage of a DC motor can be described by:
$$V_{net} = V_{supply} - \varepsilon_{back}$$
As the coil's angular velocity increases, the rate of change in flux increases, which increases the back emf as `\varepsilon = -N (∆\Phi)/(∆t)`. This means that the net voltage will decrease, causing
the current to also decrease, according to Ohm's Law (`V = IR`). Since `\tau = nIAB \sin \theta`, a decreasing current (I) causes torque to decrease.
Eventually, the back emf will equal the supply voltage and the net voltage and current will equal 0. At this point, the coil has reached it's maximum rotational velocity since torque will be 0 N m.
This is why back emf is said to limit the angular speed of a DC motor.
The graph below illustrates this phenomenon. The vertical axis is the angular velocity and the horizontal axis is time.
Practice Question 1
(a) Describe the components that make up a DC motor (3 marks)
(b) Explain how the components you described in (a) achieve the function of a DC motor (2 marks)
Practice Question 2
A rectangular wire loop is connected to a DC power supply. Side X of the loop is placed next to a magnet. The loop is free to rotate about a pivot.
When the power is switched on, a current of 20 A is supplied to the loop. To prevent rotation, a mass of 40 g can be attached to either side X or side Y of the loop.
(a) On which side of the loop should the mass be attached to prevent rotation? (1 mark)
(b) Calculate the torque provided by the 40 g mass. (2 marks)
(c) Calculate the magnetic field strength around side X. (3 marks)
Previous section: Parallel vs Radial Magnets
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Horizontal and Vertical
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The transverse waves considered up to now represent string vibration only in a single two-dimensional plane. One such plane can be chosen as being perpendicular to the top plate of a stringed musical
instrument. We will call this the vertical plane of polarization for transverse waves on a string (or simply the vertical component of the transverse vibration). To more fully model a real vibrating
string, we also need to include transverse waves in the i.e., a horizontal plane of polarization (or horizontal component of vibration). Any polarization for transverse traveling waves can be
represented as a linear combination of horizontal and vertical polarizations, and general transverse string vibration in 3D can be expressed as a linear superposition of vibration in any two distinct
Figure 6.19: Digital waveguide model of a rigidly terminated string vibrating transversally in three-dimensional space (two uncoupled planes of vibration).
If string terminations were perfectly rigid, the horizontal polarization would be largely independent of the vertical polarization, and an accurate model would consist of two identical, uncoupled,
filtered delay loops (FDL), as depicted in Fig.6.19. One FDL models vertical force waves while the other models horizontal force waves. This model neglects the small degree of nonlinear coupling
between horizontal and vertical traveling waves along the length of the string--valid when the string slope is much less than unity (see §B.6).
Note that the model for two orthogonal planes of vibration on a single string is identical to that for a single plane of vibration on two different strings.
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Adding fractions with tape diagrams and area model
This is the third of a multi-part series of posts sharing ideas for how to teach fraction concepts and operations through the use of visual representations. There is tons of evidence that students
learn math better when the math is accompanied by visuals, so let’s dig into fractions…
Suppose we want to add \(\frac{2}{5}\) and \(\frac{2}{3}\) and represent it with a tape diagram.
Modeling both \(\frac{2}{5}\) and \(\frac{2}{3}\) with tapes, we see that \(\frac{2}{5} +\frac{2}{3}\) cannot be added together because they do not yet have equal sized pieces.
We can cut the pieces into identical sizes by cutting each fifth into 3 pieces and each third into 5 pieces. This results in each tape being cut into 15 pieces…our common denominator!
Once the equal sizes are created, \(\frac{2}{5} +\frac{2}{3}\) now becomes
\(\frac{6}{15} +\frac{10}{15}\), which equals \(\frac{16}{15}\).
Since \(\frac{16}{15}\) is an improper fraction, it can be rewritten as the mixed number \(1\frac{1}{15}\).
We can see that \(1\frac{1}{15}\) makes sense because we can move pieces from one tape to to the other tape to create one full tape (1 whole).
Numerically, we would write the entire process like this…
What about using the area model? Let’s solve \(\frac{2}{3} + \frac{3}{4}\).
First, let’s model the two fractions. But…you will notice that we made vertical slices to model \(\frac{2}{3}\) and we used horizontal slices to model \(\frac{3}{4}\). This is intentional!
Then we use horizontal lines to cut each of the thirds into four pieces. We use vertical lines to cut each of the fourths into three pieces. Now both rectangles have been cut into 12ths. Also, we see
that \(\frac{2}{3} + \frac{3}{4}\) has been renamed as \(\frac{8}{12} + \frac{9}{12}\)
The final answer is \(1\frac{5}{12}\), which makes sense since we can move some of the green pieces to the left to create 1 whole and 5 green pieces left over.
Here is a video using tape diagrams for addition:
Here is a video using area model for addition:
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Does Mathway do Word Problems: Mathway Alternatives - GradesHQDoes Mathway do Word Problems: Mathway Alternatives - GradesHQ
Does Mathway do Word Problems: Mathway Alternatives
Many students have a degree of loathing toward maths. They hate algebra, equations, and anything math-related. And this adversely affects their grades because math problems are virtually in every
course. However, with the advent of technology, everything is becoming simpler.
You can turn around your grades thanks to several math apps on the market. And among these apps is Mathway, a highly user-intuitive app. But what exactly does it do? Does Mathway do word problems?
What can it solve, and how does it work?
This article explores Mathway and if it can solve word problems. Take a look.
Does Mathway do Word Problems?
Mathway is essentially an equation solver and math problem calculator app. Many consider it the number one math solver tool, and it answers your questions on the go without any hassle. You can use it
to solve one or multi-step equations and see step-by-step explanations on how to complete the problem yourself.
New Service Alert!!!
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Currently, Mathway does not do word problems. It is limited to solving integer workings, arithmetic, factors, and roots. However, other useful apps can help you with your word problems. Similarly,
Mathway’s developers are working round the clock to add a word problem-solving feature to help more students, parents, and teachers.
Also see: Is mathway worth it?
Mathway has an advanced graphing calculator and a wide range of mathematical tools, calculators, and scientific functions that you can use to calculate anything from complex to algebraic equations.
It currently supports two ways of solving math problems. You can either type onto the app or take a picture using your phone.
Apart from using it to solve problems, you can use it to check answers to problems you’ve done on paper. You can also use it to help you work out complex problems.
Is there a Math App that Solves Word Problems?
The market is developing, and it is not surprising to find a math app that can solve word problems. Several math apps are designed to solve word problems, despite many teachers desiring students to
work out math problems by hand. This is because working by hand helps students to understand the concepts behind the numbers better. Nevertheless, these math apps can go a long way in helping
students grasp the content more broadly if appropriately used.
However, if you’re looking for something that offers word solutions apart from what they are primarily made for, the market is full of them. The apps are considered two-in-one and go a long way in
helping students and teachers better understand concepts and make work easier.
Mathway Alternatives
Mathway is limited in the math problems it will solve. However, there are plenty of alternatives for anyone looking beyond Mathway.
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Among these are Microsoft Math Solver and Photomath, both of which have free and paid versions. Here is a look at these alternatives, what they provide, and the features that make them a top choice
for many.
Microsoft Math Solver
Ranked the second-best alternative is the Microsoft Math Solver. This app is more like your tutor due to its wide range of capabilities. As a learning aid, it provides you with three ways to solve
your math problems.
You can choose to draw, type, or scan the problem. However, the drawing option is not well-defined, but the developers continue to refine it to give you a great user experience. Some of its great
features are
• Provides well-detailed steps as part of the answers
• It has video tutorials to help you get started easily
• Can solve a wide range of problems
Photomath is one of the best math-solving apps. It helps you improve your math mastery by providing you with how to solve math problems. All you need is to open the app, use your camera, upload, and
the answers will be provided.
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Other key features of Photomath include
• Provides work details to help you grasp and understand the concepts
• It has an intuitive design which you can navigate and learn without much help
• It has all answers for a wide range of math word problems
• It supports both Android and iOS devices
Mathway is an excellent option if you are looking for the best tool to do math problems. Its features are great, and it offers answers to many math problems. It also gives you the feeling of having a
human tutor beside you.
The answer to the ‘does Mathway do word problems’ question is in using the discussed alternatives. However, Mathway is a great tool, and you can get help from their customer support when stuck.
Additionally, the other options can still get your math problem solved. | {"url":"https://gradeshq.com/mathway-word-problems/","timestamp":"2024-11-07T13:03:56Z","content_type":"text/html","content_length":"83278","record_id":"<urn:uuid:1dec5528-b9cf-4d38-b4ff-19b64c6b6b88>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00741.warc.gz"} |
Calculating the Value of Your Reward Points - Money We Have
Calculating the Value of Your Reward Points
**This post may contain affiliate links. I may be compensated if you use them.
Calculating the value of your reward points is essential when travel hacking. Since every flight and hotel you book has a different price, the value you get when redeeming your points will change. In
an ideal world, you’ll always be looking to maximize the value of your points, but it’s not always that simple.
Loyalty programs make it easy to redeem your points and miles, but they don’t always spell out the exact value you’re getting. As a consumer, you need to establish a base value for your points. Once
you have that, you can easily compare different redemptions to see if what you’re getting is worth it or not.
Here’s how to calculate the value of your points and what numbers you should use for the base value of one point.
How to calculate the value of your points
For many people, calculating the value of your reward points is challenging. They don’t understand the math behind it and get confused when taxes or additional fees are factored in.
Regardless of what loyalty program you’re part of, you need to figure out the value of one point or mile. Fortunately, the formula to calculate the value per point is the same for each loyalty
(Cash value of redemption – taxes paid on redemption) X 100 / points required = Cost per point
Remember that you would need to subtract any taxes or fees you need to pay out of pocket from the cash value before you make your calculations. For example, let’s say you want to book a round-trip
ticket from Ottawa to Vancouver. The cash price is $600. However, booking via Aeroplan will cost you 24,000 Aeroplan points + $75 in taxes. That means the cash value of the redemption is $525, so
your formula would be as follows:
$525 X 100 / 24,000 = 2.19 cents per point.
Since the value of this redemption is worth more than the established base value of 2 cents per point for Aeroplan, it’s worth redeeming your points.
Cents per point (or cpp) is the most common way to value redemption from points and miles.
The opportunity cost of redeeming points
Redeeming points for free flights, hotel stays or gift cards is awesome since everyone loves free stuff! What’s interesting is that many people view loyalty points as Monopoly money since the points
they earned were “free.” Any redemption they get is considered a bonus. However, since loyalty points have a real value, you should treat them similarly to how you treat real money.
For example, you can redeem 24,000 Aeroplan points for a $170 gift card. But for the same number of points, you could have used them on that $525 Ottawa to Vancouver flight mentioned above. That’s a
threefold difference in value!
As you can see, not all redemptions are equal in value, so you need to check the value of your redemptions to validate if it’s an acceptable use of points or not. The easiest way to quickly evaluate
if a redemption is worth spending points for is to compare it against your base value for the points.
The value of one point for each travel rewards program
Before you start calculating the value of your reward points for each redemption, you need to establish a base value for one point. You would do this by comparing hundreds of flights and hotels in
destinations all around the world at various times. With all of this data compiled, you’ll get an average value. That’s the base value you’ll use to compare all redemptions moving forward.
Obviously, no person in their right mind is going to do that. Who has that kind of time? Guess what? I do. As a travel writer, I’ve literally searched thousands of flights and have done detailed
comparisons with many different loyalty programs.
Generally speaking, you should use the following base values when comparing redemptions:
• Aeroplan – 2 cents per point for business seats, 1.5 cents per point for economy
• Marriott Bonvoy – .9 cents per point
• American Express Membership Rewards – 2 cents per point when transferred to Aeroplan or used for the fixed travel program. 1 cent per point otherwise
• WestJet Dollars – 1 WestJet dollar = $1
• British Airways Avion – 1.6 cents per point
• RBC Rewards – Up to 2 cents per point when used for fixed points flights or transferred to other programs. 1 cent per point otherwise.
• Scotia Rewards – 1 cent per point
• CIBC Rewards – 1 cent per point
• HSBC Rewards – .50 cents per point
• BMO Rewards – .67 cents per point
• TD Rewards – .50 cents per point
• Hilton Honors – .50 cents per point
• Air Miles – 10.5 cents per point
You only really need to worry about the base value of airlines and hotel programs since the value of your redemptions will change depending on what you’re booking. The rest of the programs have a
fixed value per point (although a few have additional redemption options), so it doesn’t matter much.
Basically, when redeeming Aeroplan points, you want them to be worth around 1.5 tp 2 cents each. For Marriott Bonvoy, aim for a value of 0.9 cents each. If the value falls below that, save your
points and pay cash. Not only paying cash will save your points for a better value redemption later, but you will also earn points when paying for a flight or a stay.
If we go back to my Ottawa to Vancouver Aeroplan example, the redemption value was 2.19 cents per point. Since the value of this redemption is worth more than the established base value of 2 cents
per point for Aeroplan, it’s worth redeeming your points.
Now let’s look at an example for Marriott. I recently searched for a stay at the Sheraton Wall Centre in Vancouver. The cash rate was $296 per night after taxes and fees. At the time of my initial
search, the points cost was 40,000 Marriott Bonvoy points for a free night.
$296 X 100 / 40,000 = 0.74 cent per point
Since I value one Marriott Bonvoy point at 0.9 cents each, this redemption represents an 18% opportunity cost. In my opinion, it wasn’t worth using my points, so I decided to pay cash.
Oddly enough, a few days before my stay, I searched the hotel again, and I noticed that the points cost had changed and getting a free night would now only cost me 35,000 points. I quickly did the
math to see if this was worth it.
$296 X 100 / 35,000 = 0.85 cent per point
While this value still fell below my target value of 0.9 cents per point, it was only a slight devaluation of 6%. I considered this a good enough deal and rebooked on points.
As you can see, once you know the formula to calculate the value of one point and have an established base value, it’s easy to determine whether a redemption is worth it.
Travel Hacking for Lazy People | {"url":"https://www.moneywehave.com/value-of-your-rewards-points/","timestamp":"2024-11-02T10:56:55Z","content_type":"text/html","content_length":"256485","record_id":"<urn:uuid:eda24a61-c2bf-4685-85d8-e2844e7a0c68>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00018.warc.gz"} |
Stuck Keyboard
You are given two strings typed and target. You want to write target, but the keyboard is stuck so some characters may be written 1 or more times. Return whether it’s possible that typed was meant to
write target.
• n ≤ 100,000 where n is the length of typed
• m ≤ 100,000 where n is the length of s
Example 1
• typed = aaabcccc
• target = abc
Each of the "a", "b", and "c" were typed
Example 2
"c" was not typed | {"url":"https://yaeba.github.io/exercises/stuck-keyboard/","timestamp":"2024-11-11T09:48:06Z","content_type":"text/html","content_length":"25801","record_id":"<urn:uuid:f95d15ac-beca-4c98-856b-372d7384f90e>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00519.warc.gz"} |
Fuzzy logic
In which the truth value of a statement can be any real number from 0 up to 1
Let us now move beyond the three-valued logics to the case of fuzzy logic, which offers up infinitely many truth values, uncountably many! In fuzzy logic, we allow the truth value of a statement to
be any real number in the unit interval [0,1]—all the real numbers from 0 up to 1. In fuzzy logic we speak of a proposition or assertion holding with truth value 0.123 or with value 1/2 or value 1/π
or with value 0 or 1 and so on. The motivating idea, of course, is that 0 represents false, 1 represents true, and the numbers between 0 and 1 represent shades or degrees of truth. And then we define
the fuzzy logical operations on these values.
Such kinds of logic were studied from the 1920s by Łukasiewicz, Tarski, and Gödel and popularized as “fuzzy logic” in the 1960s by Zadeh. Fuzzy logic is now sometimes used in control theory and
industrial applications, including household appliances—my rice cooker was advertised as using fuzzy logic in its control decision procedures. Perhaps the thermostat in your home uses fuzzy logic.
The fuzzy logic connectives
In fuzzy logic, we conceive of the fundamental logical operations as mathematical functions defined on the set of truth values, the real numbers from 0 to 1. One natural way to proceed, for example,
is to define them as follows:
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How to Convert Millimeters to Miles with the Length Conversion Formula
Converting millimeters (mm) to miles (mi) is a straightforward mathematical process involving a simple conversion factor: 1 millimeter is equal to 0.000000621371 miles. This ratio is all you need to
perform the calculation. To calculate the conversion, start with the length in millimeters you need to convert and multiply the value by 0.000000621371. This will output the corresponding length in
x millimeters * 0.000000621371 = Number of Miles
For example: If you have a measurement of 1,000,000 millimeters, multiplying it by 0.000000621371 will convert it to approximately 0.621371 miles. This formula, focusing on multiplication by
0.000000621371, is suitable for any millimeter to mile conversion, and will provide accurate and reliable results. Conversely, if you wanted to convert miles to millimeters, you would simply multiply
the number of miles by 1,609,344, as there are 1,609,344 millimeters in a mile. Converting between two units of measurement is often just a matter of division or multiplication, but keep in mind that
your answer may be a fraction or decimal, so a calculator or useing our online converter might be handy!
Common Millimeters to Miles Conversion Table
Millimeters (mm) Miles (mi)
1,000 mm 0.000621371 mi
10,000 mm 0.00621371 mi
100,000 mm 0.0621371 mi
500,000 mm 0.310685 mi
1,000,000 mm 0.621371 mi
5,000,000 mm 3.10685 mi
10,000,000 mm 6.21371 mi
In-Depth on the Millimeter!
The millimeter, a unit of length in the metric system, is equivalent to one thousandth of a meter, the base unit of length in the International System of Units (SI). Represented by the symbol 'mm',
it is widely used in precision measurements in engineering, technology, and everyday life. The mm is particularly useful in contexts where small, accurate measurements are needed, such as in
manufacturing or construction. Fun fact: A millimeter is often used to measure small distances or thicknesses, such as the diameter of a wire or the thickness of a sheet of paper.
In-Depth on the Mile!
The mile, a unit of length in the imperial system, is primarily used in the United States and the United Kingdom. One mile is equivalent to 1,609.344 meters or 5,280 feet. The mile is a key unit in
various fields and is commonly used for measuring long distances, such as in road signs and races. Fun fact: The mile has its origins in the Roman "mille passus," meaning a thousand paces, where each
pace was two steps.
Good luck, and don't forget to bookmark this mm to mi length converter to save time when you need help converting a metric system number to the imperial system. | {"url":"https://www.calculyte.com/convert/length/mm-to-mile.html","timestamp":"2024-11-13T11:10:14Z","content_type":"text/html","content_length":"17965","record_id":"<urn:uuid:27466fdf-bdc6-4ccb-ab4d-632b2a8418b6>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00867.warc.gz"} |
Best-of-five versus Best-of-seven
Suppose that when Team X and Team Y play, the probability that X will win a single game is p and the probability that Y will win is q = 1 − p.
What is the probability that X will win the majority of a series of N games for some odd number N?
We know intuitively that the team more likely to win each game is more likely to win the series. But how much more likely?
We also know intuitively that the more games in the series, the more likely the better team will win. But how much more likely?
The probability of X winning a series of N games is
It doesn’t matter that maybe not all games are played: some games may not be played precisely because it makes no difference whether they are played. For example, if one team wins the first three in
a best-of-five series, the last two games are not played, but for our purposes we may imagine that they were played.
Let’s plot the probability of X winning a best-of-five series and a best-of-seven series.
The better team is more likely to win a series than a game. The probability of the better team winning a single game would be a diagonal line running from (0, 0) to (1, 1). The curves for a series
are below this line to the left of 0.5 and above this line to the right of 0.5. But the difference between a best-of-five and a best-of-seven series is small.
Here’s another plot looking at the difference in probabilities, probability of winning best-of-seven minus probability of winning best-of-five.
The maximum difference is between 3% and 4%.
This assumes the probability of winning a game is constant and that games are independent. This is not exactly the case in, for example, the World Series in which human factors make things more
Related posts
2 thoughts on “Best-of-five versus Best-of-seven”
1. I’d love to see the first graph augmented with a best-of-three curve.
2. Of course the home advantage in three home games verses four home games may be greater than the advantage difference in seven games on neutral field. | {"url":"https://www.johndcook.com/blog/2023/10/22/best-of-n-series/","timestamp":"2024-11-02T11:37:44Z","content_type":"text/html","content_length":"53742","record_id":"<urn:uuid:46b96a4c-45f6-4619-a769-133385688243>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00668.warc.gz"} |
Data Analysis in R Archives - Page 3 of 3 - KANDA DATA
One of the assumptions required in Ordinary Least Squares (OLS) linear regression is that the variance of the residuals is constant. This assumption is often referred to as the homoscedasticity
assumption. Some researchers are more familiar with the term heteroscedasticity test.
How to Analyze Multicollinearity in Linear Regression and its Interpretation in R (Part 2)
Non-multicollinearity is one of the assumptions required in the ordinary least square (OLS) method of linear regression analysis. Non-multicollinearity assumption implies that there is no strong
correlation among the independent variables in the equation.
How to Analyze Multiple Linear Regression and Interpretation in R (Part 1)
Multiple linear regression analysis has been widely used by researchers to analyze the influence of independent variables on dependent variables. There are many tools that researchers can use to
analyze multiple linear regression. | {"url":"https://kandadata.com/category/data-analysis-in-r/page/3/","timestamp":"2024-11-12T00:18:30Z","content_type":"text/html","content_length":"173143","record_id":"<urn:uuid:0795ecab-1ff6-42ba-b85c-4cea38274d90>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00785.warc.gz"} |
Initial Value Theorem of Laplace Transform
Laplace transform is very useful in the various fields of science and technology as Laplace transform replaces operations of calculus by operation of algebra. Initial and Final value theorems are
basic properties of Laplace transform. These theorems were given by French mathematician and physicist Pierre Simon Marquis De Laplace. Initial and Final value theorem are collectively called
Limiting theorems. In this article, we will be focusing on these two theorems.
Laplace Transform
The Laplace transform is a powerful mathematical tool used in engineering, physics, and applied mathematics to analyze linear time-invariant systems, particularly in the context of differential
equations. It transforms a function of time, often a function describing a dynamic system, into a complex function of a complex variable.
Let f(t) be a given function defined for all t ≥ 0 If we multiply f(t) by e^-st, where s is a real or complex parameter, independent of t and integrate with respect to t from 0 to ∞, and if the
resulting integral exists, the result will be a function of s, denoted by F(s).
F(s)= L{f(t)} = ∫[0]^∞ e^−st f(t) dt
The function F(s) is called Laplace transform of the function f(s).
Key points about Laplace transform:
• Linearity: L{af(t)+bg(t)}=aL{f(t)}+bL{g(t)}
• Derivative property: L{f′(t)}=sF(s)−f(0^+)
• Integration property: L{∫[0]^t^f(τ)dτ}=^1/[s]F(s)
• Time-shift property: L{f(t−a)u(t−a)}=e^−^asL{f(t)}
• Convolution property: L{f∗g}=F(s)G(s)
Laplace transforms are particularly useful for solving linear ordinary differential equations (ODEs) and linear partial differential equations (PDEs) with constant coefficients. The transformed
equations often simplify into algebraic equations, making them easier to solve. After finding the solution in the Laplace domain, an inverse Laplace transform is applied to obtain the solution in the
time domain. | {"url":"https://www.naukri.com/code360/library/initial-and-final-value-theorem-of-laplace-transform","timestamp":"2024-11-06T05:34:20Z","content_type":"text/html","content_length":"387326","record_id":"<urn:uuid:f7d90850-b1a0-4cf5-8c32-66932641e307>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00814.warc.gz"} |
Matrix Notation
A matrix is written as a grid of numbers with brackets on either side and represents a linear mapping between an input space and an output space. When referring to the dimensions of the matrix, the
convention is to say the number of rows followed by the number of columns. For example, the matrix shown below would be called a “2 by 3” matrix because it has two rows and three columns.
The matrix notation is shorthand for a system of linear equations. So if you consider the matrix below which represents a linear mapping between the vector and the vector .
This can also be written as a system of linear equations like so: | {"url":"https://wumbo.net/notation/matrix/","timestamp":"2024-11-04T20:26:25Z","content_type":"text/html","content_length":"4941","record_id":"<urn:uuid:eac21a01-5164-4b3f-9054-367f7e07be0b>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00453.warc.gz"} |
13.1: Vector-Valued Functions and Space Curves
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Our study of vector-valued functions combines ideas from our earlier examination of single-variable calculus with our description of vectors in three dimensions from the preceding chapter. In this
section, we extend concepts from earlier chapters and also examine new ideas concerning curves in three-dimensional space. These definitions and theorems support the presentation of material in the
rest of this chapter and also in the remaining chapters of the text.
Definition of a Vector-Valued Function
Our first step in studying the calculus of vector-valued functions is to define what exactly a vector-valued function is. We can then look at graphs of vector-valued functions and see how they define
curves in both two and three dimensions.
A vector-valued function is a function of the form
\[\vecs r(t)=f(t)\,\hat{\mathbf{i}}+g(t)\,\hat{\mathbf{j}} \quad \text{or} \quad \vecs r(t)=f(t)\,\hat{\mathbf{i}}+g(t)\,\hat{\mathbf{j}}+h(t)\,\hat{\mathbf{k}},\]
where the component functions \(f\), \(g\), and \(h\), are real-valued functions of the parameter \(t\).
Vector-valued functions can also be written in the form
\[\vecs r(t)=⟨f(t),\,g(t)⟩ \; \; \text{or} \; \; \vecs r(t)=⟨f(t),\,g(t),\,h(t)⟩.\]
In both cases, the first form of the function defines a two-dimensional vector-valued function in the plane; the second form describes a three-dimensional vector-valued function in space.
We often use \(t\) as a parameter because \(t\) can represent time.
The parameter \(t\) may lie between two real numbers: \(a≤t≤b\), or its value may range over the entire set of real numbers.
Each of the component functions that make up a vector-valued function may have domain restrictions that enforce restrictions on the value of \(t\).
The domain of a vector-valued function \(\vecs r\) is the intersection of the domains of its component functions, i.e., it is the set of all values of \(t\) for which the vector-valued function is
State the domain of the vector-valued function \(\vecs r(t)=\sqrt{2-t} \,\hat{\mathbf{i}}+\ln(t+3)\,\hat{\mathbf{j}} + e^t \, \hat{\mathbf{k}}\).
We first consider the natural domain of each component function. Note that we list the domains in both set-builder notation and interval notation.
Function: Domain:
\(\begin{array}{ll} \sqrt{2-t} & & \big\{\,t\,|\, t \le 2\big\} \quad \text{or} \quad (-\infty, 2\big] \\
\ln(t+3) & & \big\{\,t\,|\, t \gt -3\big\} \quad \text{or} \quad (-3, \infty ) \\
e^t & & (-\infty, \infty) \end{array} \)
The domain of \(\vecs r\) is the intersection of these domains, so it must contain all values of \(t\) that work in all three, but no value of \(t\) that does not work in any one of these functions.
Hence, the domain of \(\vecs r\) is: \(\text{D}_{\vecs r}: \big\{\,t\,|\, -3 \lt t \le 2\big\}\) or \( (-3, 2\big] \).
Note that only one form of the domain of \(\vecs r\) need be given. The first, \(\big\{\,t\,|\, -3 \lt t \le 2\big\}\), is in set-builder notation, while the second, \( (-3, 2\big] \), is in interval
For each of the following vector-valued functions, evaluate \(\vecs r(0)\), \(\vecs r(\frac{\pi}{2})\), and \(\vecs r(\frac{2\pi}{3})\). Do any of these functions have domain restrictions?
1. \(\vecs r(t)=4\cos t\,\hat{\mathbf{i}}+3\sin t\,\hat{\mathbf{j}}\)
2. \(\vecs r(t)=3\tan t\,\hat{\mathbf{i}}+4 \sec t\,\hat{\mathbf{j}}+5t\,\hat{\mathbf{k}}\)
1. To calculate each of the function values, substitute the appropriate value of \(t\) into the function:
\begin{align*}\vecs r(0) \; & = 4\cos(0) \hat{\mathbf{i}}+3\sin(0) \hat{\mathbf{j}} \\[4pt] & =4\hat{\mathbf{i}}+0 \hat{\mathbf{j}}=4\hat{\mathbf{i}} \\[4pt] \vecs r\left(\frac{\pi}{2}\right) \;
& = 4\cos\left(\frac{π}{2}\right)\hat{\mathbf{i}}+3\sin\left(\frac{π}{2}\right) \hat{\mathbf{j}} \\[4pt] & = 0\hat{\mathbf{i}}+3 \hat{\mathbf{j}}=3 \hat{\mathbf{j}} \\[4pt] \vecs r\left(\frac{2\
pi}{3}\right) \; & =4\cos\left(\frac{2π}{3}\right)\hat{\mathbf{i}}+3\sin\left(\frac{2π}{3}\right) \hat{\mathbf{j}} \\[4pt] & =4\left(−\tfrac{1}{2}\right)\hat{\mathbf{i}}+3\left(\tfrac{\sqrt{3}}
{2}\right) \hat{\mathbf{j}}=−2 \hat{\mathbf{i}}+\tfrac{3 \sqrt{3}}{2} \hat{\mathbf{j}}\end{align*}
To determine whether this function has any domain restrictions, consider the component functions separately. The first component function is \(f(t)=4 \cos t\) and the second component function is
\(g(t)=3\sin t\). Neither of these functions has a domain restriction, so the domain of \(\vecs r(t)=4\cos t\,\hat{\mathbf{i}}+3 \sin t \,\hat{\mathbf{j}}\) is all real numbers.
2. To calculate each of the function values, substitute the appropriate value of t into the function:\[\begin{align*}\vecs r(0) \; & = 3\tan(0)\hat{\mathbf{i}}+4\sec(0) \hat{\mathbf{j}}+5(0) \hat{\
mathbf{k}} \\[4pt] & = 0\hat{\mathbf{i}}+4j+0 \hat{\mathbf{k}}=4 \hat{\mathbf{j}} \\[4pt] \vecs r\left(\frac{\pi}{2}\right) \; & = 3\tan\left(\frac{\pi}{2}\right)\hat{\mathbf{i}}+4\sec\left(\frac
{\pi}{2}\right) \hat{\mathbf{j}}+5\left(\frac{\pi}{2}\right) \hat{\mathbf{k}},\,\text{which does not exist} \\[4pt] \vecs r\left(\frac{2\pi}{3}\right) \; & =3\tan\left(\frac{2 \pi}{3}\right)\hat
{\mathbf{i}}+4\sec\left(\frac{2\pi}{3}\right) \hat{\mathbf{j}}+5\left(\frac{2\pi}{ 3}\right) \hat{\mathbf{k}} \\[4pt] & =3(−\sqrt{3})\hat{\mathbf{i}}+4(−2)\hat{\mathbf{j}}+\frac{10π}{3} \hat{\
mathbf{k}} \\[4pt] & =(-3\sqrt{3})\hat{\mathbf{i}}−8\hat{\mathbf{j}}+\frac{10π}{3} \hat{\mathbf{k}}\end{align*}\]To determine whether this function has any domain restrictions, consider the
component functions separately. The first component function is \(f(t)=3\tan t\), the second component function is \(g(t)=4\sec t\), and the third component function is \(h(t)=5t\). The first two
functions are not defined for odd multiples of \(\frac{\pi}{2}\), so the function is not defined for odd multiples of \(\frac{\pi}{2}\). Therefore, \[\text{D}_{\vecs r}=\Big\{t\,|\,t≠ \frac
{(2n+1)\pi}{2}\Big\},\nonumber\] where \(n\) is any integer.
For the vector-valued function \(\vecs r(t)=(t^2−3t) \,\hat{\mathbf{i}}+(4t+1) \,\hat{\mathbf{j}}\), evaluate \(\vecs r(0),\, \vecs r(1)\), and \(\vecs r(−4)\). Does this function have any domain
Substitute the appropriate values of \(t\) into the function.
\(\vecs r(0) = \hat{\mathbf{j}},\, \vecs r(1)=−2 \hat{\mathbf{i}}+5 \hat{\mathbf{j}},\, \vecs r(−4)=28 \hat{\mathbf{i}}−15 \hat{\mathbf{j}}\)
The domain of \(\vecs r(t)=(t^2−3t)\hat{\mathbf{i}}+(4t+1)\hat{\mathbf{j}}\) is all real numbers.
Example \(\PageIndex{1}\) illustrates an important concept. The domain of a vector-valued function consists of real numbers. The domain can be all real numbers or a subset of the real numbers. The
range of a vector-valued function consists of vectors. Each real number in the domain of a vector-valued function is mapped to either a two- or a three-dimensional vector.
Graphing Vector-Valued Functions
Recall that a plane vector consists of two quantities: direction and magnitude. Given any point in the plane (the initial point), if we move in a specific direction for a specific distance, we arrive
at a second point. This represents the terminal point of the vector. We calculate the components of the vector by subtracting the coordinates of the initial point from the coordinates of the terminal
A vector is considered to be in standard position if the initial point is located at the origin. When graphing a vector-valued function, we typically graph the vectors in the domain of the function
in standard position, because doing so guarantees the uniqueness of the graph. This convention applies to the graphs of three-dimensional vector-valued functions as well. The graph of a vector-valued
function of the form
\[\vecs r(t)=f(t)\, \hat{\mathbf{i}}+g(t)\,\hat{\mathbf{j}} \nonumber \]
consists of the set of all points \((f(t),\,g(t))\), and the path it traces is called a plane curve. The graph of a vector-valued function of the form
\[\vecs r(t)=f(t) \,\hat{\mathbf{i}}+g(t) \,\hat{\mathbf{j}}+h(t) \,\hat{\mathbf{k}} \nonumber \]
consists of the set of all points \((f(t),\,g(t),\,h(t))\), and the path it traces is called a space curve. Any representation of a plane curve or space curve using a vector-valued function is called
a vector parameterization of the curve.
Each plane curve and space curve has an orientation, indicated by arrows drawn in on the curve, that shows the direction of motion along the curve as the value of the parameter \(t\) increases.
Create a graph of each of the following vector-valued functions:
1. The plane curve represented by \(\vecs r(t)=4 \cos t \,\hat{\mathbf{i}}+3 \sin t \,\hat{\mathbf{j}}\), \(0≤t≤2\pi\)
2. The plane curve represented by \(\vecs r(t)=4 \cos(t^3) \,\hat{\mathbf{i}}+3 \sin(t^3) \,\hat{\mathbf{j}}\), \(0≤t≤\sqrt[3]{2\pi}\)
3. The space curve represented by \(\vecs r(t)=4 \cos t \,\hat{\mathbf{i}}+4 \sin t \,\hat{\mathbf{j}}+t \,\hat{\mathbf{k}}\), \(0≤t≤4\pi\)
1. As with any graph, we start with a table of values. We then graph each of the vectors in the second column of the table in standard position and connect the terminal points of each vector to form
a curve (Figure \(\PageIndex{1}\)). This curve turns out to be an ellipse centered at the origin.
Table \(\PageIndex{1}\): Table of Values for \(\vecs r(t)=4 \cos t \,\hat{\mathbf{i}}+3 \sin t \,\hat{\mathbf{j}}\), \(0≤t≤2\pi\)
\(t\) \(\vecs r(t)\) \(t\) \(\vecs r(t)\)
\(0\) \(4\hat{\mathbf{i}}\) \(\pi\) \(-4\hat{\mathbf{i}}\)
\(\dfrac{\pi}{4}\) \(2 \sqrt{2} \hat{\mathbf{i}} + \frac{3 \sqrt{2}}{2}\hat{\mathbf{j}}\) \(\dfrac{5\pi}{4}\) \(-2 \sqrt{2} \hat{\mathbf{i}} - \frac{3 \sqrt{2}}{2}\hat{\mathbf{j}}\)
\(\dfrac{\pi}{2}\) \(\mathrm{3\hat{\mathbf{j}}}\) \(\dfrac{3\pi}{2}\) \(\mathrm{-3\hat{\mathbf{j}}}\)
\(\dfrac{3\pi}{4}\) \( -2 \sqrt{2} \hat{\mathbf{i}} + \frac{3 \sqrt{2}}{2}\hat{\mathbf{j}}\) \(\dfrac{7\pi}{4}\) \( 2 \sqrt{2} \hat{\mathbf{i}} - \frac{3 \sqrt{2}}{2}\hat{\mathbf{j}}\)
\(2\pi\) \(4\hat{\mathbf{i}}\)
Figure \(\PageIndex{1}\): The graph of the first vector-valued function is an ellipse.
2. The table of values for \(\vecs r(t)=4 \cos(t^3) \,\hat{\mathbf{i}}+3 \sin(t^3) \,\hat{\mathbf{j}}\), \(0≤t≤\sqrt[3]{2\pi}\) is as follows:
Table of Values for \(\vecs r(t)=4 \cos(t^3) \,\hat{\mathbf{i}}+3 \sin(t^3) \,\hat{\mathbf{j}}\), \(0≤t≤\sqrt[3]{2\pi}\)
\(t\) \(\vecs r(t)\) \(t\) \(\vecs r(t)\)
\(0\) \(\mathrm{4\hat{\mathbf{i}}}\) \(\displaystyle\sqrt[3]{\pi}\) \(\mathrm{-4\hat{\mathbf{i}}}\)
\(\displaystyle \sqrt[3]{\dfrac{\ \(\mathrm{ 2 \sqrt{2} \hat{\mathbf{i}} + \frac{3 \sqrt{2}}{2}\ \(\displaystyle \sqrt[3]{\dfrac{5\ \(\mathrm{ -2 \sqrt{2} \hat{\mathbf{i}} - \frac{3 \sqrt{2}}{2}\
pi}{4}}\) hat{\mathbf{j}}}\) pi}{4}}\) hat{\mathbf{j}}}\)
\(\displaystyle \sqrt[3]{\dfrac{\ \(\mathrm{3\hat{\mathbf{j}}}\) \(\displaystyle \sqrt[3]{\dfrac{3\ \(\mathrm{-3\hat{\mathbf{j}}}\)
pi}{2}}\) pi}{2}}\)
\(\displaystyle \sqrt[3]{\dfrac{3\ \(\mathrm{ -2 \sqrt{2} \hat{\mathbf{i}} + \frac{3 \sqrt{2}}{2}\ \(\displaystyle \sqrt[3]{\dfrac{7\ \(\mathrm{ 2 \sqrt{2} \hat{\mathbf{i}} - \frac{3 \sqrt{2}}{2}\
pi}{4}}\) hat{\mathbf{j}}}\) pi}{4}}\) hat{\mathbf{j}}}\)
\( \displaystyle\sqrt[3]{2\pi}\) \(\mathrm{4\hat{\mathbf{i}}}\)
The graph of this curve is also an ellipse centered at the origin.
Figure \(\PageIndex{2}\): The graph of the second vector-valued function is also an ellipse.
3. We go through the same procedure for a three-dimensional vector function.
Table of Values for \(\mathrm{r(t)= 4 \cos t \hat{\mathbf{i}}+4 \sin t \hat{\mathbf{j}}+t \hat{\mathbf{k}}}\), \(\mathrm{0≤t≤4\pi}\)
\(t\) \(\vecs r(t)\) \(t\) \(\vecs r(t)\)
\(\mathrm{0}\) \(\mathrm{4\hat{\mathbf{i}}}\) \(\mathrm{\pi} \(\mathrm{-4\hat{\mathbf{i}}}+ \pi \hat{\mathbf{k}}\)
\(\dfrac{\pi} \(\mathrm{2 \sqrt{2} \hat{\mathbf{i}} + 2\sqrt{2} \hat{\mathbf{j}} + \frac{\pi}{4} \(\dfrac{5\pi} \(\mathrm{ -2 \sqrt{2} \hat{\mathbf{i}} - 2\sqrt{2} \hat{\mathbf{j}} + \frac{5\pi}
{4}\) \hat{\mathbf{k}}}\) {4}\) {4} \hat{\mathbf{k}}}\)
\(\dfrac{\pi} \(\mathrm{4\hat{\mathbf{j}} +\frac{\pi}{2} \hat{\mathbf{k}}}\) \(\dfrac{3\pi} \(\mathrm{-4\hat{\mathbf{j}} +\frac{3\pi}{2} \hat{\mathbf{k}}}\)
{2}\) {2}\)
\(\dfrac{3\pi} \(\mathrm{ -2 \sqrt{2} \hat{\mathbf{i}} + 2\sqrt{2} \hat{\mathbf{j}} + \frac{3\pi} \(\dfrac{7\pi} \(\mathrm{ 2 \sqrt{2} \hat{\mathbf{i}} - 2\sqrt{2} \hat{\mathbf{j}} + \frac{7\pi}
{4}\) {4} \hat{\mathbf{k}}}\) {4}\) {4} \hat{\mathbf{k}}}\)
\(\mathrm{2\pi} \(\mathrm{4\hat{\mathbf{j}} + 2\pi \hat{\mathbf{k}}}\)
The values then repeat themselves, except for the fact that the coefficient of \(\hat{\mathbf{k}}\) is always increasing ( \(\PageIndex{3}\)). This curve is called a helix. Notice that if the \(\hat
{\mathbf{k}}\) component is eliminated, then the function becomes \(\vecs r(t)=4\cos t \hat{\mathbf{i}}+ 4\sin t \hat{\mathbf{j}}\), which is a circle of radius 4 centered at the origin.
Figure \(\PageIndex{3}\): The graph of the third vector-valued function is a helix.
You may notice that the graphs in parts a. and b. are identical. This happens because the function describing curve b is a so-called reparameterization of the function describing curve a. In fact,
any curve has an infinite number of reparameterizations; for example, we can replace \(t\) with \(2t\) in any of the three previous curves without changing the shape of the curve. The interval over
which \(t\) is defined may change, but that is all. We return to this idea later in this chapter when we study arc-length parameterization. As mentioned, the name of the shape of the curve of the
graph in \(\PageIndex{3}\) is a helix. The curve resembles a spring, with a circular cross-section looking down along the \(z\)-axis. It is possible for a helix to be elliptical in cross-section as
well. For example, the vector-valued function \(\vecs r(t)=4 \cos t \,\hat{\mathbf{i}}+3 \sin t \,\hat{\mathbf{j}}+t \,\hat{\mathbf{k}}\) describes an elliptical helix. The projection of this helix
into the \(xy\)-plane is an ellipse. Last, the arrows in the graph of this helix indicate the orientation of the curve as \(t\) progresses from \(0\) to \(4π\).
At this point, you may notice a similarity between vector-valued functions and parameterized curves. Indeed, given a vector-valued function \(\vecs r(t)=f(t)\,\hat{\mathbf{i}}+g(t)\,\hat{\mathbf{j}}
\) we can define \(x=f(t)\) and \(y=g(t)\). The graph of the parameterized function would then agree with the graph of the vector-valued function, except that the vector-valued function's graph would
be traced out by vectors rather than just being a collection of points. Since we can parameterize a curve defined by a function \(y=f(x)\), it is also possible to represent an arbitrary plane curve
by a vector-valued function.
Finding a Vector-Valued Function to Trace out the Graph of a Function y = f(x)
As you can see in the examples above, a vector-valued function traces out a curve in the plane or in space. What if we wish to write a vector-valued function that traces out the graph of a particular
curve in the \(xy\)-plane?
What function's graph is traced out by the vector-valued function in Exercise \(\PageIndex{2}\) above: \(\vecs r(t)=t \,\hat{\mathbf{i}}+ t^3 \,\hat{\mathbf{j}}\)? It looks like the graph of \( y = x
^3\), doesn't it?
Remembering what was just said about the components of the vector-valued function corresponding to the parametic equations of a parameterized curve, we see that here we have:
\[\begin{align*} x &= t \\ y &= t^3\end{align*}\nonumber\]
Since \(x = t\), we can replace \(t\) in the equation \( y = t^3\) with \(x\), giving us the function: \( y = x^3\).
So we were correct in our guess.
How could we write a vector-valued function to trace out the graph of a function, \( y = f(x)\)?
Well, there are two orientations to consider: left-to-right and right-to-left.
To trace out the graph of \( y = f(x)\) from left-to-right, use: \(\vecs r(t) = t \,\hat{\mathbf{i}}+ f(t) \,\hat{\mathbf{j}}\)
Note that what's important here is to have the \(x\) component be an increasing function. Any increasing function will work. We could use \(x = t^3\), for example. But then we would need to remember
to replace \(x\) in the function \(f(x)\) with this expression \(t^3\), giving us \(y = f(t^3)\). This means the function \(y = f(x) \) could also be parameterized from left-to-right by the
vector-valued function: \(\vecs r(t) = t^3 \,\hat{\mathbf{i}}+ f(t^3) \,\hat{\mathbf{j}}\)
To trace out the graph of \( y = f(x)\) from right-to-left, use: \(\vecs r(t) = -t \,\hat{\mathbf{i}}+ f(-t) \,\hat{\mathbf{j}}\)
Again note that we could use any decreasing function of \(t\) for the \(x\) component and obtain a vector-valued function that traces out the graph of \(y = f(x) \) from right-to-left. Using \(x = -t
\) is just the simplest decreasing function we can choose.
Determine a vector-valued function that will trace out the graph of \(y = \cos x\) from left-to-right, and another one to trace it out from right-to-left.
Left-to-right: \(\vecs r(t) = t \,\hat{\mathbf{i}}+ \cos t \,\hat{\mathbf{j}}\)
Right-to-left: \(\vecs r(t) = -t \,\hat{\mathbf{i}}+ \cos (-t) \,\hat{\mathbf{j}}\)
Finding a Vector-Valued Function to Trace out the Graph of an Equation in x and y and Vice Versa
What if we wish to find a vector-valued function to trace out the graph of a circle, an ellipse, or a hyperbola, given its implicit equation?
Well, note that in Example \(\PageIndex{3}\), the vector-valued function \(\vecs r(t)=4 \cos t \,\hat{\mathbf{i}}+3 \sin t \,\hat{\mathbf{j}}\) traced out the graph of the ellipse \(\frac{x^2}{16} +
\frac{y^2}{9} = 1\).
In this vector-valued function we see that: \[x = 4\cos t \quad \text{and} \quad y = 3\sin t\]
What we need now is a way to convert this to an implicit equation involving \(x\) and \(y\). To accomplish this, remember the Pythagorean identity, \[\cos^2 t + \sin^2 t = 1\].
Now all we need to do is solve the above equations for \(\cos t\) and \(\sin t\) and we can substitute into this identity to obtain an equation in \(x\) and \(y\).
So: \[\cos t = \frac{x}{4} \quad \text{and} \quad \sin t = \frac{y}{3} \]
Substituting into the identity gives us: \[ \left( \frac{x}{4}\right)^2 + \left( \frac{y}{3}\right)^2 = 1\]
Simplifying this implicit equation gives us the implicit equation of the ellipse in Example \(\PageIndex{3}\) that we wrote above:
\[\frac{x^2}{16} + \frac{y^2}{9} = 1\]
To go the other way and find a vector-valued function that traces out an ellipse requires us to simply take these steps in the opposite direction!
Write a vector-valued function that traces out each of the following implicit curves:
1. The ellipse: \(\frac{x^2}{16} + \frac{y^2}{9} = 1\)
2. The circle: \(x^2 + y^2 = 4\)
3. The hyperbola: \(\frac{x^2}{25} - \frac{y^2}{16} = 1\)
a. Let's just use the process shown above in reverse. First, let's rewrite the implicit equation so it shows a sum of quantities squared equals one.
\[ \left( \frac{x}{4}\right)^2 + \left( \frac{y}{3}\right)^2 = 1 \nonumber\]
Now we need the identity we used above, \(\cos^2 t + \sin^2 t = 1\).
Equating the parts being squared (note we actually have a choice here about which to make \(\cos t\) and which to make \(\sin t\)), we get:
\[ \frac{x}{4} = \cos t \quad \text{and} \quad \frac{y}{3} = \sin t \nonumber\]
Now we just need to solve for \(x\) and \(y\).
\[x = 4\cos t \quad \text{and} \quad y = 3\sin t \nonumber\]
We can now write a vector-valued function that traces out this ellipse: \(\vecs r(t) = 4\cos t\,\hat{\mathbf{i}}+ 3\sin t \,\hat{\mathbf{j}}\)
Note that we could also have written \(\vecs r(t) = 4\sin t\,\hat{\mathbf{i}}+ 3\cos t \,\hat{\mathbf{j}}\), since we could have chosen to switch the \(\sin t\) and \(\cos t\) above. It will trace
out the same ellipse, but with the opposite orientation.
b. Tracing out a circle is fairly straightforward, not really needing the process we showed above, although it still may be helpful at first. Remember that all the vectors on the unit circle can be
represented in the form: \(\vecs v = \cos \theta \, \hat{\mathbf{i}} + \sin \theta \, \hat{\mathbf{j}}\).
So the vector-valued function, \(\vecs r(t) = \cos t\,\hat{\mathbf{i}}+ \sin t \,\hat{\mathbf{j}}\), will trace out the unit circle with equation, \(x^2 + y^2 = 1\).
To obtain a circle of radius \(2\) centered at the origin (which is the graph of \(x^2 + y^2 = 4\)), we just need to multiply through this vector-valued function by a scalar factor of \(2\).
Thus, a vector-valued function that will trace out this circle is: \(\vecs r(t) = 2\cos t\,\hat{\mathbf{i}}+ 2\sin t \,\hat{\mathbf{j}}\).
Note again that another possibility is: \(\vecs r(t) = 2\sin t\,\hat{\mathbf{i}}+ 2\cos t \,\hat{\mathbf{j}}\). It will trace out the same circle, but with the opposite orientation.
To use the technique above, you start by dividing each term in the equation by the square of the radius, here 4, thus putting the circle equation in "ellipse form". The rest of the steps follow the
pattern shown in part a.
c. To trace out a hyperbola of the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) or \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), we need to locate a trigonometric identity showing the difference of
two squares equals 1. If you don't already have such an identity memorized, we can obtain one from the Pythagorean identity used above. That is,
\[\cos^2 t + \sin^2 t = 1 \nonumber\]
Dividing through each term by \(\cos^2 t\)
\[\frac{\cos^2 t}{\cos^2 t} + \frac{\sin^2 t}{\cos^2 t} = \frac{1}{\cos^2 t} \nonumber\]
\[1+ \tan^2 t = \sec^2 t \nonumber\]
Rewriting this equation gives us the identity we need:
\[\sec^2 t - \tan^2 t = 1 \nonumber\]
Now, the equation of this hyperbola is:
\[\frac{x^2}{25} - \frac{y^2}{16} = 1 \nonumber\]
Rewriting the left side to show the quantities that are squared:
\[\left(\frac{x}{5}\right)^2 - \left(\frac{y}{4}\right)^2 = 1 \nonumber\]
We can then equate the squared expressions corresponding terms:
\[\frac{x}{5} = \sec t \quad \text{and} \quad \frac{y}{4} = \tan t \nonumber\]
Solving for \(x\) and \(y\), we have:
\[x = 5\sec t \quad \text{and} \quad y = 4\tan t \nonumber\]
So a vector-valued function that will trace out the hyperbola \[\frac{x^2}{25} - \frac{y^2}{16} = 1\] is \[\vecs r(t) = 5\sec t\,\hat{\mathbf{i}}+ 4\tan t \,\hat{\mathbf{j}}. \nonumber\]
Parameterizing a Piecewise Path
There are times when it is necessary to parameterize a path made up of pieces of different curves. This piecewise path may be open or form the boundary of a closed region as does the example shown in
Figure \(\PageIndex{4}\). In addition to determining a vector-valued function to trace out each piece separately, with the indicated orientation, we also need to determine a suitable range of values
for the parameter \(t\).
Note that there are many ways to parameterize any one piece, so there are many correct ways to parameterize a path in this way.
Determine a piecewise parameterization of the path shown in Figure \(\PageIndex{4}\), starting with \(t=0\) and continuing on through each piece.
Our first task is to identify the three pieces in this piecewise path.
Note how we labeled these sequentially as \(\vecs r_1\), \(\vecs r_2\), and \(\vecs r_3\). Now we need to identify the function for each and write the corresponding vector-valued function with the
correct orientation (left-to-right or right-to-left).
Determining \(\vecs r_1\): The equation of the linear function in this piece is \(y = x\).
Since it is oriented from left-to-right between \(t = 1\) and \(t = 4\), we can write:
\[\vecs r_{1a}(t) = t\,\hat{\mathbf{i}}+ t \,\hat{\mathbf{j}} \quad\text{for}\quad 1 \le t \le 4 \nonumber\]
If we wish to begin this piece at \(t = 0\), we just need to shift the value of \(t\) one unit to the left. One way to do this is to write \(\vecs r_{1a}\) in terms of \(t_1\) instead of \(t\) to
make the translation easier to see.
Thus, we have \(\vecs r_{1a}(t_1) = t_1\,\hat{\mathbf{i}}+ t_1 \,\hat{\mathbf{j}}\) for \(1\le t_1\le 4\).
Figure \(\PageIndex{4}\): A closed piecewise path
Subtracting \(1\) from each part of this range of parameter values, we have: \(0 \le t_1 - 1 \le 3\).
Now we let \(t = t_1 - 1\). Solving for \(t_1\), we obtain: \(t_1 = t + 1\).
Replacing \(t_1\) with the expression \(t + 1\) will effectively shift the range of parameter values one unit to the left.
So, starting with \(t = 0\), we have: \[\vecs r_1(t) = (t+1)\,\hat{\mathbf{i}}+ (t+1) \,\hat{\mathbf{j}} \quad\text{for}\quad 0 \le t \le 3 \nonumber\]
Double-check that this vector-valued function will trace out this segment in the correct direction before going on to \(r_2\).
Determining \(\vecs r_2\): This piece has a label showing the function whose graph it traces along. If it were oriented from left-to-right, we would have:
\[\text{Left-to-right:}\quad\vecs r_{2a}(t) = t\,\hat{\mathbf{i}}+ \left(2\sqrt{\frac{4-t}{3}}+4\right) \,\hat{\mathbf{j}} \quad\text{for}\quad 1 \le t \le 4 \nonumber\]
But since we need it to be oriented from right-to-left, we need to replace \(t\) with \(-t\) in the function and we need to divide through the range inequality by -1 to obtain the corresponding
range. Thus we obtain:
\[\vecs r_{2b}(t) = -t\,\hat{\mathbf{i}}+ \left(2\sqrt{\frac{4-(-t)}{3}}+4\right) \,\hat{\mathbf{j}} \quad\text{for}\quad -4 \le t \le -1 \nonumber\]
Check that it works!
Now we wish to have this piece start at \(t = 3\) just after the first one finishes. Again let's make this easier to see by writing \(r_{2b}\) in terms on \(t_2\).
\[\vecs r_{2b}(t_2) = -t_2\,\hat{\mathbf{i}}+ \left(2\sqrt{\frac{4-(-t_2)}{3}}+4\right) \,\hat{\mathbf{j}} \quad\text{for}\quad -4 \le t_2 \le -1 \nonumber\]
To force \(r_2\) to start with \(t = 3\) instead of \(t = -4\), we need to add \(7\) to each part of the inequality. This yields: \(3 \le t_2 + 7 \le 6\).
Let \(t = t_2 + 7\). Then solving for \(-t_2\) (since this is what we need to replace in \(r_{2b}\)), we have: \(-t_2 = 7-t\).
Replacing \(-t_2\) with \(\left(7-t\right)\) in \(\vecs r_{2b}\), we obtain:
\[\vecs r_{2}(t) = (7-t)\,\hat{\mathbf{i}}+ \left(2\sqrt{\frac{4-(7-t)}{3}}+4\right) \,\hat{\mathbf{j}} \quad\text{for}\quad 3 \le t \le 6 \nonumber\]
This can be combined with our earlier result for \(r_1\) to write a piecewise-defined vector-valued function that traces out the first two pieces, starting at \(t = 0\):
\[\vecs r(t) = \begin{cases}
(t+1)\,\hat{\mathbf{i}} + (t+1) \,\hat{\mathbf{j}}, & 0 \le t \le 3 \\
(7-t)\,\hat{\mathbf{i}} + \left(2\sqrt{\frac{t - 3}{3}}+4\right) \,\hat{\mathbf{j}}, & 3 \lt t\le 6
\end{cases} \nonumber\]
Note that one small modification was made to the second range so that when \(t = 3\), there is no confusion about which piece to evaluate.
Determining \(\vecs r_3\): To determine this last piece we need to think a little differently. This is because it is a vertical segment, which cannot be represented with a function of the form, \(y =
f(x)\). Note that it could be represented by a function of the form \(x = f(y)\). Letting \(y = t\), we can write \(x = f(t)\) and writing a parameterization in increasing \(y\) values
(bottom-to-top), we'd get: \( \vecs r(t) = f(t) \,\hat{\mathbf{i}} + t \,\hat{\mathbf{j}}\).
The equation of this line is \(x = 1\). Thus, if we wished to parameterize this segment with upward orientation (increasing values of \(y\)), we have:
\[\vecs r_{3a}(t) = 1\,\hat{\mathbf{i}}+ t \,\hat{\mathbf{j}} \quad\text{for}\quad 1 \le t \le 6 \nonumber\]
But since we wish to use a downward orientation (decreasing values of \(y\)), we need to use a decreasing function of \(t\) for \(y\). As before, the simplest case is to use \(y = -t\). Then, in the
general case, we'd trace a function \(x = f(y)\) in a downwards orientation with \(\vecs r(t) = f(-t) \,\hat{\mathbf{i}} - t \,\hat{\mathbf{j}}\).
In the case of \(r_3\), this gives us:
\[\vecs r_{3b}(t) = 1\,\hat{\mathbf{i}}- t \,\hat{\mathbf{j}} \quad\text{for}\quad -6 \le t \le -1 \nonumber\]
Note that since \(x = 1, \, f(-t) = 1\), that is, it did not change the first component since it was constant and not a variable function of the parameter \(t\).
Also note that since we negated \(t\), we also had to negate the range, dividing it through by \(-1\).
As above, to facilitate the translation, we'll replace \(t\) with \(t_3\), giving us:
\[\vecs r_{3b}(t_3) = 1\,\hat{\mathbf{i}}- t_3\,\hat{\mathbf{j}} \quad\text{for}\quad -6 \le t_3 \le -1 \nonumber\]
Now, we wish this final piece to start at \(t = 6\) where the second piece we formed above leaves off. We see that we need to add \(12\) to the range of paramater \(t\) to accomplish this, giving us
a new range of \(6 \le t_3 + 12 \le 11\).
Let \(t = t_3 + 12\). Then solving for \(-t_3\) (since this is what we need to replace in \(r_{3b}\)), we have: \(-t_3 = 12-t\).
Replacing \(-t_3\) with \(\left(12-t\right)\) in \(\vecs r_{3b}\), we obtain:
\[\vecs r_{3}(t) = 1\,\hat{\mathbf{i}} + (12 - t)\,\hat{\mathbf{j}} \quad\text{for}\quad 6 \le t \le 11 \nonumber\]
Check that this still traces out this vertical segment from top-to-bottom.
We can now state the final answer as a single piecewise-defined vector-valued function that traces out this entire path, starting when \(t = 0\).
\[\vecs r(t) = \begin{cases}
(t+1)\,\hat{\mathbf{i}} + (t+1) \,\hat{\mathbf{j}}, & 0 \le t \le 3 \\
(7-t)\,\hat{\mathbf{i}} + \left(2\sqrt{\frac{t - 3}{3}}+4\right) \,\hat{\mathbf{j}}, & 3 \lt t\le 6 \\
1\,\hat{\mathbf{i}} + (12 - t)\,\hat{\mathbf{j}} & 6 \lt t \le 11
\end{cases} \nonumber\]
Be sure to verify that this single vector-valued function does indeed trace out the entire path!
Limits and Continuity of a Vector-Valued Function
We now take a look at the limit of a vector-valued function. This is important to understand to study the calculus of vector-valued functions.
A vector-valued function \(\vecs r\) approaches the limit \(\vecs L\) as \(t\) approaches \(a\), written
\[\lim \limits_{t \to a} \vecs r(t) = \vecs L,\]
\[\lim \limits_{t \to a} \big\| \vecs r(t) - \vecs L \big\| = 0.\]
This is a rigorous definition of the limit of a vector-valued function. In practice, we use the following theorem:
Let \(f\), \(g\), and \(h\) be functions of \(t\). Then the limit of the vector-valued function \(\vecs r(t)=f(t) \hat{\mathbf{i}}+g(t) \hat{\mathbf{j}}\) as t approaches a is given by
\[\lim \limits_{t \to a} \vecs r(t) = [\lim \limits_{t \to a} f(t)] \hat{\mathbf{i}} + [\lim \limits_{t \to a} g(t)] \hat{\mathbf{j}} , \label{Th1}\]
provided the limits \(\lim \limits_{t \to a} f(t)\) and \(\lim \limits_{t \to a} g(t)\) exist.
Similarly, the limit of the vector-valued function \(\vecs r(t)=f(t) \hat{\mathbf{i}}+g(t) \hat{\mathbf{j}}+h(t) \hat{\mathbf{k}}\) as \(t\) approaches \(a\) is given by
\[\lim \limits_{t \to a} \vecs r(t) = [\lim \limits_{t \to a} f(t)] \hat{\mathbf{i}} + [\lim \limits_{t \to a} g(t)] \hat{\mathbf{j}} +[\lim \limits_{t \to a} h(t)] \hat{\mathbf{k}} , \label{Th2}\]
provided the limits \(\lim \limits_{t \to a} f(t)\), \(\lim \limits_{t \to a} g(t)\) and \(\lim \limits_{t \to a} h(t)\) exist.
In the following example, we show how to calculate the limit of a vector-valued function.
For each of the following vector-valued functions, calculate \(\lim \limits_{t \to 3}\vecs r(t)\) for
1. \(\vecs r(t)=(t^2−3t+4) \hat{\mathbf{i}}+(4t+3)\hat{\mathbf{j}}\)
2. \(\vecs r(t)=\frac{2t−4}{t+1}\hat{\mathbf{i}}+\frac{t}{t^2+1} \hat{\mathbf{j}}+(4t−3) \hat{\mathbf{k}}\)
1. Use Equation \ref{Th1} and substitute the value \(t=3\) into the two component expressions:
\[ \begin{align*} \lim \limits_{t \to 3} \vecs r(t) \; & = \lim \limits_{t \to 3} \left[(t^2−3t+4) \hat{\mathbf{i}} + (4t+3) \hat{\mathbf{j}}\right] \\[5pt] & = \left[\lim \limits_{t \to 3} (t^
2−3t+4)\right]\hat{\mathbf{i}}+\left[\lim \limits_{t \to 3} (4t+3)\right] \hat{\mathbf{j}} \\[5pt] & = 4 \hat{\mathbf{i}}+15 \hat{\mathbf{j}} \end{align*}\]
2. Use Equation \ref{Th2} and substitute the value \(t=3\) into the three component expressions:
\[ \begin{align*} \lim \limits_{t \to 3} \vecs r(t) \; & = \lim \limits_{t \to 3}\left(\dfrac{2t−4}{t+1}\hat{\mathbf{i}}+\dfrac{t}{t^2+1}\hat{\mathbf{j}}+(4t−3) \hat{\mathbf{k}}\right) \\[5pt] & = \
left[\lim \limits_{t \to 3} \left(\dfrac{2t−4}{t+1}\right)\right]\hat{\mathbf{i}}+\left[\lim \limits_{t \to 3} \left(\dfrac{t}{t^2+1}\right)\right] \hat{\mathbf{j}} +\left[\lim \limits_{t \to 3}
(4t−3)\right] \hat{\mathbf{k}} \\[5pt] & = \tfrac{1}{2} \hat{\mathbf{i}}+\tfrac{3}{10}\hat{\mathbf{j}}+9 \hat{\mathbf{k}} \end{align*}\]
Calculate \(\lim \limits_{t \to 2} \vecs r(t)\) for the function \(\vecs r(t) = \sqrt{t^2 + 3t - 1}\,\hat{\mathbf{i}}−(4t-3)\,\hat{\mathbf{j}}− \sin \frac{(t+1)\pi}{2}\,\hat{\mathbf{k}}\)
Use Equation \ref{Th2} from the preceding theorem.
\[\lim \limits_{t \to 2} \vecs r(t) = 3\hat{\mathbf{i}}−5\hat{\mathbf{j}}+\hat{\mathbf{k}}\]
Now that we know how to calculate the limit of a vector-valued function, we can define continuity at a point for such a function.
Let \(f\), \(g\), and \(h\) be functions of \(t\). Then, the vector-valued function \(\vecs r(t)=f(t) \hat{\mathbf{i}}+g(t)\hat{\mathbf{j}}\) is continuous at point \(t=a\) if the following three
conditions hold:
1. \(\vecs r(a)\) exists
2. \(\lim \limits_{t \to a} \vecs r(t)\) exists
3. \(\lim \limits_{t \to a} \vecs r(t) = \vecs r(a)\)
Similarly, the vector-valued function \(\vecs r(t)=f(t) \hat{\mathbf{i}}+g(t)\hat{\mathbf{j}}+h(t)\hat{\mathbf{k}}\) is continuous at point \(t=a\) if the following three conditions hold:
1. \(\vecs r(a)\) exists
2. \(\lim \limits_{t \to a} \vecs r(t)\) exists
3. \(\lim \limits_{t \to a} \vecs r(t) = \vecs r(a)\)
• A vector-valued function is a function of the form \(\vecs r(t)=f(t) \hat{\mathbf{i}}+ g(t) \hat{\mathbf{j}}\) or \(\vecs r(t)=f(t) \hat{\mathbf{i}}+g(t) \hat{\mathbf{j}}+h(t) \hat{\mathbf{k}}\),
where the component functions \(f\), \(g\), and \(h\) are real-valued functions of the parameter \(t\).
• The graph of a vector-valued function of the form \(\vecs r(t)=f(t) \hat{\mathbf{i}}+g(t) \hat{\mathbf{j}}\) is called a plane curve. The graph of a vector-valued function of the form \(\vecs r
(t)=f(t)\hat{\mathbf{i}}+g(t)\hat{\mathbf{j}}+h(t) \hat{\mathbf{k}}\) is called a space curve.
• It is possible to represent an arbitrary plane curve by a vector-valued function.
• To calculate the limit of a vector-valued function, calculate the limits of the component functions separately.
Key Equations
• Vector-valued function
\(\vecs r(t)=f(t) \hat{\mathbf{i}}+g(t) \hat{\mathbf{j}}\) or \(\vecs r(t)=f(t) \hat{\mathbf{i}}+g(t) \hat{\mathbf{j}}+h(t) \hat{\mathbf{k}}\),or \(\vecs r(t)=⟨f(t),g(t)⟩\) or \(\vecs r(t)=⟨f
• Limit of a vector-valued function
\(\lim \limits_{t \to a} \vecs r(t) = [\lim \limits_{t \to a} f(t)] \hat{\mathbf{i}} + [\lim \limits_{t \to a} g(t)] \hat{\mathbf{j}}\) or \(\lim \limits_{t \to a} \vecs r(t) = [\lim \limits_{t \
to a} f(t)] \hat{\mathbf{i}} + [\lim \limits_{t \to a} g(t)] \hat{\mathbf{j}} + [\lim \limits_{t \to a} h(t)] \hat{\mathbf{k}}\)
component functions
the component functions of the vector-valued function \(\vecs r(t)=f(t)\hat{\mathbf{i}}+g(t)\hat{\mathbf{j}}\) are \(f(t)\) and \(g(t)\), and the component functions of the vector-valued function
\(\vecs r(t)=f(t)\hat{\mathbf{i}}+g(t)\hat{\mathbf{j}}+h(t)\hat{\mathbf{k}}\) are \(f(t)\), \(g(t)\) and \(h(t)\)
a three-dimensional curve in the shape of a spiral
limit of a vector-valued function
a vector-valued function \(\vecs r(t)\) has a limit \(\vecs L\) as \(t\) approaches \(a\) if \(\lim \limits{t \to a} \left| \vecs r(t) - \vecs L \right| = 0\)
plane curve
the set of ordered pairs \((f(t),g(t))\) together with their defining parametric equations \(x=f(t)\) and \(y=g(t)\)
an alternative parameterization of a given vector-valued function
space curve
the set of ordered triples \((f(t),g(t),h(t))\) together with their defining parametric equations \(x=f(t)\), \(y=g(t)\) and \(z=h(t)\)
vector parameterization
any representation of a plane or space curve using a vector-valued function
vector-valued function
a function of the form \(\vecs r(t)=f(t)\hat{\mathbf{i}}+g(t)\hat{\mathbf{j}}\) or \(\vecs r(t)=f(t)\hat{\mathbf{i}}+g(t)\hat{\mathbf{j}}+h(t)\hat{\mathbf{k}}\),where the component functions \(f
\), \(g\), and \(h\) are real-valued functions of the parameter \(t\).
• Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.
• Edited by Paul Seeburger (Monroe Community College)
• Paul Seeburger created Example \(\PageIndex{1}\), Exercise \(\PageIndex{1}\), and the subsections titled: Finding a Vector-Valued Function to Trace out the Graph of a Function \(y = f(x)\),
Finding a Vector-Valued Function to Trace out the Graph of an Equation in \(x\) and \(y\) and Vice Versa, and Parameterizing a Piecewise Path. | {"url":"https://math.libretexts.org/Courses/Montana_State_University/M273%3A_Multivariable_Calculus/13%3A_Vector-valued_Functions/13.1%3A_Vector-Valued_Functions_and_Space_Curves","timestamp":"2024-11-08T01:31:36Z","content_type":"text/html","content_length":"197418","record_id":"<urn:uuid:7b02d15c-f119-4bb5-a938-6a7b6defe122>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00138.warc.gz"} |
How to analyze haplotype block structure in genetic association studies and biostatistics? | Hire Some To Take My Statistics Exam
How to analyze haplotype block structure in genetic association studies and biostatistics? Interdisciplinary research in genetics. {#s11} 2.1. Algorithms and design methods {#s12} ———————————– The
proposed methods are divided in two parts. One part sets out our goal of extracting the haplotype blocks from a complex multilodel probabilistic population consisting of alleles that are randomly
distributed among several independent haplotypes. The other part looks up haplotypes from multiple independent haplotypes using the Bayesian framework and the Schüttner-Welcher method to assess the
degree of heritability, and this section describes a selected step in the read this of a haplotype block from multiple loci. First we add a set of 10 different values $\epsilon_0$ into $\delta_{\
epsilon_0}$ in order to obtain the final block. Denote each $H_0$ a set of $M$ loci $X$ with distinct genetic frequencies $\chi_0$. Then let $\left\{h_i, \chi_0\right\}_{i=1}^M$ be a sequence of the
haplotype blocks $h_i$ of $X$, and only those five haplotypes of $X$ with scores $\left\{1_{H_0},\, 2_{H_0},\,1_{H_0….H_0}\right\}$ that are within $\left\{h_i\right\}$ are included in the
corresponding block. Denote the set of points in $\left\{h_i\right\}$ before the first base visit this site $(\epsilon_0)$ by $X_p=X_{h_0}$, the set of points in $\left\{h_i\right\}$ after the first
base change $(\epsilon_0)$ by $X_{\chi_0}=X_{h_0}$, and the set of points in $\left\{h_i\right\}$ after the first base change by $X_{\chi_0}$. Define the distribution $\mid h_i\mid$ by $$\mathbb{P}\
left(x, H_0\mid h_i\right)=\pi_0\left\{ h_i\mid h_i\geq \lambda_0\right\}$$ where $\lambda_0=\sum h_i$, $h_i$ is the haplotype coefficient of $X_X$’s base change $(\epsilon_0)$ and $H_0$ (not
included) is the true haplotype of $X$. Then we have $h_i=\left\{h_i,\, 1_{H_0}\right\}$, where $1_H=\How to analyze haplotype block structure in genetic association studies and biostatistics? Vetal
cells from two populations can be compared by means of two-dimensional images. A recent study by several authors (Fitzford and look at this web-site 2002; Muhlenberg and Kleinig, 2002 [2010]) has
used different three-dimensional reconstruction techniques to find similar morphological blocks/defects in pairs of genetic populations. This work has focused on comparing two populations and several
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This type of image procedure is often referred to as genetic image set-based morphological reconstruction (ge-IMR) (Buck et al., 2010). (2) Two-dimensional reconstruction, which uses mathematical
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active field of research and one can find details in K. Fukuyama et al. (2013). More about the author The purpose of this work is to identify the project help blocks in a disease-pathway association
study each genotype can be compared with multiple genetic population samples. For a more quantitative comparison between the two populations, a comprehensive and simplified map based on genetic
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several species such as avian and bird as well as in arbovirus models which have been targeted with different methods for genetic association studies. However, there areHow to analyze haplotype block
structure in genetic association studies and biostatistics? Non-null explanation haplotype blocks in association studies are composed of several polymorphisms of various genotypes influencing
haplotype blocks in genetic association studies (IA) and biostatistics, are described. Most (70%) described haplotype blocks are composed of three to five polymorphisms. With the exception of a
single polymorphism, several polymorphisms can be selected among each other in these haplotype blocks (8%), in order to concentrate the hypotheses on their functional effect on haplotype basis. In
this paper, the main method proposed by Choi et al. (1987), which are suitable in the literature is by constructing a haplotype block by analyzing haplotype blocks from nucleotide sequence analysis
related to various polymorphisms. A “data partitioning” is performed for each polymorphism when the polymorphism data is analyzed at the level of haplotype block partitions (1).
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In Kimura’s statistical analyses in molecular genetics, the mean allele frequencies of haplotype blocks are used as a measure of different hypotheses and haplotype blocks are analyzed by using the
Markov process of Fischer’s Markov Chain model (MP; Sarma et al, 1996, 1996). For purposes of using these “data partitioning” we adopted a combination of three different statistics and three
different haplotype blocks of a pair of variables of several biological groups: genetic capacity theory, haplotype block size, and the nonnull allele block size in association studies, as our main
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combine our three statistical tests for haplotype information in a few | {"url":"https://hireforstatisticsexam.com/how-to-analyze-haplotype-block-structure-in-genetic-association-studies-and-biostatistics","timestamp":"2024-11-07T04:22:04Z","content_type":"text/html","content_length":"170004","record_id":"<urn:uuid:c1c04ecf-0368-4246-985b-5291de2ea57e>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00179.warc.gz"} |
CGS (Collisionless Galactic Simulator) uses Fourier techniques to solve the Possion equation ∇^2Φ = 4πGρ, relating the mean potential Φ of a system to the mass density ρ. The angular dependence of
the force is treated exactly in terms of the single-particle Legendre polynomials, which preserves accuracy and avoids systematic errors. The density is assigned to a radial grid by means of a
cloud-in-cell scheme with a linear kernel, i.e., a particle contributes to the density of the two closest cells with a weight depending linearly on the distance from the center of the cell
considered. The same kernel is then used to assign the force from the grid to the particle. The time step is chosen adaptively in such a way that particles are not allowed to cross more than one
radial cell during one step. CGS is based on van Albada's code (1982) and is distributed in the NEMO (ascl:1010.051) Stellar Dynamics Toolbox. | {"url":"https://ascl.net/1904.003","timestamp":"2024-11-11T05:04:34Z","content_type":"text/html","content_length":"9087","record_id":"<urn:uuid:a02ac581-c5cf-46f9-8ebb-bb52d322c74c>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00318.warc.gz"} |
Boyle's Law: J-Tube
The relationship between the volume of a gas and the pressure it exerts, known as Boyle's Law, is shown with a J-tube.
The relationship between the volume of a gas and the pressure it exerts can be studied by adding mercury to a J-tube until the height of the mercury is equal on both sides and a sample of gas is
trapped. The pressure in the closed portion of the tube now equals atmospheric pressure. The cylinder of trapped air is 34 centimeters high, and its volume is proportional to its height.
The first addition of mercury raises its height on the left to 8.6 centimeters. The pressure of the trapped air goes up and prevents the mercury from rising above 2.2 centimeters on the right. Adding
25.9 centimeters of mercury on the left gives a height of 6.7 centimeters on the right, and 45.9 centimeters corresponds to 10.7 centimeters.
The difference between heights of mercury plus atmospheric pressure equals the pressure on the gas. By plotting the experimental values for the pressure vs. the reciprocal of the volume of air, we
get a straight line. The inverse proportionality between volume and pressure is known as Boyle’s law.
• Design and Demonstration
□ Kristin Johnson University of Wisconsin - Madison, Madison, WI 53706
• Text
□ John W. Moore University of Wisconsin - Madison, Madison, WI 53706
□ Kristin Johnson University of Wisconsin - Madison, Madison, WI 53706
• Voice
□ Jerrold J. Jacobsen University of Wisconsin - Madison, Madison, WI 53706
• Video Production and Editing
□ Greg Minix College of Engineering, University of Wisconsin - Madison, Madison, WI 53706
□ Jerrold J. Jacobsen University of Wisconsin - Madison, Madison, WI 53706 | {"url":"https://www.chemedx.org/video/boyles-law-j-tube","timestamp":"2024-11-02T17:48:09Z","content_type":"text/html","content_length":"29335","record_id":"<urn:uuid:1d25e95b-3403-424c-b2cf-b37cef3380cc>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00033.warc.gz"} |
A Study of Stellar Orbits in a Rotating, Gaseous Bar.
Document Type
Degree Name
Doctor of Philosophy (PhD)
Physics and Astronomy
First Advisor
Joel E. Tohline
In an effort to better understand the formation and evolution of barred galaxies, the properties of orbits in the effective potential of one specific model of a rapidly rotating, steady-state
gas-dynamical bar that has been constructed via a self-consistent hydrodynamical simulation have been examined. This bar is used to test the following idea. If primordial galaxies evolve to a rapidly
rotating, bar-like configuration before a significant amount of star formation has taken place, and then stars form from the gas that makes up the bar, the initial stellar distribution function
should be much different than those used in previous bar formation studies. As a first step towards understanding such a distribution function, orbits in the two-dimensional, equatorial slice of the
above mentioned bar are studied. Orbits that result from a systematic search of initial conditions are compared to orbits that have initial conditions determined by the Restriction Hypothesis. The
Restriction Hypothesis is the implementation of the idea that stars are forming from the gas that makes up the bar. Specifically, the initial velocities of Restriction Hypothesis orbits are set equal
to the known gas velocities at the points of formation. It is found that Restriction Hypothesis orbits are a subset of all possible orbits and that the most important regular orbit family has a "bow
tie" shape. These orbits are vastly different than the main family of orbits previously thought to sustain bar shaped distributions. Extending the Restriction Hypothesis to the fully
three-dimensional bar potential, a method of characterizing the resulting orbits based on the number of conserved quantities respected by the orbits has been utilized. These conserved quantities are
known as integrals of motion and are related to the number of dimensions that a phase space orbit exists in. This technique is found to be robust and provides a straightforward way of categorizing
orbits. Using this technique, it has been determined that a large percentage of examined three-dimensional Restriction Hypothesis orbits respect at least two integrals of motion.
Recommended Citation
Barnes, Eric Ian, "A Study of Stellar Orbits in a Rotating, Gaseous Bar." (2001). LSU Historical Dissertations and Theses. 263. | {"url":"https://repository.lsu.edu/gradschool_disstheses/263/","timestamp":"2024-11-08T07:20:34Z","content_type":"text/html","content_length":"39083","record_id":"<urn:uuid:428bc78d-5b08-4051-92ef-2dacacd19fe1>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00806.warc.gz"} |
Approximate and compare irrational numbers | Stage 3 Maths | HK Secondary S1-S3
As we have seen before, surds can be manipulated into simpler forms. For example, $\sqrt{250}$√250 can be rewritten as $5\sqrt{10}$5√10. When we rewrite a surd in this way we do not lose any
accuracy, we are simply changing the way we have expressed the surd by finding a factor of $250$250 that is a square number, and then evaluating it. Therefore, we can say that $5\sqrt{10}$5√10 still
has the exact value of $\sqrt{250}$√250.
However, sometimes we want to give an answer in decimal form to understand how big it is more easily. When we do this, we will need to use a calculator. Try typing $\sqrt{250}$√250 or $5\sqrt{10}$5√1
0 into a calculator, and you will get $15.8113883$15.8113883. You should note that this is not the exact value! The calculator has given you a rounded number, as it can only display a certain number
of digits. In reality, $5\sqrt{10}$5√10 is an irrational number! We may not need to be this accurate as the calculator though, and we are often asked to round to 2 decimal places. So $\sqrt{250}$√250
is almost equivalent to $15.81$15.81. We could write this as $\sqrt{250}$√250 ≈ $15.81$15.81.
There're also some ways to approximate surds without using a calculator directly. One way is to consider the nearest integer value as a way to check our workings. This is thanks to a rule that
if $aa<b,
then $\sqrt{a}<\sqrt{b}$√a<√b
This is pretty obvious, so let's see how it can help us to approximate. Let's say we have a surd $\sqrt{40}$√40. If we ask ourselves what are the closest square numbers that are bigger and smaller
than $40$40, then we'll find that they're $36$36 and $49$49. So then we have $36<40<49$36<40<49, which leads us to say that $\sqrt{36}$√36$<$<$\sqrt{40}$√40$<$<$\sqrt{49}$√49. And if we evaluate that
further we get $6$6$<$<$\sqrt{40}$√40$<$<$7$7, so we've managed to narrow this surd down to somewhere between $6$6 and $7$7!
What if we wanted to approximate it further and see if we can get a decimal? There's a method for that as well! Once you know what integers the surd lies between, you can find the decimal part by
using the following formula:
decimal part of approximation = $\frac{\text{number inside surd - closest smaller square}}{\text{closest bigger square - closest smaller square}}$number inside surd - closest smaller squareclosest
bigger square - closest smaller square
This means that, still using our $\sqrt{40}$√40 example, the decimal part would be equal to $\frac{40-36}{49-36}$40−3649−36≈$0.3$0.3 so $\sqrt{40}$√40 can be approximated to $6.3$6.3. If you plug
this surd into a calculator, you'll see that it is indeed rounded to $6.3$6.3! However this method only works well on larger numbers, and bigger they are the better they'll work! Try and see the
difference between using this on say, $\sqrt{2}$√2 and $\sqrt{300}$√300
Question 1
The value of $\sqrt{74}$√74 lies between two consecutive integers.
1. Between which two consecutive perfect square numbers does $74$74 lie between?
Complete the inequality.
2. Between which two consecutive integers does $\sqrt{74}$√74 lie?
Complete the inequality.
question 2
Find the largest value out of the following
A) $2\pi$2π B) $\sqrt{50}$√50 C) $4.21$4.21 D) $\sqrt{49}$√49
Think: Let's convert all the numbers to decimals or approximate them using decimals, so that we can compare them. Remember that $\pi$π is approximately $3.14$3.14.
$2\pi$2π ≈ $2\times3.14$2×3.14
$=$= $6.28$6.28
$\sqrt{50}$√50 is bigger than $\sqrt{49}$√49 but smaller than $\sqrt{64}$√64 so we can say it's between $7$7 and $8$8.
$\sqrt{49}$√49 is not actually a surd in that we can evaluate it to $7$7 exactly.
Therefore the biggest value is $\sqrt{50}$√50.
Question 3
Approximate $\sqrt{95}$√95 to the nearest hundredth without using a calculator.
Think: Hundredths are represented by the second decimal place, so we need two decimal places. Since $95$95 is a large number, we can use the formula above to approximate its value.
So now we know the surd is equal to $9$9 point something.
For the decimal part:
$\frac{\text{number inside surd - closest smaller square}}{\text{closest bigger square - closest smaller square}}$number inside surd - closest smaller squareclosest $=$ $\frac{95-81}{100-81}$95−811
bigger square - closest smaller square = 00−81
$=$ $\frac{14}{19}$1419
≈ $0.74$0.74
Therefore $\sqrt{95}$√95 is approximately $9.74$9.74.
question 4
Andrea needs fences for all $4$4 sides for each of her $3$3 paddocks. If each paddock is square and has an area of $27$27 m^2, how many metres of fencing would she need (to the nearest whole number)?
Think: We need to figure out what the perimeter is for one paddock and then multiply it by the number of paddocks.
One square paddock is $27$27 m^2 so $x^2=27$x2=27 where $x$x is the length of one of its sides. That means $x=\sqrt{27}$x=√27.
Using our calculator, this surd is approximately $5.2$5.2.
That means one paddock needs $4\times5.2=20.8$4×5.2=20.8 m of fencing.
Therefore she needs $3\times20.8=62$3×20.8=62 m (nearest whole number) of fencing in total. | {"url":"https://mathspace.co/textbooks/syllabuses/Syllabus-98/topics/Topic-1472/subtopics/Subtopic-17348/?activeTab=theory","timestamp":"2024-11-12T09:27:21Z","content_type":"text/html","content_length":"471560","record_id":"<urn:uuid:a74dc139-8c52-4029-b61f-4b3c695cd0de>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00628.warc.gz"} |
On some weighted average values of l-functions
Let q≥2 and N≥1 be integers. W.Zhang recently proved that for any fixed ε>0 and q^ε≤N≤q^1/2-ε, \[ ∑ χ ≠ χ 0} ∑ ^N[n=1] χ (n)^2 |L(1, χ )|^2 = (1 + o(1)) α[q] q N, where the sum is taken over all
nonprincipal characters modulo q, L(1,) denotes the L-functions corresponding to χ, and α[q]=q^o(1) is some explicit function of q. Here we improve this result and show that the same asymptotic
formula holds in the essentially full range qNq^1.
Bibliographical note
Copyright 2009 Cambridge University Press. Article originally published in Bulletin of the Australian Mathematical Society, Vol. 79 No. 2, pp 183-186. The original article can be found at http://
Dive into the research topics of 'On some weighted average values of l-functions'. Together they form a unique fingerprint. | {"url":"https://researchers.mq.edu.au/en/publications/on-some-weighted-average-values-of-l-functions","timestamp":"2024-11-11T01:20:54Z","content_type":"text/html","content_length":"51748","record_id":"<urn:uuid:cb7a33e5-f188-4c19-947a-ba631bf7a2a1>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00282.warc.gz"} |
the STACK logo
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The mathematics of the STACK logo
The STACK logo is based on the following problem.
If you stack identical blocks one on top of the other, how far can it lean before it falls over? You have a potentially unlimited supply of blocks.
The answer is that there is no limit to how far it can lean! A physical model is shown below.
This model was made by Dr John Bryant.
The mathematics of this problem
To see why the STACK of blocks doesn't fall over assume that the width of each domino is \(2\) "units". Our strategy is this: at each stage we consider an existing balancing stack of \(n\) dominoes
which has its centre of mass a distance \(c_n\) from its left-hand edge.
Obviously \(c_n\leq 2\) for all \(n\) as the centre of mass is to be above the bottom domino! We then place this stack on top of a new domino a distance \(\delta_n\) from the left of the domino.
There will clearly be no toppling if
\[ \delta_n+c_n\leq 2.\]
That is to say we maintain balance if we don't displace the top stack so far that the displacement plus the distance of the centre of mass from the left pushes the centre of mass over the edge of the
bottom domino - which has width \(2\). The new centre of mass of the whole stack of \(n+1\) dominoes will be \(c_{n+1}\) from the left of the bottom domino where
\[ c_{n+1} = \frac{(\delta_n+c_n)n+1}{n+1} \text{ with } c_1=1.\]
Using the first inequality the maximum displacement without toppling is \(\delta_n=s-c_n\). Combining this with the formula for \(c_{n+1}\) and solving for \(\delta_n\) gives
\[ \delta_{n+1} = 2-c_{n+1} = 2-\frac{(\delta_n+c_n)n+1}{n+1} = 2-\frac{(\delta_n+2-\delta_n)n+1}{n+1} = \frac{1}{n+1} \text{ with } \delta_1=1. \]
So that for all \(n\), \(\delta_n = \frac{1}{n}\).
How big does the displacement become?
The question becomes, what is the value of
\[ 1 +\frac{1}{2} +\frac{1}{3} +\frac{1}{4} + \cdots +\frac{1}{N}= \sum_{n=1}^N \frac{1}{n} \]
for large N? If \(N\rightarrow \infty\), this is a particularly famous infinite series - the harmonic series. Actually, this diverges. That is to say it is possible to make the sum as large as one
would wish. To see this, we group the terms as follows,
\[ \sum_{n=1}^N \frac{1}{n} = \left(1+\cdots+\frac{1}{9}\right) + \left(\frac{1}{10}+\cdots+\frac{1}{99}\right) + \left(\frac{1}{100}+\cdots+\frac{1}{999}\right) + \cdots \]
\[ \geq 9\frac{1}{10} + 90\frac{1}{100} + 900\frac{1}{1000} + \cdots \]
\[ = \frac{9}{10}+\frac{9}{10}+\frac{9}{10}+\cdots\]
which shows that the series keeps getting larger as we continue to add terms. That is to say it does not converge.
In terms of the domino problem: we can choose displacements \(\delta_n\) so that (i) the stack does not topple over, and (ii) we can produce an arbitrarily large horizontal displacement. Bizarre
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2 Times Table Worksheets - Worksheets Day
Two times table worksheets are a helpful tool for children to learn and practice multiplication. These worksheets focus on the multiplication facts of the number 2. By using these worksheets,
children can develop a strong foundation in multiplication, which is essential for their mathematical skills.
Understanding the concept of multiplication
Multiplication is a mathematical operation that combines equal groups or sets to find the total quantity. For example, if we have 2 groups of 3 apples each, we can find the total number of apples by
multiplying 2 and 3, which gives us 6 apples in total. This concept of repeated addition helps us understand multiplication better.
Benefits of using 2 times table worksheets
1. Visual representation: These worksheets provide visual representations of multiplication, making it easier for children to understand the concept.
2. Practice and reinforcement: Regular practice with 2 times table worksheets helps children memorize the multiplication facts of 2 and reinforces their understanding of the concept.
3. Improved mental calculations: By repeatedly solving problems on these worksheets, children can enhance their mental calculation skills, which are useful in everyday life.
4. Preparation for higher-level math: Understanding multiplication lays the foundation for more complex mathematical concepts, such as division, fractions, and algebra. Using 2 times table worksheets
is a stepping stone towards mastering these advanced topics.
Content of 2 times table worksheets
These worksheets typically include a variety of exercises to engage children in different ways. Some common types of questions found in 2 times table worksheets are:
1. Multiplication sentences: Children are given a number and are asked to write the multiplication sentence for that number. For example, if the number is 4, they would write “4 x 2 = 8.”
2. Filling in the missing numbers: In these exercises, children are given a partial multiplication sentence and need to fill in the missing number. For instance, if they are given “_ x 2 = 10,” they
would write “5” in the blank space.
3. Word problems: Word problems help children apply their multiplication skills to real-life situations. These problems involve scenarios where multiplication is needed to find the solution. For
example, “If John buys 2 packs of pencils, each containing 6 pencils, how many pencils does he have in total?”
4. Coloring activities: Some worksheets incorporate coloring activities to make the learning process more enjoyable. Children may need to solve multiplication problems and then color the
corresponding numbers or shapes based on the answers.
Two times table worksheets are a valuable resource for children to learn and practice multiplication. They provide a structured and engaging way to develop multiplication skills, enhance mental
calculations, and prepare for more advanced mathematical concepts. By using these worksheets regularly, children can build a strong foundation in multiplication, which will benefit them throughout
their academic journey.
The Two Times Table Worksheet
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Md2.pdf [yl4z4jm5j7qr]
MACHINE DESIGN -II Mechanical Engineering, Semester –VII, University of Mumbai
PROF. SANJAY W. RUKHANDE FR. C. RODRIGUES INSTITUTE OF TECHNOLOGY, VASHI NAVI MUMBAI
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
It gives me immense pleasure to present this compilation on Machine Design –II. This content has been compiled specially for Final Year Semester VII students of Mechanical Engineering in University
of Mumbai. Numerous solved design have been added for the benefit of student community and teaching faculty. Design Data Book by PSG is referred for design. Despite my best efforts, should some
mistakes have crept in, these may kindly be brought to my notice. I welcome constructive criticism for further improvement of this compilation. I would like to express my thanks to all my students
whose continuous feedback is the source of inspiration. A feedback in the form of suggestion and comments from the readers will be highly appreciated.
Sanjay W. Rukhande Fr. C. Rodrigues Institute of Technology, Vashi, Navi Mumbai
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
SR. No.
Page No.
Module 1. GEAR DESIGN
Spur Gear
Helical Gear
Bevel Gear
Worm and worm wheel
Gear box
Design problem 1 (Spur Gear)
Design problem 2 (Spur Gear)
Design problem 3 (Helical Gear)
Design problem 4 (Bevel Gear)
Design problem 5 (Worm and worm wheel)
Module 2. ROLLING CONTACT BEARING Numerical ( 1 to 7) Module 3. SLIDING CONTACT BEARING Numerical (1 to 3) Module 4 : CAM AND ROLLER FOLLOWER MECHANISM
Numerical 1.
Numerical 2.
Module 5. BELT AND CHAIN DRIVE
Flat Belt design
Numerical (1and 2)
V-Belt design
Numerical 1
Chain Drives
Numerical (1 and 2)
Module 6. CLUTCH DESIGN
Numerical (1 to 3)
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Course Code: MEC701/Subject: Machine Design - II/Credits: 4+1 Objectives 1. To study functional and strength design of important machine elements 2. To study selection of rolling element bearing and
design of hydrodynamic bearing. Outcomes: Learner will be able to… 1. Select appropriate gears for power transmission on the basis of given load and speed. 2. Design gears based on the given
conditions. 3. Select bearings for a given applications from the manufacturers catalogue. 4. Select and/or design belts for given applications. 5. Design cam and follower and clutches Modules 01
Detailed Content
Design of spur, helical, bevel and worm gears with strength, wear and thermal considerations. Two stage Gear box with fixed ratio consisting of spur, helical and bevel gear pairs: gear box housing
layout and housing design. Types of bearing and designation, Selection of rolling contact bearings based on constant / variable load & speed conditions (includes deep groove ball bearing, cylindrical
roller, spherical roller, taper roller, self-aligning bearing and thrust bearing). Design of hydro dynamically lubricated bearings (Self-contained) Introduction to hydro static bearings Types and
selection of Mechanical Seals Design of cam and roller follower mechanisms with spring and shaft.
Design and selection of Belts: - Flat and V belt with Pulley 08 construction. Design and selection of standard roller chains. Design of single plate, multiplate and cone clutches, with spring, lever
08 design and thermal, wear considerations.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
MODULE: 1 GEARS Introduction to Spur Gear: Spur gears or straight-cut gears are the simplest type of gear. They consist of a cylinder or disk with teeth projecting radially. Though the teeth are not
straight-sided but usually of special form to achieve a constant drive ratio, mainly involute but less commonly cycloidal, the edge of each tooth is straight and aligned parallel to the axis of
rotation. These gears mesh together correctly only if fitted to parallel shafts. No axial thrust is created by the tooth loads. Spur gears are excellent at moderate speeds but tend to be noisy at
high speeds. Spur gear is the most common type of gear used. It is used for transmitting power between two parallel shafts. In this type of gear, the teeth cut on the cylindrical face are parallel to
the axis of the gear. Though several types of profiles can be used for cutting teeth on a spur gear, we commonly use involute teeth profile. A spur gear drive is a combination of two spur gears
properly meshed with each other. It is used for transmitting rotational motion between parallel shafts. It offers a definite velocity ratio.
If the driving gear is smaller the than the driven gear, then rotating velocity is reduced in the driven gear. If the driving gear is larger than the driven gear, rotational velocity is increased in
the driven gear. If both the driving and driven gears have the same number of teeth, there is no change in the rotational velocity. In a spur gear drive, the smaller gear is called pinion and the
larger one is called gear. When designing a spur gear drive, the pinion is made harder than the gear because the pinion has to run more no. of cycles as compared to the gear.
Advantages and Disadvantages of Gear Drive over Belt and Chain Drive: Advantage:
Transmission efficiency is high. Large power can be easily transmitted. Maintenance is easy. Gear drives are compact. They have good durability and precision. 4
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Disadvantage: Gear drives are more costly as compared with chain drive. Installation is difficult. Manufacturing of gears is complex and expensive. Tooth wear may occur during power
transmission. Inaccuracies in gear teeth causes noise and vibrations.
Classification of Gears:
Gears can be classified into many types based on several criteria. The classification of gears is listed below: 1. Based on the peripheral velocity of gears A. Low velocity gears – Gears with
peripheral velocity < 3 m/s B. Medium velocity gears – Gears with peripheral velocity = 3-15 m/s C. High velocity gears – Gears with peripheral velocity > 15 m/s 2. Based on the position of axes of
revolution A. Gears with parallel axes I. Spur gear II. Helical Gear a) Single Helical Gear, b) Double Helical Gear (or) Herringbone Gear B. Gears with intersecting axes a. Bevel Gear i. Straight
bevel gear ii. Spiral bevel gear iii. Zerol bevel gear iv. Hypoid bevel gear b. Angular gear c. Mitre gear C. Gears with non-parallel and non-intersecting axes a. Worm gear i. Non-throated worm gear
ii. Single-throated worm gear iii. Double-throated worm gear b. Hypoid gear c. Screw gear (or crossed helical gear) 3. Based on the type of gearing a. Internal gear, b. External gear, c. Rack and
Pinion 4. Based on the tooth profile on the gear surface 5
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
a. Gears with straight teeth b. Gears with curved teeth c. Gears with inclined teeth Conjugate action: The gears must be designed such that the ratio of rotational speeds of driven and driver
gear is always constant. When the tooth profiles of two meshing gears produce a constant angular velocity during meshing, they are said to be executing conjugate action. That is (ω1 / ω2 ) =
constant, where ω1 = Angular velocity of the driver. ω2 = Angular velocity of the driven. Gears are mostly designed to produce conjugate action. Theoretically, it is possible to selection arbitrary
profile for one tooth and then to find a profile for the meshing tooth, which will give conjugate action. One of these solutions is involute profile. The involute profile is universally used for
constructing gear teeth with few exceptions.
Law of gearing:
The law of gearing states that the angular velocity ratio of all gears of a meshed gear system must remain constant also the common normal at the point of contact must pass through the pitch point.
Example: if ω1 and ω2 are the angular velocities and D1 and D2 are the diameters of two gears meshed together then
Different Pressure angle systems:
Pressure angle in relation to gear teeth, also known as the angle of obliquity, is the angle between the tooth face and the gear wheel tangent. It is more precisely the angle at a pitch point between
the line of pressure (which is normal to the tooth surface) and the plane tangent to the pitch surface. The pressure angle gives the direction normal to the tooth profile. The pressure angle is equal
to the profile angle at the standard pitch circle and can be termed the "standard" pressure angle at that point. Standard values are 14.5 and 20 degrees. Earlier gears with pressure angle 14.5 were
commonly used because the cosine is larger for a smaller angle, providing more power transmission and less pressure on the bearing; however, teeth with smaller pressure angles are weaker. To run
gears together properly their pressure angles must be matched. The table below shows the value of addendum, dedendum, working depth & tooth thickness in terms of module for standard pressure angle. 6
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Gear Terminology
14.50 full depth involute system
200 full depth involute system
200 stub involute system
0.8 m
1.157 m
1.25 m
0.157 m
0.25 m
0.2 m
Working Depth
1.6 m
Whole Depth
2.157 m
2.25 m
1.8 m
Tooth thickness
1.5708 m
1.5708 m
1.5708 m
Forms of teeth: There are three standard systems for the shape of teeth as shown in figure below.
14.50 full depth
200 full depth
200 stub
i. 14.50 Full Depth Involute system: The basic rack for this system is composed of straight sides except for the fillet arcs. In this system, interference occurs when the number of teeth on the
pinion is less than 23. This system is satisfactory when the number of teeth on the gears is large. If the number of teeth is small and if the gears are made by generating process, undercutting is
unavoidable. ii. 200 Full Depth Involute system: The basic rack for this system is also composed of straight sides except for the fillet arcs. In this system, interference occurs when the number of
teeth on the pinion is less than 17. The 200 pressure angle system with full 7
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
depth involute teeth is widely used in practice. Increasing pressure angle improves the tooth strength but shortens the duration of contact. Decreasing pressure angle requires more number of teeth on
the pinion to avoid undercutting. The 200 pressure angle is a good compromise for most of the power transmission as well as precision gearboxes. The 200 pressure angle system has the following
advantage over the 14.50 pressure angle system. It reduces the risk of undercutting. It reduces interference. Due to the increased pressure angle, the tooth become slightly broader at the root.
This makes the tooth stronger and increases the load carrying capacity. It has greater length of contact. iii. 200 Stub Involute system: The gears in this system have shorter addendum and shorter
dedendum. The interfering portion of the tooth, that is, a part of the addendum, is thus removed. Therefore, these teeth have still smaller interference. This also, reduces the undercutting. In this
system the minimum number of teeth on the pinion, to avoid interference, is 14. Since the pinion is small, the drive becomes more compact. Stub teeth are stronger than full depth teeth because of the
smaller moment arm of the bending force. Therefore, the stub system transmits very high load. Stub teeth results in lower production cost, as less metal must be cut away. The main drawback of this
system is that the contact ratio is reduced due to short addendum. Due to insufficient overlap, vibrations are likely to occur.
Interference in involute gear:
A gear teeth has involute profile only outside the base circle. In fact, the involute profile begins at the base circle. In some cases the dedendum is so large that it extends below this base circle.
In such situations, the portion of the tooth below the base circle is not involute. The tip of the tooth on the mating gear, which is involute, interferes with this non-involute portion of the
dedendum. This phenomenon of tooth profile overlapping and cutting into each other is called ‘interference’. In this case, the tip of the tooth overlaps and digs into the root section of its mating
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Interference is non-conjugate action and results in excessive wear, vibrations and jamming. When the gears are generated by involute rack cutters, the interference is automatically eliminated because
the cutting too remove the interfering portion of the flank. This is called ‘undercutting’.
Gear nomenclature terminology:
Pitch surface: The surface of the imaginary rolling cylinder (cone, etc.) that the toothed gear may be considered to replace. Pitch circle: A right section of the pitch surface. Addendum circle: A
circle bounding the ends of the teeth, in a right section of the gear. Root (or dedendum) circle: The circle bounding the spaces between the teeth, in a right section of the gear. Addendum: The
radial distance between the pitch circle and the addendum circle. Dedendum: The radial distance between the pitch circle and the root circle. Clearance: The difference between the dedendum of one
gear and the addendum of the mating gear.
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Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Face of a tooth: The part of the tooth surface lying outside the pitch surface. Flank of a tooth: The part of the tooth surface lying inside the pitch surface. Face width (b): Face width is the width
of the tooth measured parallel to the axis. Fillet radius: The radius that connects the root circle to the profile of the tooth is called fillet radius. Circular thickness (Tooth thickness): The
thickness of the tooth measured on the pitch circle. It is the length of an arc and not the length of a straight line. Tooth space: The distance between adjacent teeth measured on the pitch circle.
Backlash: The difference between the circle thickness of one gear and the tooth space of the mating gear. Backlash=Space width – Tooth thickness Circular pitch (p): The width of a tooth and a space,
measured on the pitch circle. Diametral pitch (P): The circular pitch, equals the pitch circumference divided by the number of teeth. The diametral pitch is, by definition, the number of teeth
divided by the pitch diameter. Module (m): Pitch diameter divided by number of teeth. The pitch diameter is usually specified in inches or millimetres; in the former case the module is the inverse of
diametral pitch. Velocity ratio: The ratio of the number of revolutions of the driving (or input) gear to the number of revolutions of the driven (or output) gear, in a unit of time. Pitch point: The
point of tangency of the pitch circles of a pair of mating gears. Common tangent: The line tangent to the pitch circle at the pitch point. Base circle: An imaginary circle used in involute gearing to
generate the involutes that form the tooth profiles.
Gear material and Heat treatment: It is essential to select proper materials and heat treatments in accordance with the intended application of the gear. Since gears are applied for various usages,
such as industrial machinery, electric/ electronic devices, household goods and toys and composed of many kinds of materials, typical materials and their heat treatment methods are introduced.
Requirements of material : High ultimate tensile strength to control bending failure. High hardness to control pitting. Low density to control moving mass. Good anti wear properties Appropriate
toughness to absorb shocks. Less thermal expansion and high conductivity (in case of worm & worm wheel). 10
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Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Good machinability. Availability. Cost.
Heat treatment: Quenching: Quenching is a treatment performed on steel, applying rapid cooling after heating at high temperature (Approximate 800C). Quenching is applied to adjust the hardness of
steel. There are several types of quenching in accordance with cooling conditions; oil quenching, water quenching, and spray quenching. After quenching, tempering must be applied to give toughness
back to the steel that might become brittle. Thermal Refining: Thermal Refining is a heat treatment applied to adjust hardness / strength / toughness of steel. This treatment involves quenching and
tempering. Since machining is applied to products after thermal refining, the hardness should not be raised too high in quenching. Induction Hardening: Induction Hardening is a heat treatment
performed to harden the surface of the steel containing carbon more than 0.35%, such as S45C or SCM440. For gear products, induction hardening is effective to harden tooth areas including tooth
surface and the tip, however, the root may not be hardened in some cases. The precision of gears declines by induction hardening. To encourage the gear accuracy, grinding must be applied.
Carburizing: Carburizing is a heat treatment performed to harden only the surface of low-carbon steel. The surface, in which carbon is present and penetrated the surface, gets especially hardened.
Inner material structure (with low-carbon C=0.15%) is also hardened by some level of carburizing, however, it is not as hard as the surface. The precision of carburized gears declines by 1 grade or
so, due to deformation (dimensional change) or distortion. To encourage the gear accuracy, grinding is essential. Nitriding Nitriding is a heat treatment performed to harden the surface by
introducing nitrogen into the surface of steel. If the steel alloy includes aluminium, chrome, and molybdenum, it enhances nitriding and hardness can be obtained.
Gear tooth failures and remedies: There are two basic modes of gear tooth failure, breakage of the tooth due to static and dynamic loads and the surface destruction. The complete breakage of the
tooth can be avoided by adjusting the parameters in the gear design, such as the module and the face width, so that the 11
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
beam strength of the gear tooth is more than the sum of static and dynamic loads. The principal types of gear tooth wear are as follows:
Bending failure: Every gear tooth act as a cantilever beam. If the total repetitive dynamic load acting on the gear teeth is greater than the beam strength, then gear tooth will fail in bending. To
avoid such failure the module and the face width of the gear is adjusted so that the beam strength is more than dynamic load. Pitting failure: Pitting is known as the failure of surface fatigue in
the gear tooth. Mainly it occurs due to the frequent loading of the tooth surface and the contact stress is above the surface fatigue strength of the material. In the fatigue region material is
removed and there is a formation of pit in the material. Due to the presence of the pit, which causes the stress concentration plus the pitting is spread over to the adjacent region. The pitting
covers the whole surface. Scoring failure: Due to high temperature, particles of gear material are separated and are rewelded specially observed when pinion and gear are made of same material. To
avoid scoring design the parameter such as speed pressure and proper flow of lubricant to maintain temperature within permissible limits. Use different materials for gear and pinion. The bulk
temperature of the lubricant can be reduced by providing fins on the outside surface of the gear box and a fan for forced circulation of air over te fins. Abrasive wear: Foreign particles in the
lubricant such as dust, rust, spatter or metallic debris can scratch or brinell the tooth surface. Remedies against this type of wear are provision of oil filters, increasing surface hardness and use
of high viscosity oils. A thick lubricating film developed by these oils allows fine particles to pass without scratching. Corrosive wear: The corrosion of the tooth surface is caused by corrosive
elements, such as, extreme pressure additives present in lubricating oils and foreign materials due to external contamination. These elements attack the tooth surface, resulting in fine wear
uniformly distributed over the entire surface. Remedies against this type of wear are providing complete enclosure for the gears free from external combinations, selecting proper additives and
replacing the lubricating oil at regular intervals.
Gear material selection For ferrous and nonferrous metals Cast iron: for low speed and very low power operation Carbon steel: HB < 350 – Normalised/ Annealing. HB >350 – Case hardened. Alloy
steel: low/high speed 12
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
For medium- high torque and speed other material combination is used. The constrain for steel material combination is HB ≤ 350, 60≥HB1- HB2 ≥ 30. Material selection: Cast Iron: PSG 1.4 Cast steel:
PSG 1.9 Alloy steel: PSG 1.14 - 1.15 Recommended combinations: PSG 8.4 - 8.5 (with permissible stresses)
Lewis Bending Strength Equation Lewis Assumptions:
Load is shared by only one pair of teeth. Sliding friction is neglected. The tangential force is spread uniformly along the width of the gear. The effect of radial force i.e. direct compression &
bending are neglected. Direct shear by force Ft is neglected. Tooth is considered as a cantilever beam. Tooth profile is involute. Beam theory is applied under static condition. Inscribed parabolic
tooth based on uniform strength beam is assumed.
Beam strength equation:
Considering gear tooth as cantilever and uniformly distributed load as shown in figure. It is known that, 13
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Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
= Z
= =
σb =
= σb =
Ft = σb × y=
= σb × b × ---- Lewis form factor.
Ft = σb.b.π.m.y Y = π.y Ft = Fs = σb.Y.m.b
….. Lewis beam strength equation
Modified equation of Lewis: From Lewis beam strength equation, Ft = σb.Y.m.b ψm = width factor =
= m ≥ 1.26
…….. Modified equation of Lewis.
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Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Derivation for contact stress equation: Lewis wear load is given by Fw = d1.Q.k.b Where d1- P.C.D of pinion Q - ratio factor =
+ for external - for internal i = gear ratio
k - elastic factor B – face width Modified wear load equation for pinion Fw1 = d1.b.k.Q FN = d1.b.
α = 200 for full depth
( )
centre distance a =
d1 =
Contact stress equation for pinion.
Barth velocity factor: Since higher velocity gear operation results in increased stresses due to impacts at initial contact, a velocity-based factor is commonly included in tooth bending stress.
The Barth velocity factor for carefully cut wheels is given from PSG 8.51 as Where,
Cv =
Cv = Barth velocity factor, Vm = Pitch line velocity. 15
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Lewis dynamic load: Lewis Dynamic load is given by Fd = Ft × Cv …….. PSG 8.51 Where Fd = Lewis dynamic load, Ft = Lewis beam strength, Cv = Barth velocity factor
Buckingham’s dynamic load:
According to Buckingham, small machining error and deflection of teeth under load cause periods of acceleration, inertia forces, and impact loads on the teeth similar to variable load superimposed on
a steady load. The Buckingham’s dynamic load is given as Fd = Ft +
….. PSG 8.51
Where Fd = Lewis dynamic load, Ft = Lewis beam strength, Vm= Pitch line velocity Wear load: Replacing an entire gear set or gearbox is a costly affair. Hence to enable the gearbox to work for a
long period of time, the wear strength has to be greater than the dynamic load. Lewis wear load is given by Fw = d1.Q.k.b Where d1 - P.C.D of pinion Q - Ratio factor =
(+ for external, - for internal, i = gear ratio)
b – Face width, k - elastic factor k=
The surface contact strength can be used for checking, ……PSG 8.13A
AGMA relations:
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
American Gear Manufacturing Association (AGMA) is the global network for technical standards, education, and business information for manufacturers, suppliers, and users of mechanical power
transmission components. The design contact stress according to AGMA relation is given by [σc] = 2.8*BHN – 70
Gear tooth proportions:
No of teeth Pitch circle diameter Addendum diameter Dedendum diameter
Pinion Z1 d1 = m.z1 da = d1 + 2m dd = d1 – 2d
Refer PSG 8.22 Gear Z2 d2 = m.z2 da = d2 + 2m dd = d2 – 2d
Constructional details: The construction of pinion and gear is given by the relation: n = 0.55 Where n - number of arms Pc - Circular pitch in cm Z - Number of teeth Conditions: n < 3 - Integral
shaft construction 3 < n < 7 - Web type construction n > 7 - Arm type construction
Fig: Representation of spur gear pair
Force Analysis of Spur Gear:
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Design procedure of spur gear pair:
As per given specification the design procedure can be adopted. For open end problems the generalized procedure is given as below:
Selection of drive Deciding number of stage Selecting pressure angle system Checking for variation in V.R Material selection Lewis form factor Module calculation Checking for bending failure Checking
for pitting failure Gear proportions Constructional details
Comparison of Spur Gear and Helical Gear: SPUR GEAR
1) In spur gears, the teeth are parallel to the axis 1) In helical gears the teeth are inclined to the axis of of rotation. rotation 2) In spur gear, the initial contact line extends 2) The initial
contact of helical gear teeth is point all the way across the tooth face. which changes into a line as the teeth come into more engagement. 3) There is comparatively less gradual 3) There is gradual
engagement of the teeth and the engagement of the teeth and the smooth transfer smooth transfer of load from one tooth to another of load from one tooth to another occur. occur. 18
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
4) The operation is not as smooth as in case of 4) This gradual engagement makes the gear operation helical gears. smoother and quieter than with spur gears. 5) Comparatively higher dynamic factor,
5) Comparatively lower dynamic factor, Kv.
Helix Angle: A helix angle is the angle between any helix and an axial line on its right, circular cylinder or cone. It ranges between 15o and 45o. Normal Pitch: Circular pitch measured in normal
plane is called normal circular pitch. Normal Module: It is the module in the normal plane. Transverse Circular Pitch: measured in transverse plane.
It is the Circular pitch
Transverse Module: It is the module in the transverse plane. Representation of Helical Gear Pair is shown in figure. Fig: Representation of helical gear pair
Relation between normal module and transverse module. Normal section and transverse section of the helical gear is as shown in figure.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Virtual No. of Teeth on Helical Gear:
Following figure shows the section in normal plane and plane of rotation. In plane of rotation section is circle and in the normal plane the section is ellipse.
The equivalent number of teeth (also called virtual number of teeth, Z𝐯) is defined as the number of teeth on a gear of radius Re. Semi minor axis of the ellipse = d/2 Semi major axis of the ellipse
= d/ (2cos ψ) The shape of the tooth in the normal plane is nearly the same as the shape of a spur gear tooth having a pitch radius equal to radius Re. Re = d/(2cos2 ψ) . Z𝐯 = (Circumference of the
equivalent spur gear)/(Circular pitch) Z𝐯 = (𝟐π𝑹𝒆. /π𝒎𝒏) 𝒁𝒗= 𝟐𝑹𝒆./𝒎𝒏 = 𝒅 / 𝒎𝒏𝒄𝒐𝒔𝟐𝝍 . , Substituting mn = mt cosψ, and d = Z mt, 𝒁𝒗 = 𝒁/ 𝒄𝒐𝒔3 𝝍.
Strength of Helical Gear:
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Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
When the bending strength of helical teeth is computed, values of the Lewis form factor Y are the same as for spur gears having the same number of teeth as the virtual number of teeth (Zv) in the
helical gear and a pressure angle equal to αn. Determination of geometry factor J is also based on the virtual number of teeth.
Lewis Dynamic Load:
Fs = [σb].b.y.P Fd = Ft ×Cv Fs > Fd for safe design under lewis dynamic load, Where, Fd = dynamic load acting on the gear, Fs = static load, σb = bending strength of the gear, b = width of the gear,
Cv = Velocity factor
Buckingham Dynamic Load :
(According to Buckingham, the incremental dynamic load is given by the following equation; Fd = Ft + Fi Where, Fi
V = pitch line velocity, m/s c = deformation factor, (PSG 8.53, Table 41) e = sum of errors between meshing teeth, mm, (PSG 8.53, Table 42 ) b = face width of teeth, mm = helix angle, deg.
Wear Strength:
The wear strength of spur gear is Fw = b.Q.d1.K, For a helical gear, the component of (Fw)n in the plane of rotation is related as Fw . cos β = (Fw)n . Further, for a helical gear, face width along
the tooth width is b. cos β and the pitch circle diameter for a formative pinion is d1 /cos2 β. 21
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Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Substituting these values, the equation for wear strength of a helical gear is (Fw) n= b. cos β.Q. d1. cos2 β. K Fw. cos β = b. Q. d1. cos3 β. K Fw = b. Q. d1. K / cos2 β This is the Buckingham’s
equation for wear strength in the plane of rotation. Therefore, Fw is the maximum tangential force that the tooth can transmit without pitting failure. It may be recalled that the virtual number of
teeth Zv is given by Zv = Z/ cos3 β. Therefore, Z1v = Z1/ cos3 β and Z2v = Z2/ cos3 β. And Q = 2 .Z2v/ (Z2v + Z1v)
Q = 2 Z2/ (Z2 + Z1)
(Similarly for a pair of internal gears Q = 2.Z2/ (Z2 - Z1) Where, Z1 and Z2 are the actual number of teeth in the helical pinion and gear, respectively) The pressure angle αn = 20o is in a plane
normal to the tooth element. Thus the K factor is given by , Where, [
= Surface endurance strength (N/mm2)
E1, E2 = moduli of elasticity of materials for pinion and gear, respectively, (N/mm2) αn = pressure angle in a plane normal to the tooth element
Tooth Proportions:
In helical gears, the normal module mn should be selected from standard values, the first preference values are, mn (in mm) = 1, 1.25, 1.5, 2, 2.5, 3, 4, 5, 6, 8 and 10 The standard proportions of
tooth (Ref. PSG 8.22) Helix angle, β = Normal module:
Transverse module:
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Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Centre distance: Height factor:
; = 1 for full depth and 0.8 for stub teeth;
Bottom clearance: c = 0.25
, for stub teeth c = 0.3
Pitch diameter:
Tip diameter:
; and
Root diameter:
; and
Force Analysis for helical gear:
Helix angle ;
Contact Stresses:
In the case of spur gears of contact ratio less than 2, the theoretical length of tooth contact is 1.0b. With helical gears, the length of contact per tooth is b/cosβ and the helical action causes
the total length of tooth contact to be approximately b/cosβ times the contact ratio (CR) at all times. The AGMA recommends that 95% of this value be taken as the length of contact when computing
contact stress. For checking,
….. Ref. PSG 8.13
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Design Procedure:
For generalised open end problem the design procedure can be followed as given below. 1) Selection of drive and decision of no. of stages. 2) Assumption, selection of profile of gear, helix angle and
finding number of teeth. 3) Material selection ( PSG1.9 & 8.5) 4) Determination of weaker gear : Lewis form factor (PSG 8.50) 5) Design for the weaker gear based on bending criteria (PSG 8.13A) 6)
Checking failure in dynamic loading (Lewis and Buckingham) 7) Checking failure for the wear / pitting ( PSG 8.13) 8) Gear tooth proportions 9) Construction details 10) Force analysis 11) Shaft design
based on torsion and bending BEVEL GEAR:
Introduction to Bevel Gear:
Bevel gears are gears where the axes of the two shafts intersect and the tooth-bearing faces of the gears themselves are conically shaped. Bevel gears are most often mounted on shafts that are 90
degrees apart, but can be designed to work at other angles as well. The pitch surface of bevel gears is a cone. Two important concepts in gearing are pitch surface and pitch angle. The pitch surface
of a gear is the imaginary toothless surface that you would have by averaging out the peaks and valleys of the individual teeth. The pitch surface of an ordinary gear is the shape of a cylinder. The
pitch angle of a gear is the angle between the face of the pitch surface and the axis. The most familiar kinds of bevel gears have pitch angles of less than 90 degrees and therefore are cone-shaped.
This type of bevel gear is called external because the gear teeth point outward. The pitch surfaces of meshed external bevel gears are coaxial with the gear shafts; the apexes of the two surfaces are
at the point of intersection of the shaft axes. Bevel gears that have pitch angles of greater than ninety degrees have teeth that point inward and are called internal bevel gears.
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Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Bevel gears that have pitch angles of exactly 90 degrees have teeth that point outward parallel with the axis and resemble the points on a crown. That's why this type of bevel gear is called a crown
Gear Terminology
1. Shaft angle: Angle between the intersecting axes of gear generally is 90. 2. Face Width: Length of the tooth measured along the pitch cone generator b = (8 to 10) m. The angle subtended at apex by
axis and pitch cone angle. 3. Cone Distance (R): The length of pitch cone generator from pitch circle to apex. 4. Back Cone The cone generated by line which is perpendicular to pitch cone generator
at point on a pitch circle. 5. Pitch cone angle ( δ1, δ2 ): The angle subtended at the apex by the axis and the pitch cone generator. 6. Back cone angle: The angle between the back cone generator and
axis of geometry is known as back cone angle. It is a compliment of pitch cone angle i.e. (90 - δ1+ δ2 )
Characteristics of Bevel Gear:-
1. Power transmission to intersecting shaft. 2. Intersecting shaft angle Σ = δ1 + δ2. 3. Pitch surfaces, frustum of cone with common apex to get pure rolling. 4. Nature of contact is line contact. 5.
Kinematics and dynamics are same as spur gear in back cone. 6. All terminology is referring to a larger cone. 7. Force analysis refers to midpoint of face width of the tooth. 8. Maximum Velocity
Ratio = 5. 9. Difficult in manufacturing & assembly for VR > 2. 10. Nature of load is impact to backlash. 11. Tooth thickness varies along the generator for straight bevel gears. It is maximum at the
large cone and minimum at the smaller cone. 25
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Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
12. Due to variation in tooth thickness manufacturing is difficult so for Multistaging bevel gearing stage is selected at high speed side.
Gear ratio:-
O o
δ1 δ2
RP1 PP
δ2 δ1 RP2
Let Pitch cone angles are δ1 , δ2 and Intersecting Shaft angle is Σ Velocity Ratio ,
; ;
and 26
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Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Virtual number of teeth: Bevel gear has equivalent spur gear in back cone. RPv is back cone radius. In Δ AMD, , ,
, ,
Relation between Average module (ma) & Transverse Module (mt) ,
; Normally, b= 10 ; PSG … 8.38
General steps in Bevel Gear Design:-
1. Calculate design power [P]. 2. Assumptions, 27
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Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
i. Tooth profile ii. Type of Meshing iii. Cutting Method 3. Standard Velocity Ratio 4. Pitch angles 5. Virtual no. of teeth 6. Lewis form factor 7. Material Selection 8. Module calculation 9.
Checking for Dynamic Load 10. Checking for Wear Load 11. Tooth proportion 12. Gear Bodies 13. Force analysis 14. Shaft design
Force analysis: 𝛼 – pressure angle 𝛾 – helix angle
The force (Pn) is resolved into three components, tangential, (Pt), and radial component (Pr) and axial components (Pa) which are related to the pressure angle as above. 28
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Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Gear Bodies:
Depending upon the size, material, and type of application, cost and other deciding factors, such as anticipated stress conditions and operational parameters, the gears may have different condition
structural designs and shapes. Gear blanks may be machined from a solid raw-stock or may be manufactured by casting, forging and fabricating by welded construction. Often the designer has to reckon
with such considerations as machining facility and availability of heattreatment measures. Small pinions are often made integral with the shaft. In such a design, which is normally referred to as a
pinion shaft, the key is dispensed with and the provision of an axiallocating device is also eliminated. Gears are also made by drop-forging and die-casting. Steel gears with diameters up to 500 mm
are usually made full without recess. Large gears are generally of cast construction. Very large and wide gears are usually of two-walled variety and are either of cast or welded construction. For
saving costly materials, composite designs of gears are sometimes resorted to. In such designs, the gear rim of quality steel is press-fitted or shrinkfitted on to the gear-hub which is made of
comparatively inferior material. Grub screws are sometimes fitted between the rim and the hub for extra securing. To avoid fatigue failure, gear teeth are often chamfered sideways or are rounded off
laterally. To effect reduction in weight, the gear crown or the rim may be joined to the central hub through arms or spokes. , where Pc is circular pitch (𝛱m) in cm.
If, n ≤ 3 then construct Integral Shaft type, 3 ≤ n ≤ 7 then Web type Construction and if n ≥ 7 then Arm type Construction.
Shaft Design
In order to design the shaft on which the gear is mounted, many diverse factors have to be taken into account. Stress concentration is a major factor. The main determining factors for the calculation
of the diameter of the shaft are the maximum bending moment and the torque to which it is subjected. Besides strength considerations, shafts are sometimes checked for stiffness, deflection and
critical speed. For long shafts, a certain angle of twist must not be exceeded. For such cases this is the deciding factor for the ultimate selection of the shaft diameter. Shafts which 29
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
have to with stand bending forces caused by machine elements mounted on them should be checked against the deflection which results from such loadings. Elements carried by the shaft should be mounted
close to the bearings to reduce deflection. The maximum deflection must be kept within the allowable limit which is normally 0.001 to 0.003 times the span, i.e. the distance between the bearings. In
case of high speed shafts (greater than 1500 rpm) of certain types of machines, the critical speed should be checked. Imbalance in the system causes vibration. The vibration amplitude may reach such
values as to cause ultimate failure of the shaft. Resonance occurs when at a critical shaft velocity, the frequency of vibration of external forces and of the shaft system coincide. The speed at
which resonance sets in is called the “critical speed”. To avoid the disturbance caused by this phenomenon, the operating speed must be widely away from the critical speed which should lie at least
10% above or sometimes widely below the operating speed. An indication of the impending failure of the shaft is its excessive vibration. However, the shaft does not fail all of a sudden. Proper
balancing, therefore, is of utmost importance for high speed shafts together with the machine elements mounted on them.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
WORM AND WORM WHEEL A worm drive is a gear arrangement in which a worm (which is a gear in the form of a screw) meshes with a worm gear (which is similar in appearance to a spur gear). The two
elements are also called the worm screw and worm wheel. This type of arrangement is used to transmit power between non-intersecting and non-parallel shafts.
Worm Wheel
Worm gears are used when large gear reductions are needed. It is common for worm gears to have reductions of 1:20, and even up to 1:100 or greater.
Many worm gears have an interesting property that no other gear set has: the worm can easily turn the gear, but the gear cannot turn the worm. This is because the angle on the worm is so shallow that
when the gear tries to spin it, the friction between the gear and the worm holds the worm in place. This feature is useful for machines such as conveyor systems, in which the locking feature can act
as a brake for the conveyor when the motor is not turning. One other very interesting usage of worm gears is in the Torsen differential, which is used on some high-performance cars and trucks.
Terminology in Worm and Worm Wheel:
Diametral Pitch: Diametric pitch (also referred to as pitch) is the relationship between the number of teeth in a gear and each inch of the gear’s pitch diameter (PD). For example, a worm gear with
16 teeth (T) and a one-inch pitch diameter is a 16-diametral. Circular Pitch: With a worm, circular (also referred to as linear) pitch is a distance measured along the pitch line of the gear. It can
be determined by measuring – with an ordinary scale – the distance between any two corresponding points of adjacent threads parallel to the axis. With a worm gear, circular pitch is a distance
measured along the pitch circle of the gear. It can be determined by measuring – with an ordinary scale – the distance between any two corresponding points of adjacent teeth. As noted above, this
measurement should be taken on the pitch circle, which is approximately halfway down a tooth. 31
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Figure – Worm Circular Pitch
Thread Dimensions:
The following terms are used when describing the dimensions of a worm-thread. 1. Dedendum – the part of the thread from the pitch line of the worm to the bottom of the thread. The dedendum is equal
to one addendum plus the working clearance (defined below). 2. Working Clearance – the distance from the working depth (defined below) to the bottom of the thread. 3. Working Depth – the space
occupied by the mating worm gear tooth. It is equal to twice the addendum. 4. Whole Depth – the distance from the bottom of the thread to its outside diameter.
Figure: Drawing of Worm showing cross section and full view of the thread
Pitch Diameter: The pitch diameter of a worm is the diameter of the pitch circle (the “imaginary” circle on which the worm and worm gear mesh). There is no fixed method for determining the pitch
diameter of a worm.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Figure – Pitch Diameter of Worm
Hand of Gear: Worms and worm gears are manufactured with right or left-hand threads and teeth. The hand of a worm or worm gear may be determined by noting the direction in which the threads or teeth
lean when the worm or worm gear is held with the hole facing up. In a worm gear set, the worm and gear must have the same hand, pitch, number of threads, and tooth dimensions. They also must have the
same pressure angle and lead angle. Leads and Lead Angle: The lead of a worm is the distance any one thread advances in a single revolution. The lead angle of a worm is the angle formed by the worm
thread and a line perpendicular to the worm axis.
Figure– Lead Angle Pressure Angle: The pressure angle is the angle at which a force is transmitted from the worm thread to the worm gear tooth. It determines the relative thickness of the base and
top of the thread.
Figure– Pressure Angle
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Center Distance: The center distance of a worm and worm gear in mesh is the distance between the centres of the two shafts. When mounted on the proper center distance, the worm and worm gear will
mesh correctly.
Figure – Centre Distance
Nature of Bodies of Worm and Worm Wheel: A worm gear is used when a large speed reduction ratio is required between crossed axis shafts which do not intersect. A basic helical gear can be used but
the power which can be transmitted is low. A worm drive consists of a large diameter worm wheel with a worm screw meshing with teeth on the periphery of the worm wheel. The worm is similar to a screw
and the worm wheel is similar to a section of a nut. As the worm is rotated the worm wheel is caused to rotate due to the screw like action of the worm. The size of the worm gear set is generally
based on the center distance between the worm and the worm wheel.
Figure– Nature of bodies for worm and worm wheel.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
If the worm gears are machined basically as crossed helical gears the result is a highly stress point contact gear. However normally the worm wheel is cut with a concave as opposed to a straight
width. This is called a single envelope worm gear set. If the worm is machined with a concave profile to effectively wrap around the worm wheel the gear set is called a double enveloping worm gear
set and has the highest power capacity for the size. Single enveloping gear sets require accurate alignment of the worm-wheel to ensure full line tooth contact. Double enveloping gear sets require
accurate alignment of both the worm and the worm wheel to obtain maximum face contact.
Efficiency of Drive: The worm and worm gear drive is never 100% efficient as there is always some power loss due to the friction (rubbing action) between the worm and worm gear. The following
factors have an impact on the friction and, therefore, the efficiency of a drive: • Lubrication, • Speed of worm, • Material of worm and gear, • Load, • Finish of surface on worm thread, • Accuracy
of cutting worm and gear, • Lead angle of worm Figure shows the Efficiency with respect to number of threads and Efficiency of Drive with respect to
lead angle. Figure: Efficiency with respect to number of threads and lead angle.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Material Selection:
Gear manufactures standard stock worms made from high quality steel (both hardened and unhardened). Depending on pitch, hardened worms are available with polished only threads as well as with ground
and polished threads. Standard stock worm gears are available – depending on pitch – in fine grain cast iron and bronze. Material
Worm Low Cost, low duty
Toys, domestic appliances, instruments
Excellent machinability, medium friction.
Used infrequently in modern machinery
Carbon Steel
Low cost, reasonable strength
Power gears with medium rating.
Hardened Steel
High strength, good durability
Power gears with high rating for extended life
Acetyl / Nylon Cast Iron
Worm wheel Low Cost, low duty
Toys, domestic appliances, instruments
Phosphor Bronze
Reasonable strength, low friction and good compatibility with steel
Normal material for worm gears with reasonable efficiency
Cast Iron
Excellent machinability, medium friction.
Used infrequently in modern machinery
Acetyl /Nylon
Design Criteria: Transverse module is axial module of worm. The axial module can be calculated based on wear failure and checked for bending failure under static and dynamic conditions.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Force Analysis of worm and worm wheel:
Figure - Force Analysis of Worm and Worm Wheel.
The tangential, axial, and radial force components acting on a worm and gear are illustrated in the Figure. For the usual shaft angle, the worm tangential force is equal to the gear axial force and
vice versa. F =F 1t
, F =F 2t
The worm and gear radial or separating forces are also equal, F1r = F2r
The radial force: If the power and speed of either the input or output are known, the tangential force acting on this member can be found from equation, 1. In the Figure above, the driving member is
a clockwise-rotating right hand worm. 2. The force directions shown can readily be visualized by thinking of the worm as a right hand screw being turned so as to pull the “nut” (worm gear tooth)
towards the “screw head”. 3. Force directions for other combinations of worm hand and direction of rotation can be similarly visualized. The thread angle λ of a screw thread corresponds to the
pressure angle of the worm. One can apply the force, efficiency, and self-locking equations of power screw directly to a worm and gear set. These equations are derived below with reference to the
worm and gear geometry. 37
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Figure below show in detail the forces acting on the gear. Components of the normal tooth force are shown solid. Worm driving condition and forces acting on the worm gear tooth are shown below. The
friction force is always directed to oppose the sliding motion. The driving worm is rotating clockwise: F2t = F1a =
cos λ -
Thermal Analysis of Worm and worm : The efficiency of a worm gear drive is low and the work done by friction is converted into heat. When the worm gears operate continuously, considerable amount of
heat is generated. The rate of heat generated ( is by by,
, kW = power transmitted by gears
The heat dissipated through lubricating oil to the housing wall and finally to the surrounding air. The rate of heat dissipated by the housing walls to the surrounding is given by,
where, k = overall heat transfer co-efficient of housing walls 38
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
= temperature of lubricating oil
A = effective surface area of housing The heat generated should be less than equal to the heat dissipated. In limiting condition the above two equation can be written as,
The above equation gives the power carrying capacity based on thermal considerations.
General Design Procedure for worm and worm wheel drive: 1. Determine the design power. 2. Determine the layout foe worm and worm wheel. 3. Decide the number of stars based on the gear ratio. 4.
Determine the number of teeth on worm wheel. 5. Decide the diametrical factor (q) from PSG [8.45]. 6. Determine the lead angle. 7. Determine the helix angle individually for worm and worm wheel. 8.
Find the virtual number of teeth. 9. Determine the Lewis form factor. 10. Perform material selection from PSG [8.45]. 11. Determine the weaker element. 12. Decide the Axial module of worm i.e.
transverse module of worm wheel from PSG [8.44] 13. Select a standard module from the found value. 14. Check for dynamic failure. 15. Determine the efficiency of drive. 16. Write the tooth
proportions. 17. Calculate the heat dissipation area of our design, 18. Determine the heat generated and heat dissipated. 19. Decide locational dimensions. 20. Perform force calculation. 21. Analyse
forces and find bearing reaction. 39
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
22. Select the lubrication method.
Worm Gear Speed Reducers: Worm gear speed reducers are comprised of the terms “gearbox” and “speed reducer” that are used interchangeably in the world of power transmission and motion control.
Gearboxes are used for speed reduction and torque multiplication. The term speed reducer became vernacular when gearboxes were first implemented in industry. Speed reduction was an important function
for the gearbox, to replace more cumbersome belts and pulleys technology. Demand for worm gear speed reducers is increasing as more mechanical applications in several industries require speed
reduction, ranging from rock crushers to robots.
Figure – Typical Speed Reducer for Worm and Worm Wheel.
For the case of a worm wheel as the driver, the forces are as in Figure Ft2 = Fn (cosα n cos γ + μ sin γ) Fx2 = Fn (cosα n sin γ – μ cos γ) Fr2 = Fn sinα n Note that the direction of F depends on the
direction of rotation of the worm. The three force components, F, FR and FTW must be taken up by both worm and gear bearings. The directions acting on the worm gear and worm are opposite. Total
bearing force on each member is the vector sum of these three forces. With the worm as driver and the gear rotating as shown in Figure, the direction of these forces on each member are shown in
Figures. With the aid of these figures the radial bearing loads for shafts with two bearings can be obtained from Cases (a) and (b). Once again both thrust and radial forces need to be taken up by
the bearings. 40
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
GEARBOX A gearbox is a set of gears for transmitting power from one shaft to another. They are used in a wide range of industrial, automotive and home machinery applications. The shape of the tooth
is an involute helicoid as if a paper piece of the shape of a parallelogram is wrapped around a cylinder, the angular edge of the paper becomes the helix. If the paper is unwound, each point on the
angular edge generates an involute curve. In spur gear, the initial contact line extends all the way across the tooth face. The initial contact of helical gear teeth is point which changes into a
line as the teeth come into more engagement. Herringbone or double helical gear are two helical gears with opposing helix angle stacked together. As a result, two opposing thrust loads cancel and the
shafts are not acted upon by any thrust load. The advantages of elimination of thrust load in Herringbone gears, is offset by considerably higher machining and mounting costs. This limits their
applications to heavy power transmission.
Figure: 3 stage helical gear box
Housing of Gears: The function of housing is to support the transmission elements like gears, shafts and bearings in correct position and to take up all the forces developed in the speed reducer,
during its operation. Construction of housing: The housing for the gearboxes is usually split into two parts; lower part and upper part or cover that are joined in a plane passing through the axes of
shaft. This plane is usually horizontal.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Common elements of housing: 1. Wall of housing 2. Bosses of bearings 3. Flanges for lower part of housing and cover 4. Flanges for connecting lower part of housing to base plate or frame 5.
Inspection opening 6. Seat for nuts or screw heads for joining lower part of housing to base plate of frame 7. Inspection opening 8. Boss for drain plug 9. Boss with threaded hole for oil gauge 10.
Threaded hole for drain gauge 11. Holes for puller bolts 12. Holes for dowel pins 13. Threaded holes for screws or studs joining lower part of housing to cover 14. Threaded holes for screws or studs
joining lower part of housing to base plate or frame 15. Grooves foe cap collars of bearing 16. Lifting eyes 17. Threaded hole for air vent
Figu re: Housing view of gear box
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Proportions of housing: t tc tf1 tf2 tr dst d1 d2 d3 C C1 K K1
Thickness of housing wall Thickness of cover wall Thickness of flanges between housing and cover Thickness of foundation flanges Thickness of rib Diameter of the foundation bolt Diameter of the
bearing bolt Diameter of the bolts for securing cover and housing Diameter of the bolts for bearing cap Distance of the foundation bolt axis from housing wall Distance of bolt housing wall Width of
the foundation flange Width of flanges between housing and cover
t = 0.0025a + 5 tc = 0.02a + 5 tf1 = 1.5 t tf2 = ( 2 - 2.4) t tr = 0.85 t dst = 0.036a +12 d1 = 0.75 dst d2 = (0.5 - 0.6 ) dst d3 = (0.4 - 0.5 ) dst C = 1.2 dst + 5 C1 = 1.2 d2 + 5 K = C + dst + (2
to 5) K1 = C + d2 + (2 to 5)
Sectional view of gear box are shown in following figure:
Figure: Cross section of single stage Gear box and two stage gear box.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Gear bodies: Integral type construction Advantages: 1. It reduces the amount of machining since there is no need to cut keyways on the shaft and the pinion. 2. It reduces the number of parts since
there is no key. This reduces the cost. 3. It increases the rigidity of the shaft and also increases the accuracy of contact. Disadvantages: 1. The shaft has to be fabricated from the same material
as that of the pinion, which is often of the higher quality and costly. 2. When the pinion is to be replaced because of wear or tooth break down, the shaft has to be discarded as well.
Fig : Integral Type Construction
Web Type Construction Advantages: 1. Used for small gears up to 250 mm pitch circle diameter. 2. Capable of transmitting the shaft without shearing of the hub. Disadvantages: 1. Cannot used for large
sized gears. Approximate proportion are as follow , , , ; ;
F ig: Web Type construction 44
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Rim Type Construction Advantage: 1. Rimmed gears save costly high strength materials. Disadvantage: 1. More expensive to manufacture.
Forged Gears Advantages:
Fig: Rim Type
1. In case of forged gears, material utilization factor is (2/3), which is more. 2. Gears forged with integral teeth normally have longer fatigue and wear life. 3. Precision forged gears have more
load carrying capacity. 4. Less or no material loss. 5. With the elimination of internal discontinuities, low rejection rates, better machinability, less machining allowances--cost of production of
aggregates is reduced considerably. 6. Forged gears have light weight construction which reduces the inertia and centrifugal forces. Disadvantages: 1. Difficult to maintain good surface finish of
teeth without grinding. 2. Forging involves significant capital expenditure for machinery, tooling, facilities and personnel. 3. Forged gears become economical only when they are manufactured on
large scale.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Problem 1 Design a single stage gearbox using spur gear for following specifications Power= 10 KW, Input speed = 1440 rpm, Output speed =360 rpm. Solution: 1. Drive selection and no. of stages:
Selecting open drive for given power of 10KW: Velocity Ratio, i =
= 4 < 5 hence, selecting single stage with velocity ratio as 4
2. Assumption: i. Selecting tooth profile as 20° involute, Full Depth. ii. Meshing of gear Sn gearing, where, X1=X2=0 iii. Quality of gear – Precision cut to control dynamic load 3. Number of teeth
calculation Minimum no. of teeth on pinion Z1= 18 Minimum no. of teeth on gear Z2= i
Z1 = 4 18 =72
Adding 1 hunting tooth on gear, Z2 = 73 4. Checking for percentage variation in VR iact =
= 4.055
% deviation =
= 1.375% < 3%
Hence, hunting tooth addition is valid. 5. Material selection (PSG 1.9, 8.4)
[ ]
[ ]
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
6. Lewis form factor y1 = 0.154 –
……20 FD involute (PSG 8.50)
y1 = 0.1033 y2 = 0.154 –
= 0.1415 = 116.829 N/mm2 = 159.952 N/mm2
Fs1 = [ ] Y1 = Fs2 = [ ] Y2 = Fs1 < Fs2 Therefore, pinion is weaker. Therefore, design pinion. = 140 N/mm2
7. Permissible stress, [
8. Design Criteria: Design of pinion is based on strength and checking for wear and dynamic load. m> P= 10KW [P]= 1.1 10 = 11KW [Mt]
(assuming minor shock working for 8hrs)
= 72.94
[ ] = 140 N/mm2 Y1 = = 0.3245 z1 = 18 1
= 10
….. (PSG 8.14)
9. Module calculation m>
, m>
, m > 3.763
Increasing module by 20% to take care of pitting failure, m = 1.2 3.763 = 4.515, From PSG 8.2 selecting std. module as m = 5 mm Therefore, the width of the gear tooth, b = 10 5 = 50 mm 47
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
10. Checking for Lewis dynamic load i. Static strength is given as , Fs = [ ] Fs = = 11.35 KN ii. Lewis dynamic load is given by , Fd = Ft Cv Ft =
= 1609.77 N
Finding pitch line velocity,
Vm =
= 6.78 m/s
For carefully cut gear (PSG 8.51), Cv =
= 2.13
Therefore, Fd = Ft Cv = 1609.77 2.13 = 3.428 KN Since, Fs > Fd , design is safe in Lewis dynamic load. 11. Checking for pitting, Induced contact stress is given as, ( )= Where centre distance, a =
……. (PSG 8.13, eq 1.4) =
= 227.5 mm
Design contact stress, [ ] = 500 N/mm2 Velocity ratio, i = 4 Modulus of elasticity, E = 2 105 N/mm2 ( )= ( ) = 310.590 N/mm2 Since, [ ] > ( ), design is safe in wear. 12. Gear proportions Addendum, a
= m = 5 mm Dedendum, d = 1.25 m = 6.25 mm Clearance, c = 0.25 m = 1.25 mm Working depth, hw = a + d - c = 10 mm Pinion
Gear 48
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
No. of teeth, z
Pitch circle diameter
d1 = 90 mm
d2 = 365 mm
Addendum diameter , da
d1+2m = 100 mm
d2+2 m = 375 mm
Dedendum diameter, dd
dd1 = (z1-2f0)m-2c = 77.5 mm
dd2 = (z2 -2f0) m-2c = 352.5 mm
13. Construction details i. Pinion n = Therefore, Integral shaft construction. ii. Gear
= 2.612 < 3
n = Therefore, Web type construction.
= 5.26 < 7
14. Shaft Design Pinion shaft Let material for shaft = C-40 with [Mt] = 72.94 × 103 N/mm2 [т] =
, ds =
[𝞽] = 40N/mm2
= 21.019mm
Lets select, shaft diameter as ds = 30mm Gear shaft Let material for shaft = C-40 with [𝞽] = 40N/mm2 [Mt] = [Mt] = 291.78× 103 N/mm2 ds =
= 33.36mm, Lets select shaft diameter as ds = 60mm
15. Gear Box Housing dimensions
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Thickness of housing wall
t =2
Thickness of covering wall
tc = 0.9t = 23.52mm ≥ 6
Thickness of flanges between housing and cover
tf1 = t = 26.13mm
Thickness of foundation flanges
tf2 = 1.5dst= 125.34mm
Thickness of ribs
tr = t =26.13mm
Diameter of foundation bolts
dst =
Diameter of bolts for securing cover and housing
d2 =
Width of foundation flange
K = (2.1 to 2.5)dst =208.9mm
Width of flanges between housing and cover
K1 =(2.1 to 2.5)d2 = 165.8mm
Height of shaft axes from lower surface of H =(1 to 1.12)a = 254.8mm housing
Axial clearance between gear side and protruding Δ1 =0.8t = 20.90mm inner elements of housing
Radial clearance between gear face from bottom Δ2 = 1.2t =31.35mm of housing
=26.13mm ≥ 6
83.56mm ≥ 12 66.32mm ≥ 10
………………………………………………………………………………………. Problem 2: Design a spur gear pair for the first stage of gear box having following specifications Power = 15kW Input speed = 1440 rpm Output speed = 90 rpm
Solution: 1. Deciding no of stages Overall velocity ratio, iO/A =
, iO/A = 16 50
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
is =
, is =
selecting, i1 = 3.98 and i2 = 4.02 2. Assumptions: i. Selecting pressure angle system as 20°, Full Depth , involute profile . ii. Meshing of gear Sn gearing, where, X1=X2=0 iii. Quality of gear –
Precision cut to control dynamic load 3. Number of teeth: Minimum no of teeth on pinion (z1) = z1 =
, z1 = 18
Adding 1 hunting tooth
z2 = i1 z1 = 3.98 18 = 72 z2 = 73
Checking for variations in VR istd = 3.98
iact =
= 4.05
%VR = %VR = 1.758 < 3%, Therefore, adding hunting tooth is valid. 4. Material selection (PSG 1.9, 8.4) Material
[ ]
5. Lewis form factor y1 = 0.154 –
……20 FD
y1 = 0.1033 y2 = 0.154 –
= 0.1415 51
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
= 129.81 N/mm2 = 142.251 N/mm2
Fs1 = [ ] Y1 = Fs1 = [ ] Y2 = Fs1 < Fs2 Therefore, pinion is weaker. Therefore, design pinion. = 140 N/mm2
6. Permissible stress, [
7. Design Criteria: Design of pinion is based on strength and checking for wear and dynamic load. m>
P = 15 kW [P] = 1.1 15 … (Assuming minor shock working for 8 hours) [P] = 16.5 kW [Mt] [
= 109.419
] = 400 N/mm2
Y1 =
= 0.3245 z1 = 18
= 10
….. (PSG 8.14)
8. Module calculations m≥
Increasing module by 25% to take care of pitting failure, m = 1.25 2.726 = 3.407 From PSG 8.2, m = 5 mm Therefore, width of gear tooth, b = 10 5 = 50 mm 9. Checking for Lewis dynamic load Static
strength is given as Fs = [ kN Dynamic load, Fd = Ft Cv Where, Ft =
= 32.45
= 2341.33 N
Finding pitch line velocity, Vm =
= 6.78 m/s
For carefully cut gear (PSG 8.51), 52
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Cv =
= 2.13
Therefore, Fd = Ft Cv = 2431.33 2.13 = 5.178 kN Fs > Fd , Therefore, design is safe in Lewis dynamic load. 10. Checking for pitting, Induced contact stress, ( ) =
…. (PSG 8.13, eq 1.4)
Where, centre distance, a =
= 227.5 mm
[ ] = 1100 N/mm2, velocity ratio, i = 4.05 Modulus of elasticity , E = 2 105 N/mm2 ( )= ( ) = 383.72 N/mm2 [ ] > ( ) , Therefore, design is safe in wear. 11. Gear proportions Addendum, a = m = 5 mm
Dedendum, d = 1.25 m = 6.25 mm Clearance, c = 0.25 m = 1.25 mm Working depth, hw = a + d - c = 10 mm Parameter
No. of teeth, z
Pitch circle diameter
d1 = 90 mm
d2 = 365 mm
Addendum diameter , da
d1+2m = 100 mm
d2+2m = 375 mm
Dedendum diameter, dd dd1 = (z1-2f0)m-2c = 77.5 mm
dd2 = (z2-2f0)m-2c = 352.5 mm
12. Construction details i. Pinion 53
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
n =
= 2.612 < 3 , Therefore, Integrated shaft construction.
ii. Gear n =
= 4.97 < 7 , Therefore, Web type construction.
NUMERICAL 3: A single stage helical gear box is used to transmit 12.5 KW power at 1440 rpm of pinion. The desire transmission ratio is 5:1 Assume 20 degree FD involute profile and material C50 for
pinion and gear. i) Find the module ii) Check gear for Lewis dynamic load iii) Check gear for contact stresses. iv) Write constructional details. Solution: Step 1: Selecting open drive Step 2:
Deciding stages and Selecting pressure angle system i = 5, single stage Selecting 20° FD involute system and Assuming meshing of gear Sn gearing, where, X1=X2=0 and quality of gear as Precision cut
to control dynamic load Step 3: Number of teeth calculation Minimum number of teeth on pinion , 54
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Minimum no. of teeth on pinion Z1= 18 Minimum no. of teeth on gear Z2= i Z1 =5 18 =90 Adding 91 hunting tooth on gear, Z2 = 91 Step 4: Checking for variation in VR iact =
= 5.055
% deviation =
= 1.111% < 3%
Hence, approximation is valid Step 5: Material selection (PSG 1.9, 8.4)
Material Pinion Gear
and C50
[ ] in Mpa
[ ] in Mpa
[ ] in Mpa
Step 6: Virtual no teeth Assuming Helix angle,
Step 7: Lewis form factor y1 = 0.154 –
……(for 20 FD )
y1 = 0.1096 y2 = 0.154 –
= 0.1467
Since y1< y2 and material for pinion and gear is same, Therefore, pinion is weaker. Therefore, designing pinion. Step 8: Permissible stress 55
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
…. PSG 8.18
Hence, [
Step 9: Design criteria: Design of pinion is based on strength and checking for wear and dynamic load.
P= 12.5 KW, [P] = 1.1 [Mt]
12.5 = 13.75 KW =
….. (Assuming minor shock working for 8hrs)
= 91.1825
[ ] = 152 N/mm2 Y1 = = 0.3443 z1 = 18 Assuming, width factor,
= 10 …. (PSG 8.14)
2.3437 Increasing module by 20% to take care of pitting failure, = 2.3437 1.2 = 2.81 mm, From PSG 8.2 selecting std. module as =3 Therefore, width of the tooth, b = 10 3 = 30 mm Transverse module, mt
= 3.13 mm
Step 10: Checking for Dynamic load, Gear tooth strength, Fs = [ Fs = Lewis Dynamic load,
Fd = Ft Cv 56
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Ft =
= 3377.129 N
Finding pitch line velocity, Cv =
Vm =
= 4.07 m/s
= 1.67, for carefully cut gear (PSG 8.51),
Therefore, Fd = Ft Cv = 3377.129 x 1.67 = 5639.8 N Fs < Fd Therefore, design is not safe in bending. Hence modifying face width to 12 times module , hence width b = 36mm Hence, Static strength, Fs =
Step 11: Checking for pitting, Induced contact stress, ( ) =
Where, centre distance , a =
…. (PSG 8.13, Eq.. 1.4)
= 170.97 mm
Velocity ratio, i = 5 , Assuming , Modulus of elasticity as E = 2
..........(PSG 8.22) 105 N/mm2
( )= ( ) = 605.675 N/mm2 [ ] < ( ),
Design contact stress, [ ] = 604 Mpa,
Therefore, design is not safe in wear, hence increasing BHN to 250 Now modified contact stress of material by AGMA relation, [ ] = 2.8 X 250 -70 = 630 Mpa, > ( ) , safe in wear.
Step 12: Gear proportions (PSG 8.22) Centre distance, a = 170.97 mm 57
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Tooth depth, h = 2.25 mn = 6.75 mm Bottom clearance c = 0.25 mn = 0.75 mm Height factor fo = 1,
No. of teeth
d1 = mt x Z1 = 56.34 mm
d2 = 284.83 mm
Tip diameter, da
Root diameter, dd
= 62.467 mm
291.474mm dd1 =
dd2 =
= 48.9673 mm
= 277.9738 mm
Step 13: Construction details iii. For Pinion Number of arm , n = Therefore, Using Integral type construction. iv. For Gear n = Therefore, Using Web type construction.
= 2.32 < 3
= 5.22 > 3
and < 7
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Numerical 4)
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Numerical 5) Design a gear box for worm and worm wheel drive for a following specifications, Power = 12KW, Worm speed =900RPM, Velocity ratio =30. Solution: a) Design power Assuming Service Factor:
b) Velocity Reduction ratio i= 30; N2= N1 / 30 = 30 rpm c) Step 4: Layout Selection of layout with worm up position as shown in figure. d) Assumptions i. Selecting tooth profile is 20° involute for
Worm and Worm Wheel ii. Meshing of gear Sn gearing, where, X1=X2=0 iii.Quality of gear – Precision cut to control dynamic load iv. Selecting overhauling type Worm and Worm wheel, efficiency, η > 50 %
v: Selecting nature of bodies as single throated. vi : Selecting RH- RH, nature of Helix. e) Selection of starts (Z) Selection is based on overall gear ratio, Z (increase number of starts =2) Z+ z
≥40 ……PSG 8.52 Z(1+i) = 40 Z = 40 /31 Z = 1.29 , selecting no. of start as 2 No. of tooth on gear are z= i × Z = 30 × 2 = 60 f) Diameter factor ‘q’ , …..
PSG 8.44/8.45
let q = 11 72
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
g) Lead angle and Helix angle on WW & W …. (PSG 8.44) Lead angle,
= 10.3°
Helix angle on WW =
= 10.3°
Helix angle on Worm =
= 90 -10.3 = 79.7°
g) Calculation of Virtual number of teeth = 349.87 Zv2=
= 63
h) Lewis form factor Yv1 = π ×
……PSG8.50 ….. For 20° involute
Yv1 = 0.4756 Yv2= π ×
, Yv2 = 0.438
i) Material selection Pa Element Material
[σb] in MPa
Worm ( steel)
Worm Wheel
Bronze chilled, σu >390 MPa
[σc] in MPa
(For mx>6)
N/ High σu for bending strength, High BHN to control pitting, Heat generation causes scoring. In WW scoring is predominant than pitting hence to avoid scoring dissimilar material are selected. j)
Checking for weaker element Strength factor for worm, fs1= [σb1] × Yv1= 135 × 0.4756 = 64.206 Strength factor for worm wheel, fs2 = [σb2] × Yv2 =110 × 0.438 = 48.18 Worm wheel is weaker, design the
worm wheel. k) Design criteria Transverse module, mt of worm wheel is axial module of worm. The axial module is calculated based on wear failure under static and checked for bending failure under
static and dynamic condition. 73
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Where, z = 60, q=11, =
=149 Mpa = 4965.634 N.m
Calculating Centre distance, a = 38.91cm = 389.1mm,
mx = 10.96, Selecting standard mx =12 …….PSG8.2 l) Checking for bending (static) Induced bending stress, <
……E11.4/ …PSG 8.44 …. Safe in bending
m) Check for dynamic load Static strength, Where,
n) Checking for Lewis dynamic load Lewis dynamic load, Pitch line velocity, ,
, ….. PSG8.52 74
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
....hence design is safe in Lewis dynamic load. o) Constructional details, ….Ref: PSG 8.43 Addendum, a = mx =12mm Clearance, c = 0.25mx = 0.25×12 = 3mm Dedendum, d = a + c =15mm Parameter Pitch
circle diameter
Worm d1= mx × q =12×11=132mm
Worm Wheel d2 = z × mx =720mm
Addendum diameter
da1=d1+2×fo ×mx =156mm
da2=(z+2fo+2X). mx = 744
Dedendum diameter
df1 =d1 - 2fo.mx - 2c =102mm
df2=d2 - 2fo.mx -2c = 690 mm
p) Thermal check Due to high sliding velocity heat generation is major problem is Worm and Worm wheel. The thermal conditions are checked only after study state conditions. After steady condition
heat transfer will takes place by convection and radiation. Estimated projected area by AGMA,
Heat generated: Hg = (1-η) × [P] Pitch line velocity at worm, Sliding velocity, (From PSG 8.49) , From graph using extrapolation, μ =0.03 for Vs =6.32 m/s , Efficiency, Heat generated, Hg =
(1-0.8549)*15600 = 2263.56 watt 75
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Heat Dissipation: Hd = Hc +Hr Where, Hc = heat transfer by convection = Hr = heat transfer by radiation = Where, C1 = (11 to 15) W/m2K (still air – normal breeze), C1 = 35 W/m2K (forced convection),
and C2 =5.67 x 10-8 Assuming, Temperature of housing, , Equating heat generated with heat dissipated and calculating required area, Hg = Hd = + 2263.56 = + Hence, A= 3.08 m2 Required area without
blower = 3.08 m2 Required area with blower = 1.625 m2 Since available projected area , blower has to be provided. OR the balance area can be provided by fins or fan can be mounted on the worm shaft.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Chapter 2 Rolling Contact Bearing Introduction Bearing is a machine member whose function is to support a second member, preventing its motion in the direction of an applied load but at the same
time allowing its motion in another predetermined direction.
Classification of Bearings 1. According to friction applied. (a) Sliding Bearing: In sliding bearings, a lubricant is introduced between mating surfaces. Based on type of lubrication, sliding
bearings are further classified as: Hydrodynamic lubricated bearings: If lubricant is a fluid and completely separated the moving surface during normal operating conditions, then such a bearing is
called hydrodynamic lubricated bearing. Type of lubrication is called fluid film/thick film or perfect lubrication. Partial lubricated bearings: If layer of fluid lubricant does not completely
separate the moving surfaces so that partial metal to metal contact exist, then such bearings are called bearings with boundary lubrication. (b) Rolling contact bearing: Rolling contact bearings are
bearing in which balls, rollers or needles are used to separate two mating surfaces and to support applied load. 2. According to the direction in which the applied load is supported by bearing
relative to axis of shaft (a) Journal bearing: If bearing supports a load in radial direction, then it is called as journal or sleeve bearing. They are further classified based on angle of contact of
bushing with journal. 1) Full journal bearing: Contact angle of bushing with journal is 360° Bearing completely surrounds journal Shaft member-journal, Cylindrical body around
journal-bearing. Commonly used in machinery especially when load varies in radial direction. 2) Partial journal bearing: 77
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Angle of contact of bushing with journal is 180° or less. Used in limited application when load acts in a constant direction or when there is little change in direction of load. (b) Thrust bearing:
If bearing supports the shaft in axial direction, then it is called thrust bearing.
Comparison of rolling contact bearing and sliding contact bearing: Sr.
Rolling contact Bearing
Sliding contact Bearing
No. 1.
Require lower starting torque and suitable for Require higher starting torque due to metal applications where there are frequent starts. to metal contact in the beginning and during reverse of
Suitable for low load and speed application.
For precise location of journal axis these are Journal moves eccentrically with bearing; preferable. eccentricity varies with load.
Better from space consideration.
They are not cost affected since they are Cost is affected with quantity. Quiet high standard usually made is mass scale. for few pieces but compatible for larger pieces.
Maintenance cost is less.
Maintenance cost is high.
Can be lubricated once for bearing life
Require alternate or continuous lubrication.
For finite life load carrying capacity decreases Load carrying capacity with increase in speed. proportions to speed.
Lower resistance to shock loads.
Not suitable for low load and speed application.
They require pump, filter, sump and other accessories. Hence consume more space.
It is better suited for shock loads. Ex. Connecting rod, crankshaft.
Noisier at high speed.
Less noisy at high speed.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Life of bearing is finite and decreases with It can adjust load and speed within increase in load and speed. prescribed limit.
Low initial cost.
High initial cost.
Mounting of bearing is difficult.
Mounting of bearing is relatively simple
With increase in speed coefficient of friction As speed increases and hydrodynamic increases. action takes place, coefficient of friction reduces.
Rolling element bearings types: Rolling Element Bearings, also known as Rolling Contact Bearings or simply Rolling Bearings, dissipate the load on the bearing by transmitting the load to the
rolling elements inside the bearings. Depending on type, shape, and size of these roller elements, Rolling Contact Bearings are further classified into the following types:
Fig: (a) Cylindrical roller bearings with cage. (b) Deep groove ball bearing. (c) Needle roller bearing. (d) Taper roller bearing. (e) Spherical roller bearing. (f) Roller thrust bearing. (g)
Self-aligning ball bearing.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Designation of RCB RCB is designated normally by a 4-digit number, let’s consider ABCD where, i. ‘A’ represents the type of bearing 6 – DGBB 2 – Self aligning bearing 3 – Double Row Angular roller
bearing 30, 31, 32 – Taper Roller bearing NU2 – Cylindrical Roller bearing ii. ‘B’ represents type of load 0- Extra light 2- Light 3- Medium 4- Heavy iii. ‘CD’ represent the diameter of the bore of
the bearing
Dimension Series The last two digits of the designation of RCB represents the bore diameter. These are termed as the dimension series. The diameters are represented in the following manner. 00 –
Diameter = 10 mm 01 – Diameter = 12 mm 02 – Diameter = 15 mm 03 – Diameter = 17 mm From 04 onwards the diameter is determined by multiplying the designated last two-digit number by 5, for e.g. 04 –
Diameter = 20 mm 05 – Diameter = 25 mm 06 – Diameter = 30 mm and so on.
Static load carrying capacity It is defined as load carrying capacity on bearing when shaft is stationary. It produces permanent deformation in balls and races which increases with increase in
load. Permissible static load depends upon permissible deformation.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
The static load carrying capacity of a bearing is defined as the static load which corresponds to a total deformation of balls and races at the most heavily stressed point of contact, equal to 0.0001
of the ball diameter.
Static equivalent load for RCB The static equivalent load is a hypothetical load that produces contact stress equal to maximum stress under actual conditions, while bearing is stationary, in the
area of contact between the most heavily stressed rolling element and bearing raceway. The static radial load passing through center of the bearing is taken as the static equivalent load for radial
bearings, while the static axial load in the direction coinciding with the central axis is taken a static equivalent load for thrust bearing. P0 = (X.V.Fr + Y.Fa)*S.F P0 = Static equivalent load; X =
Radial load factor; Y = Axial load factor, V= race rotation factor, Fr = Radial load, Fa = Axial load, S.F = Service Factor.
Dynamic load carrying capacity The Dynamic Load capacity can be defined as the load that will give a life of one million revolutions of the inner race. The dynamic load rating hence plays a vital
role for the bearing life. The relation between the bearing life and the dynamic load capacity is expressed as follows: L = (C/P)3- for Ball bearing L = (C/P) 10/3 – for Roller bearings Here L =
Bearing Life in millions of revolution, P = Equivalent Radial Load, C = Dynamic Load capacity of the bearing
Basic Life or ‘L10’ is the life that 90% of bearings can be expected to reach or exceed the theoretically calculated life under conventional operating conditions. L10 indicates that only 10% of the
bearing sample will fail to reach or exceed the theoretically calculated life. L10 is expressed in terms of ‘million revolutions’ or ‘mr’.
Median Life ‘L50’ is the life of the bearing at 50% reliability i.e. only 50% of the bearings can be expected to reach the rated life. Median Life is 5 times of the Basic Life.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Load Life Relationship: This expression gives a direct relationship between the dynamic capacity of the bearing and the basic life of the bearing. It is expressed as
Here, C = Basic Dynamic Carrying Capacity P = Equivalent Dynamic Load, which depends on the radial and axial loads on the bearing. k = Bearing constant which varies as:
For ball bearings, k = 3 For roller bearings, k = 3.333
It is important to note that the terms ‘C’ and ‘P’ are considered under standard conditions. The load life relationship can be modified for operating conditions as follows:
Where, L is the desired life in million revolutions C and P are the actual operating Basic Dynamic Carrying Capacity and Equivalent Dynamic Load respectively.
Dynamic Load Rating for RCB under variable loads
The rating life of bearing is based on fundamental equation,
Consider a bearing is subjected to variable loads. Let P1, P2, P3…….be the loads on the bearing for successive N1, N2, N3………revolutions respectively. If the bearing is operated at constant load P1,
then the life is given by, L1 The fraction of life consumed with load P1 for N1 revolution is given by, 82
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Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Similarly, fraction of life consumed with load P2 for N2 revolution is given by,
Hence, total life is given by,
If equivalent load is acting on N revolutions then,
N = N1 + N2 + N3……
Bearings, in general are subjected to a variety of loads and operating conditions. Thus, there are different causes of bearing failure. Essentially, bearing failure due to structural or strength
factors are widely considered during the design of bearings. Bearing failure can occur due to one or more of the following reasons: 1. 2. 3. 4. 5. 6. 7. 8.
Abrasion Corrosion Fatigue Cycles Extreme Loads or Induced Pressures Overheating Misalignment True and False Brinelling Improper Lubrication 83
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Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
9. Contamination 10. Loose or Tight Fittings Life of bearings is generally expressed in terms of reliability and probability. Life of bearing is not a definitive value but an estimate presented on
the basis of a sample calculation.
Selection of Bearing Life:
Bearing life is generally expressed in the number of revolutions is sustains under operating conditions before undergoing some form of failure as mentioned previously. However bearings of a certain
type i.e. Deep Groove Ball Bearings, Self-Aligning Ball Bearings, etc. can have different life under identical operating conditions. Thus, it is necessary to calculate bearing size along with the
bearing life calculation. The life of bearing is expressed in terms of reliability of the bearing. For instance, if the life of bearing is expressed as 500 million revolutions at 95% reliability,
then it means that the 95% of the bearing population of this category will sustain 500 million revolutions before failing. This expression of bearing life is a calculative prediction method. Bearing
life based on the type of machinery: Sr. No.
Class of machines
Rarely used machines; eg: Demonstrative machines
Intermittent operations; secondary machines; eg: hand tools, domestic appliances
4000 to 8000
Intermittent operations; primary machines; eg: Machinery used in conveyor plants, lifts
8000 to 12000
Machine use 8hrs/day and not always fully used; eg: Stationary electric motors, general purpose machine
12000 to 20000
Fully utilized machines for 8hrs/day; eg: cranes, industrial material handling systems
20000 to 30000
Continuous use machines for 24hrs/day; eg: Pumps, compressors, mine hoists, etc.
40000 to 60000
Machines with high degree of reliability with 24hrs/day full utilization; eg: Paper making
100000 to 200000
Bearing hours
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
machinery, on-board merchant ship machines
Reliability of RCB:
Reliability of the Rolling Contact Bearings is the proportional estimate of the bearings that will reach or exceed the rated life as provided by the manufacturer. If the reliability of a bearing is
80% then 80% of the bearings from the sample population can be expected to reach or exceed the rated life. Relation between the life and probability of survival is given by following equation.
Where P is the desired reliability and P10 is the basic reliability, P10 is always 90%. L10 is the basic life and L is the desired life. b - Constant 1.34 for DGBB.
Selection of RCB from catalogue:
Bearings are subjected to essentially two forces namely; Radial force and Axial force. The extent and combination of these two forces on the bearings help in basic determination of the type of
bearing. The following is the list of different bearings classified according to the nature of the subjected load: Sr. No.
Type of Bearing
Application (Load Nature)
Deep Groove Ball Bearing
Both radial and thrust loads – at high speeds
Self-Aligning Ball Bearings
Insensitive to shaft misalignment
Single Row Angular Ball Bearings
For heavy axial loads
Double Row Bearings
Spherical Roller Bearings
High radial and bidirectional axial loads
Cylindrical Roller Bearings
Heavy radial loads at high speeds
Taper Roller Bearings
Combined radial and axial loads
Single Thrust Ball Bearings
Unidirectional axial loads only
Ball Radial and bidirectional heavy thrust loads
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Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Double Thrust Ball Bearings
Bidirectional axial loads
Spherical Roller Thrust Bearing
Heavy axial loads at high speeds; self-aligning
Factors Influencing Life of RCB
1. Manufacturing process: The standard rolling contact bearings have 4 major parts: inner and outer races, rolling element, and cage. Both the races are manufactured by steel tubing, which may
involve drawing, extrusion, and turning operations. The rolling elements are generally produced by the cold forging process. The cage is produced by stamping operation on the sheet metals. All these
components are heat treated before their assembly. Any miscalculation, deviations, errors, limitations, and incompetence in the manufacturing process may cause a reduction in both expected bearing
life and performance. 2. Variable loading: Bearings are designed, manufactured, and optimized for certain set of working conditions. Any change in the loading pattern on the bearing may cause damage
to the bearing resulting in loss of work, efficiency, time, energy, resources, and money. If the loading patterns on the bearing during its operation are unknown or random, it is difficult to
calculate and estimate the efficiency and durability of the bearing. Also, the bearings should be utilized as per the conditions prescribed by the manufacturer or according to the conditions
mentioned in the design data book for proper and efficient performance. 3. Temperature: The temperature of the working environment around the bearing also affects the life of bearing. If the bearings
are designed without considering proper temperature coefficients, the resultant bearing life may be less than the required or expected bearing life as the thermal expansion of the roller elements
causes improper functioning of the bearings. 4. Miscellaneous Other factors that influence the bearing life is also affected by abrasion of the cage, corrosion of rolling elements, improper
lubrication, faulty assembly and alignment etc. The effect of these factors can be reduced by proper maintenance of the bearings.
Applications of RCB
1. 2. 3. 4. 5. 6. 7. 8.
Dental Drills Wind Turbines Home appliances such as fans, washing machines Machines tools such as Lathe machine, Drilling machine, milling machine etc. Bicycles and motorbikes Front and rear axle of
automobiles Conveyor belt drive systems Robotic systems
The following are some of the applications of the Rolling Contact Bearings:
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
9. Material handling systems such as cranes, automated guide vehicles (AGVs) 10. Pumps, Motors, and Compressors
Steinbeck’s Equation of Basic Static Capacity of Bearing
1. 2. 3. 4.
The races are rigid and retain their circular shape even after loading. The balls are equally spaced. The balls in the upper half do not support any load. There is a single row of balls.
Basic Static Carrying Capacity [C’]: It is that load on the bearing that produces a permanent deflection of ‘0.0001d’ in the roller elements such as balls, rollers, races, etc. Richard Stribeck
proposed a relation to calculate the Basic Static Carrying Capacity depending upon the number of balls or rolling elements in the bearing based on the assumptions:
Figure : Stribeck's Basic Static Carrying Capacity Let C0 be the basic static carrying capacity of the ball bearing in Newton, N Let P1, P2, P3… be the radial forces acting at the contact between the
inner race and respective balls in Newton, N. Let ∂1, ∂2, ∂3… be the radial deflections at the contact between the inner race and respective balls, in millimetres, mm. Consider the equilibrium of the
forces in the vertical direction, 87
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Now, the radial deflections at the contact between the inner race and respective balls is related to the radial forces acting at the contact between the inner race and respective ball. ,
Where, k0 is the constant of proportionality. Now, ,
Thus, ,
Equation (1) can be written as
As the races are rigid, the displacement of inner race with respect to outer race will be due to the deformation of balls. ,
Thus, we get, ,
Thus, Equation (2) becomes;
Now, 88
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Where, Z = No. of balls
The value of M and M/Z can be found out from the table
It is seen that ‘Z/M’ is practically constant and Stribeck suggested the value of 5 for ‘Z/M’ , Substituting this value in Equation (5)
From experiment it is found that to produce the permanent deformation of a ball is maximum load at the most heavily stressed ball is given by
Where, d = diameter of the balls K = constant depending on radius of curvature at point of contact and material of the balls
Where; C0 is the Basic Static Carrying Capacity, Z is the number of rolling elements; in this case number of balls, d is the diameter of the balls, K is known as the Stribeck’s stress constant For
straight roller bearings the relation is modified incorporating the roller length (l)
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
NUMERICALS: Problem I: What change in the loading of a ball bearing will cause the expected life to be doubled? Solution: Let initial load = (Peq)i Initial expected life = Li Now as mentioned in the
question the expected life doubled New expected life (Ln) = 2* Li The relation between the expected life and load on bearing is given by ……………………………………………………PSG (4.2) For ball bearing, k = 3 For
initial conditions the equation will be, For the new conditions the equations will be, As Ln=2*Li, From equation 1 substituting value of Li
= 0.7936 90
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Therefore, for life to be doubled, the load is reduced by 20.64%.
Problem II Select suitable DGBB for following specifications. a) Radial Load = 4000 N; Thrust Load = 2000 N; Life = 8000 Hrs., Shaft Speed = 1000 RPM; Reliability = 99% b) Radial Load = 3000 N;
Thrust Load = 2000 N; Life = 3 years at 10 hrs. per day, shaft Speed = 1200 RPM, Shaft Diameter = 50 mm; Service Factor = 1.2; Solution: a) i) Lhr = 8000 Hrs. Fr = 4000 N; [Lmr] =
Fa = 2000 N;
N = 1000 RPM
= 480 mr
For 99% reliability,
, b = 1.34 for DGBB
……………………. PSG (4.2)
= L0.99 = 83.11 mr Suitable bearings for 50 mm diameter are 6010, 6210, 6310 and 6410 1) Let’s check life of DGBB 6010 Dynamic Load Capacity ‘C’ = 17000 N, Static Load Capacity ‘C0’= 13700 N……PSG
; e = 0.31;
> e;
Therefore, x = 0.56 and y = 1.4 ……………..PSG (4.4) 91
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Peq = (X.V.Fr + Y.Fa) S = (0.56*1*4000+1.4*2000) 1.2 = 6048 N………..…PSG (4.2) Lmr =
= 22.0208 mr...……….………...…PSG (4.2)
Since, [Lmr] required > Lmr,
Bearing 6010 is not suitable.
2) Let’s check life of DGBB 6210 Dynamic Load Capacity ‘C’ = 27500 N, Static Load Capacity ‘C0’= 21200 N……PSG (4.13)
; e = 0.27;
> e;
Therefore, x = 0.56 and y = 1.6
…………………………………………………PSG (4.4)
Peq = (X.V.Fr + Y.Fa) S = (0.56*1*4000+1.6*2000)1.2 = 6528 N………………..…PSG (4.2) Lmr =
= 74.75 mr…………………………………………………...PSG (4.2)
Since, [Lmr] required > Lmr,
Bearing 6210 is not suitable.
3) Let’s check life of DGBB 6310 Dynamic Load Capacity ‘C’ = 48000 N, Static Load Capacity ‘C0’= 35500 N…….PSG (4.14)
; e = 0.24;
> e;
Therefore, x = 0.56 and y = 1.8;....……………………………………….………PSG (4.4) Peq = (X.V.Fr + Y.Fa) S = (0.56*1*4000+1.8*2000)1.2 = 7008 N……………..…PSG (4.2) Lmr =
= 321.322 mr……………………………………………...PSG (4.2)
Since, [Lmr] required < Lmr, Therefore, Bearing 6310 is suitable for given specification.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Solution: (b): Given Data: Lhr = 3 years and 10 hrs. a day; Fr = 3000 N; Fa = 2000 N; N= 1200 RPM [Lmr] =
= 788.4 mr
Suitable bearings for 50 mm diameter are 6010, 6210, 6310, and 6410
For, DGBB 6010 Dynamic Load Capacity ‘C’ = 17000 N, Static Load Capacity ‘C0’= 13700 N……PSG (4.13)
; e = 0.31;
> e;
Therefore, x = 0.56 and y = 1.4; ………………………………………………….PSG (4.4) Peq = (X.V.Fr + Y.Fa) S = (0.56*1*3000+1.4*2000)1.2 = 5376 N……………..…PSG (4.2) Lmr =
= 31.62 mr………………………………………......…..…PSG (4.2)
Since, [Lmr] > Lmr Bearing 6010 is not suitable.
For, DGBB 6210 Dynamic Load Capacity ‘C’ = 27500 N, Static Load Capacity ‘C0’= 21200 N……PSG (4.13)
; e = 0.27;
> e;
Therefore, X = 0.56 and Y = 1.6;....………………………………….……PSG (4.4) Peq = (X.V.Fr + Y.Fa) S = (0.56*1*3000+1.6*2000)1.2 = 5856 N……..…PSG (4.2) Lmr =
= 103.56 mr…………………….………...…..…PSG (4.2)
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Since, [Lmr] > Lmr, Bearing 6210 is not suitable.
For, DGBB 6310 Dynamic Load Capacity ‘C’ = 48000 N, Static Load Capacity ‘C0’= 35500 N…….PSG (4.14)
; e = 0.27;
> e;
Therefore, x = 0.56 and y = 1.6; ....…………………………………….……….…PSG (4.4) Peq = (X.V.Fr + Y.Fa) S = (0.56*1*3000+1.6*2000)1.2 = 5856 N………………...…PSG (4.2) Lmr =
Since, [Lmr] > Lmr,
= 550.7 mr…………………….………………………....…PSG (4.2) Bearing 6310 is not suitable.
For, DGBB 6410 Dynamic Load Capacity ‘C’ = 70000 N, Static Load Capacity ‘C0’= 53000 N……PSG (4.15)
; e = 0.24;
> e;
Therefore, X = 0.56 and Y = 1.8...……………………………………………….…. PSG (4.4) Peq = (X.V.Fr + Y.Fa) S = (0.56*1*3000+1.8*2000)1.2 = 6336 N………………..…PSG (4.2) Lmr =
= 1348.49 mr……………………….………….………..…PSG (4.2)
Since, [Lmr] < Lmr Therefore, Bearing 6410 is suitable for given specification.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Problem III A 6207 DGBB is subjected to following repeated load cycle. Determine the expected life of the bearing in hours with probability of survival of 95%. Phase, Radial Load (KN), Axial Load
(KN), Speed N (RPM), Fraction of cycle (%) Phase
Radial Load
Axial Load
Speed N (RPM)
Fraction of
cycle (%)
Solution: Given Bearing: 6207 Type: DGBB; Medium Shock; Diameter = 35 mm. Consider cycle for 1 min, N = Speed (RPM) x Fraction of cycle. Therefore, N1 = 500*0.15 = 75, N2 = 600*0.2 = 120, N3 =
400*0.3 = 120, N4 = 800*0.35 = 280 Hence, N1 + N2 + N3 + N4 = 595 95
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
For bearing DGBB 6207, Dynamic Load Capacity ‘C’ = 20000 N, Static Load Capacity ‘C0’= 13700 N………PSG (4.13) Consider, Service Factor ‘S’ = 1.2 and V= 1 and from PSG (4.4), Phase
P = (X.V.Fr + Y.Fa)s (kN)
Dynamic Load Rating for Rolling Contact Bearing is given by,
Peq = Peq =
= 3.786 KN
Lmr =
= 147.41 mr ,
Therefore, Lhr = =
Lmr =
= 4129.33 hrs.
, For 95% probability p = 0.95, b = 1.34 for DGBB……...PSG (4.2)
= 96
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
PROBLEM IV: Find the rating life of a single row angular contact ball bearing 7310 which carry the radial load of 2500N and axial load 1500N. Assume service factor = 1.5 Solution: Data: Fa = 1500N,
Fr = 2500N, S = 1.5 Given Bearing Designation: 7310 Bearing Details: Single Row Angular Contact Bearing (PSG 4.4) Basic Dynamic Carrying Capacity, C = 5300kgf = 53000N (PSG 4.19) Basic Static
Carrying Capacity, C = 4050kgf = 40500N (PSG 4.19) , From PSG 4.4, , Thus, X = 0.56, Y = 1.8 Equivalent Load, P,
Load Life Relationship,
For Ball Bearings, k = 3
Thus, the rating life of the given bearing is 640 million revolutions. 97
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
PROBLEM V: Select a suitable spherical roller bearing for the following specifications: Radial load = 5kN, Axial load = 1.5kN, Speed = 960rpm, Expected life = 5000hrs at Reliability of 92%. Solution:
Given data, Fr = 5000N, Fa = 1500N, N = 960rpm, T = 140°C, Lhr = 5000hrs, p = 92% Let the service factor = 1.6
To find the rating life, L10
Here p = 92%, p10 = 90%, k = 1.17 for spherical bearings
Thus, L10 = 351.57 mr Since required life is high, Assume the bearing to be ‘22212C’ (PSG4.32) C0 = 8500kgf = 85000N, C = 10000kgf = 100000N , 98
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
e = 0.23,
……. PSG 4.4,
Therefore, X = 0.67, Y = 4.4 …….. (PSG 4.4) Now, Equivalent Load on Bearing on inner race,
Calculating Basic Life for 22212C , Assuming temperature effect at T= 140o, C actual = 0.9C = 90,000N
Since, the basic life (204.94 mr) is less than the required life (351.57mr), hence bearing is not suitable . Selecting the next bearing as 22214C C0 = 10800kgf = 108000N, C = 12200kgf = 122000N ,
e = 0.23
……… PSG 4.4
, Therefore, X = 0.67, Y = 4.4 ….. (PSG 4.4) Now, Equivalent Load on Bearing on inner race,
Calculating Basic Life for 22212C, At temperature , T= 140o, C actual = 0.9x122000 = 1,09,800N
Since, the basic calculated life (370.32 mr) is greater than the required life (351 mr), the bearing ‘22214C’ can be suitably selected. 99
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
PROBLEM VI: Select suitable Taper Roller bearing for following specification: Shaft diameter = 35mm, Radial load = 800N, Axial load = 750N, Speed = 760rpm, Expected life = 8000hrs at reliability of
92%. Solution: Diameter = 35mm, Fr = 800N, Fa = 750N, N = 760rpm, Lhr = 8000hrs, p = 92% Assume service factor, S = 1.5
Basic life, L10 , Here P = 92%, P10 = 90%, k = 1.17 for taper bearings
Now, Equivalent Load on Bearing on inner race, Available bearings; 32207A and 32307A based on shaft diameter of 35mm For 32207A; C0 = 5050kgf = 50800N, C = 5760kgf = 57600N
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
For 32206-208, e = 0.37 ,
, Thus, X = 0.4, Y = 1.6,
Thus, Equivalent load at the inner race,
, Thus, the Required Basic Life (446 mr) is less than the Rated Bearing Life (16123.63 mr) Hence bearing is suitable, Selecting Bearing: 32207A
PROBLEM VII: Select a suitable ball bearing to operate on a following work cycle for life of 5 years at 8hrs/day for 300 days per year. Radial Load, N
Speed, rpm
% of time
Solution: Consider cycle for 1 minute; N1 = 400*0.25 = 100 rpm N2 = 500*0.35 = 175rpm N3 = 600*0.40 = 240rpm N = N1 + N2 + N3 = 515rpm Consider Deep Groove Ball Bearing, From PSG 4.4, when Y = 0, X =
1 Thus, Equivalent load on the bearing at the inner race with a service factor of 1.5,
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
To find out cubic mean load,
Now, required life, Lhr = 8*300*5 = 12000hrs Thus, rated life in million revolutions,
Load Life Relationship;
For ball bearings, k = 3
Now, from PSG 4.12, 4.13, 4.14, 4.15, We can select the following Deep Groove Ball Bearings; 6007 and above; 6205 and above; 6304 and above; 6403 and above
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Module 3 SLIDING CONTACT BEARING: A sliding contact bearing is any bearing that works by sliding action, with or without lubricant. This group encompasses essentially all types other than
rolling-element bearings. Also referred to as sleeve bearings or thrust bearings, terms that designate whether the bearing is loaded radially or axially.
Lubricants: Lubrication is the science of reducing friction by application of a suitable substance that is lubricants between the two mating surfaces of the bodies with relative motion. Types of
Lubricants 1. Liquid lubricants e.g. Minerals , vegetable oils 2. Semi solid lubricants e.g. Grease 3. Solid lubricants e.g. Graphite, molybdenum disulphide Lubricants is a substances inserted or
introduced between the two contacting surfaces having relative motion, so as to reduce the friction and wear. The lubricants perform the following functions: 1. To reduce the friction between the
contacting surfaces; 2. To reduce the wear; 3. To carry away the frictional heat; 4. To protect the surfaces against corrosion; 5. To carry away the worn-out particles; and 6. To prevent the entity
of foreign particles, like dirt and dust, to the contact zone.
Viscosity: The internal frictional resistance offered by a fluid to change its shape or relative motion of its parts. An oil film placed between two parallel plates is shown in figure. The lower
plate is stationary and the upper plate is moved with a velocity U by means of a force P. The intermediate layers will move with velocities which are proportional to their distance from the
stationary plate. Hence According to Newton’s law of viscosity, the shear stress is proportional to the rate of shear at any point in the fluid. Hence, 103
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
When the velocity distribution is nonlinear with respect to h, Then,
Physical and Chemical Properties of Lubricants: Following properties are desirable in lubricants used in bearings: 1. It should have an optimum viscosity for the given application. 2. It should
have high viscosity index 3. It should have good oiliness property. 4. It should have high specific heat. 5. It should have high thermal conductivity. 6. It should have high flash point 7. It should
have high fire point. 8. It should have low pour point. 9. It should have anti-foaming property 10. It should have high oxidation stability 11. It should not be acidic. 12. It should have good
demulsibility i.e. it should have low demulsibility number. 13. It should be chemically stable with bearing material and atmosphere over the range of temperatures encountered in the operation. 14. It
should be commercially available at reasonable cost. Basic Modes of Lubrication: The basic modes of lubrications are 1. Thick film lubrication 2. Thin film lubrication 3. Zero film bearing which
operates without any lubricating oil film In thick film lubrication two surfaces of the bodies in relative motion are completely seperated by a film of lubricant. These are further classified to 1.
Hydrodynamic lubrication 2. Hydrostatic lubrication 3. Elasto-hydrodynamic lubrication 4. Solid-film lubrication In thin film lubrication the lubricant film is relatively thin and there is partial
metal to metal contact. These lubrication can be observed in machine tool slides and door hinges. 104
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Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Hydrodynamic lubrication: In hydrodynamic lubrication, the two surfaces in relative motion are separated by a relatively thick film of fluid, so as to prevent the metal to metal contact. The
hydrodynamic lubrication is also called full film or thick film lubrication. In hydrodynamic lubrication, the load supporting high pressure fluid film is created due to: a) The shape of the zone
between the contacting surfaces. b) The relative motion between the contacting surfaces. Hydrostatic lubrication: In hydrostatic lubrication, the load supporting high pressure fluid film is created
by an external source, like pump. The lubricant, which is pressurized externally, is supplied between the two surfaces. Elasto-hydrodynamic lubrication: This lubrication occurs when a lubricant is
introduced between the surfaces which are in rolling contact. The lubricant is entrapped in deformation zone between the contacting surfaces. Solid-film lubrication: When the bearings are operated
at extremely high temperatures, the lubricating oils are not suitable as they may be lose their properties at such high temperatures. In such cases, a solidfilm lubricant, such as graphite or
molybdenum disulphide is used.
Hydrodynamic Bearing: Applications of hydrostatic bearings are Crank shaft, hydraulic turbine, centrifugal pumps etc. Hydrodynamic Journal bearing is a sliding contact bearing working on
hydrodynamic lubrication and support the radial load. Journal bearing are classified as, 1. Full Journal Bearing In full journal bearing the angle of contact of the bearing with the journal is 360o
2. Partial Journal Bearing In Partial journal bearing the angle of contact of the bearing with the journal is less than 360o
Fig: Full Journal Bearing
3. Fitted Journal Bearing In fitted journal bearing there is no clearance between the bearing and the journal. The diameter of journal and bearing are same. Fig: Partial journal bearing with 120o
angle of contact.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Following figure shows the position of journal at rest, at start and at full speed when it rotates in clockwise direction.
In hydrodynamic bearing initially the journal is at rest. As the journal starts to rotate, sufficient pressure is developed in the clearance space. During this period there is partial metal to metal
contact and a partial lubricant film. That is there is thin film lubrication. As the speed is increased, more lubricant is forced into the wedge shaped clearance space and sufficient pressure is
built up, separating the surfaces of the journal and the bearing, which is thick film lubrication. There is transition from thin film lubrication to thick film lubrication as the speed increases.
This is McKee’s investigation.
Thrust Bearing: A thrust bearing is a particular type of rotary rolling-element bearing. Like other bearings they permit rotation between parts, but they are designed to support a predominately
axial load i.e. pressure on bearing is parallel to the axis of the shaft. There are two types of thrust bearing: a) Footstep bearing: It is a thrust bearing in which the end of the shaft is in
contact with the bearing surface. b) Collar bearing: It is a thrust bearing in which collar integral with the shaft is in contact with the bearing surface. The shaft can be with single collar or
multiple collars.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Pressure distribution in hydrodynamic bearing: The following figure demonstrates a hydrodynamic journal bearing and a journal rotating in a clockwise direction. Journal rotation causes pumping of
the lubricant (oil) flowing around the bearing in the rotation direction. If there is no force applied to the journal its position will remain concentric to the bearing position. However a loaded
journal displaces from the concentric position and forms a converging gap between the bearing and journal surfaces. The pumping action of the journal forces the oil to squeeze through the wedge
shaped gap generating a pressure. The pressure falls to the cavitation pressure (close to Fig: Oil pressure distribution in hydrodynamic bearing the atmospheric pressure) in the diverging gap zone
where cavitation forms. The oil pressure creates a supporting force separating the journal from the bearing surface. The force of oil pressure and hydrodynamic friction force counterbalance the
external load F. The final position of the journal is determined by the equilibrium between the three forces.
Hydrostatic Bearing (Externally pressurised bearing): In hydro static bearings, the load supporting high pressure fluid film is created by an external source, like pump. The lubricant, which is
pressurised externally, is supplied between the two surfaces. So, hydrostatic bearings do not require motion one surface relative to another. Figure: Elements of hydrostatic bearing
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Advantages of hydrostatic bearings: 1. 2. 3. 4. 5.
They can take heavy loads even at exceptionally low speeds. The load carrying capacity of bearings is independent of the speed. They have very low friction loss, even at the starting. They have high
positional accuracy. In hydrostatic bearings, any fluid which is already present in a device lubricant. For example, liquid oxygen in rocket engines, water hydraulic machinery and kerosene in
aircraft engine serve the purpose of lubricant.
Limitations of hydrostatic bearings: 1. They required auxiliary equipment like, pump, filter, oil supply line, etc. the system is more complicated, expensive, and liable to failure. 2. They have
high initial as well as maintenance cost. 3. Overall power loss is not necessary low.
Applications of hydrostatic bearings: The hydrostatic bearings are used in: 1. Vertical turbo generators 2. Ball mill 3. Gyroscopes 4. High speed dental drills, 5. Spindles of internal grinding
machines. 6. Large telescopes, 7. Machine tools, 8. Centrifugal, 9. Rolling mills, 10. Rocket engines.
Squeeze film bearings: In wedge film journal bearing, the bearing carries a steady load and the journal rotates relative to the bearing. But in certain cases, the bearing oscillates or rotate so
slowly that the wedge film cannot provide a satisfactory film thickness. 108
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Fig: Squeeze film bearing
If the load is uniform or varying in magnitude while acting in a constant direction, this becomes a thin film or possibly zero film problem. But if the load reverse its direction, the squeeze film
may develop sufficient capacity to carry the dynamic loads without contact between the journal and the bearing. Such bearings are known as Squeeze Film Journal Bearing.
Difference of Hydrodynamic bearing and hydrostatic Bearing Hydrodynamic bearing
hydrostatic Bearing
1. The load supporting fluid film is created by 1. The load supporting fluid film is created by the shape and relative motions of the sliding an external source like a pump. surfaces. 2. These
bearing are also called as self-acting 2. The pressure is created by external source bearing because the pressure is created within hence also called as externally pressurised the system. bearing. 3.
These bearings are simple in construction.
3. These bearings are complex in construction.
4. Low initial cost and maintenance cost.
4. Initial cost and maintenance cost is high.
5. Less load carrying capacity at low speed.
5. High load carrying capacity even at low speed.
6. Used in Engine and centrifugal pumps.
6. Used in turbo generator, ball mills and centrifuges,
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Properties of Bearing Materials: 1. Compressive strength 2. Fatigue strength: Material should have high endurance strength 3. Conformability: Accommodate shaft deflection and bearing inaccuracy
without wear and pitting. 4. Embedability: Accommodates small particles of dust, dirt etc. without scoring of material. 5. Bondability: Many high capacity bearing are made by bonding one or more thin
layers of bearing materials. 6. Corrosion resistance: It should be highly corrosion resistant. 7. Thermal conductivity: It should have high thermal conductivity to carry heat generated due to
friction. 8. Thermal expansion: It should have low thermal coefficient of expansion to keep constant eccentricity. 9. Cost and availability: Material of bearing should be readily available and cost
Materials used for Bearings: 1. Babbits 2. Bronze 3. Copper-lead alloy 4. Aluminium alloy 5. Cast-iron 6. Silver 7. Sintered metal 8. Non-metallic materials
Important Parameters used in hydrodynamic journal bearing: 1. Sommerfeld number: It is most important parameters. It contains all the variables which
are controlled by the designer. It has been used as an abscissa in all the charts except the viscosity chart. It is a function of characteristic number and r/c ratio. The Sommerfeld number is given
Where, S – Sommerfeld number, Z – Viscosity of the lubricant (MPa-s) N- Journal speed (RPS) P – Unit bearing pressure (MPa) 110
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Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
2. Length to diameter ratio (l/d) 3. Minimum film thickness variable (ho/c) 4. Eccentricity ( ):
5. Coefficient friction variable: It gives the coefficient of friction of hydrodynamic bearing.
6. Flow variables: It gives the flow rate of the lubricant in hydrodynamic bearing. , where, l – length of the bearing (mm), Q – Flow of the lubricant (mm3/s). 7. Flow ratio (Qs/Q): It gives amount
of axial flow in terms of total flow rate of lubricants. 8. Maximum film pressure ratio (P /Pmax ) 9. Position of maximum film pressure angle (θpmax): It gives an angular position of maximum pressure
in fluid film. 10. Zero pressure angle (θpo): It gives the terminating position of the high pressure fluid film. 11. Position of minimum film thickness angle (Ф): It gives an angular position of the
minimum fluid film thickness. 12. Temperature rise variable : (
Frictional power or the heat generated is given by, (kW) = f W V = Where, f- coefficient of friction, W –Radial load acting on the bearing, N- Journal speed in RPM. The heat carried away by oil flow
is given by, Where, m –mass of the lubricating oil passing through the bearing (Kg/s),
Specific heat of
lubricating oil (KJ/Kg ), – Temperature rise ( ) The mass of the lubricating oil, Equating the heat generated and the heat carried away by oil flow, change in temperature can be calculated. For most
lubricating oils, The average temperature of the lubricating oil,
Minimum film thickness (ho): The minimum oil film thickness is a very important variable which governs the load carrying capacity and the frictional loss of the bearing. It is the oil film
thickness along the line joining the centers of the journal and bearing. It depends upon the surface finish, viscosity of the oil and nature of load. 111
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Bearing characteristics number (ZN/P) and bearing modulus for journal bearing. The transition from thin film lubrication to thick film hydrodynamic lubrication can be better visualized by means of
a curve called
A bearing characteristic number is a dimensionless parameter given by, (
Where, Journal speed, p – Unit bearing pressure.
Figure show bearing characteristics number curve. Thin film lubrication in the region BC and thick film lubrication in the region CD. These two modes of lubrication are divided by AC. The coefficient
of friction is minimum at C or at the transition between these two modes. The value of the bearing characteristics number corresponding to this minimum coefficient is called the Bearing Modulus. It
is denoted by K. The bearing should not be operated near the critical value K at the point C. A slight drop in the speed (N) or a slight increase in the load (P) will reduced the value of
resulting in boundary
lubrication. That is a slight increase in pressure will make the part of a shaft or axle that rests on bearings operate in partial lubrication state resulting in high friction, heating and wear. In
the design of fluid bearings, the Sommerfeld number (S), or bearing characteristic number, is a dimensionless quantity used extensively in hydrodynamic lubrication analysis. The Sommerfeld number is
very important in lubrication analysis because it contains all the variables normally specified by the designer. Variation of coefficient of friction with ZN/P: By experimentation, it is observed
that the coefficient of friction µ is the function of µ = ø (ZN/P). Bearing Modulus is a dimensionless parameter on which the coefficient friction in a bearing depends. In the region to the left of
Point C, operating conditions are severe and mixed 112
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Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
lubrication occurs. Small change in speed or increase in load can reduce ZN′ /p and a small education in ZN′/p can increase the coefficient of friction drastically. This increases heat which reduces
the viscosity of the lubricant. This further reduces ZN′ /p leading to further increase in friction. This has a compounding effect on the bearing leading to destruction of Oil film and resulting in
metal to metal contact. In order to prevent such conditions, the bearing should operate with a ZN′/p at least three times the minimum value of the bearing modulus (K). Suppose it is operating to the
right of the line BA and there is an increase in lubricating temperature. This results in lower viscosity and hence a smaller value of the ZN′/p. The coefficient of friction decreases, and
consequently the lubricating temperature drops. Thus the region to the right of line BA defines “stable lubrication” because the variations are selfcorrecting. The guidelines foe hydrodynamic
lubrication 1. In order to avoid seizure, the operating value of the bearing characteristic number should be at least 5 to 6 times bearing modulus. 2. If the bearing is subjected to fluctuating load
the operating value of bearing characteristic number should be at least 15 times the bearing modulus. 3. When the viscosity of the lubricant is very low, the value of bearing characteristic number
will be low and boundary lubrication will result. Hence if the viscosity of the lubricant is very low then the lubricant will not separate the surfaces of the journal and the bearing. 4. Metal to
metal contact will occur resulting in excessive wear at the contacting surfaces.
Petroff’s Equation or Bearing Friction:
In 1883, Petroff published his work on bearing friction based on following simplified assumptions. 1) No eccentricity between bearings and journal and hence there is no “Wedging action” as shown in
figure. 2) Oil film is unable to support load. 3) No lubricant flow in the axial direction
Unloaded bearing /FCRIT, b) Laminar fluid in clearance space Prepared by: a) Prof. SanjayJournal W. Rukhande Vashi, flow NaviofMumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
With reference to Figure, an expression for viscous friction drag torque is derived by considering the entire cylindrical oil film as the “liquid block” acted upon by force F. From Newton’s law of
Viscosity: F = μ AU /h Where F = friction torque/shaft radius = 2 T f / d A= π d l U = π d n (Where n is in rps and d is in m) h = c (Where c = radial clearance = 0.5(D-d)) r = d /2 Substituting and
solving for friction torque: T f = 4 π2 μ n l r3/c ----------(1) If a small radial load W is applied to the shaft, then the frictional drag force f w and the friction Torque will be: Tf = f w = 0.5 f
(d l p) d ---------- (2) Equating eon. (1) and (2) and simplifying,
Where r = 0.5 d and u in Pa. This is known as Petroff’s equation for bearing friction. It gives reasonable estimate of co-efficient of friction of lightly loaded bearings. The first quantity in the
bracket stands for bearing modulus and second one stands for clearance ratio. Reynolds’s Equation: Assumption made by Reynolds to derive differential equation for hydrodynamic lubrication. 1. The
lubricant obeys Newton’s law of viscosity. 2. The lubricant is incompressible. 3. The inertia forces in the oil film are negligible. 4. The viscosity of the lubricant is constant. 5. The effect of
curvature of the film with respect to film thickness is neglected. It is assumed that the film is so thin that the pressure is constant across the film thickness. 6. The shaft and the bearing are
rigid. 7. There is a continuous flow of lubricant.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
The Reynolds’s equation is
There is no exact solution to Reynolds’s equation for a journal bearing having a finite length. However Raimondi and John Boyd solved this equation using the iteration technique. The results of this
equations are available in the form of charts and tables. In the Raimondi and John Boyd method, the performance of the bearing is expressed in terms of dimensionless parameters.
The distance between the centres of the bearing and the journal in operating condition is called as eccentricity. The radial clearance c is given by, c = R-r The eccentricity ratio ( is the ratio of
R-r eccentricity to radial clearance. From figure, R = e + r + ho Where, ho is minimum film thickness in mm. Radial clearance c = R-r = e + ho = Hence, Where,
Bearing design parameter: 1. Lengh to diameter ratio (l/d) : The shaft diameter is determined by strength or rigidity consideration. As l/d ratio increases oil film pressure also increases. Hence
long bearing has more load carrying capacity. But it is difficult to get sufficient oil flow through the passage between the journal and the bearing therefore while designing l/d is taken as 1 or
less than 1. 115
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Short bearing has more side flow which improves heat dissipation. When shaft and the bearing are precisely aligned, the shaft deflection does not present a serious problem, then l/d can be taken as
more than 1. In practice, l/d ratio varies from 0.5 to 2, but for most of the applications, l/d is taken as 1. When l/d ratio is more than 1, the bearing is known as long bearing, when l/d ratio is
less than 1, the bearing is known as short bearing and when l/d ratio is equal to 1, the bearing is known as square bearing.
Fig. Effect of l/d ratio on average bearing pressure
2. Unit bearing pressure: The unit bearing pressure is the load per unit of projected area of the bearing in running condition. It depends on many factor such as bearing material, operating
temperature, the nature and frequency of load and service conditions. The permissible bearing pressure can be selected from PSG 7.31 corresponding to application. 3. Start-up load: The unit bearing
pressure for starting condition should not exceed 2MPa. Start-up load is load on shaft when shaft is stationary. It consist of dead weight of shaft and attachments. Start-up load used to determine
minimum length of the bearing on the basis of starting condition. 4. Radial Clearance (c) : It should be small to provide necessary velocity gradient. It required costly finishing operations, rigid
mounting and clean lubricating oil. It increases initial cost and maintenance cost. C =(0.001)r 5. Minimum oil film thickness (ho): Surface finish of the journal and bearing is controlled by minimum
oil film thickness. Below the lower value of ho, there is metal to metal contact and hydrodynamic film breaks. 116
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
6. Maximum oil film temperature: Lubricating film oxidized when operating temperature exceeds 120 degree. Babbitt bearing tend sto soften at 125 degree celcius for bearing presure of 7MPa and at 190
degree celcius for bearing pressure of 1.4 MPa. There fore bearing temerature should be kept minimum. The Limiting temperature for Babbit bearing is 90 degree celcius.
Types and selection of mechanical seals. Seals are used to prevent leakage of fluids through gaps between moving and stationary parts of various machines. The design of a sealing device will depend
upon the type and properties of the sealed medium, i.e. pressure and temperature, as well as by the speed and direction of movement of the seals part. Seals are classified into; (i) Static seals (ii)
Dynamic seals
Static seals:
Static seats exist where then is no relative motion between the two surfaces being sealed. Application requirements involve keeping liquid, gas or dust out. The 0-ring is usually the first line of
defence in preventing leakage of gas, fluids or preventing environmental contamination. The O-ring is considered generally as a static seal however in a broad range of applications, environments and
limited dynamic applications. Static Seals include: Rubber O-Rings, Military O-Rings and Metal O-Rings Static Radial Seals: Static radial seals are forms when squeeze is applied to the inside
diameter and outside diameter of the O-ring. Cap and plug type configurations commonly utilise radial seals. Static Axial Seals: They are formed when squeeze is applied to the top and bottom
surfaces of the O-ring. Axial seals are most often used in flange type designs where O-ring seats against the groove’s lowpressure side. Dynamic Seals: Dynamic seals create a barrier between moving
and stationary surfaces in applications such as rotating shafts and piston rings. Dynamic seals include: V Cup-Packing, Pump Seal , Labyrinth Seal ,Oil Seal, Radial Lip Seal , Gasko-Seal , Hydraulic
& Pneumatic Seal , Bearing Isolators Exclusion Seal, Dynamic Shaft Seals. 117
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
These seals are used in more dynamic applications than all other types and these can accommodate rather high speed and continuous shaft rotation. The most common type of dynamic shaft seal is the
"Radial lip seal". In this type, a flange or lip attached to the housing is held against the shaft. Sealing is due to the interference fit between the flexible sealing element and the shaft. These
seals are primarily used for retaining lubricants in equipment having rotating, reciprocating, or oscillating shafts. These seals are also commonly known as "oil seals" or 'shaft seals'. The rotating
shaft application is most common. The lip seals have the following advantages: Fig: Construction of oil seal
(1) Small space necessary. (2) Relative low cost for high effectiveness. (3) Ease of Installation. (4) Ability to handle many variables while seating.
NUMERICALS: 1) For a full journal bearing having the following data i) Radial load: 2800N, ii) Journal speed: 1440rpm, iii) Journal diameter: 50mm iv) L/D=1, v) Radial clearance: 0.05mm, vi)
Viscosity: 25cp Find, Sommerfeld number, Co-efficient of friction, Minimum film thickness, Temperature rise of oil, Oil flow rate, Heat generated and heat dissipated, Mass of lubricated oil for
cooling Solution: Given data: Load, W = 2800N, Speed N = 1440RPM, Journal diameter D=50mm, L/D=1, Radial clearance, C=0.05mm, Viscosity, Z=25cp , Bearing pressure (p) = 7 – 14kgf /
, assuming p =12 kgf /
... (PSG 7.31)
1. Sommerfeld number ………………………………………………….... (PSG 7.31) , 118
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
= 0.5096
From PSG (7.36), taking the non-dimensional values corresponding to S = 0.5096 𝛜
3. Co-efficient of friction Using linear interpolation, = 5.70
4. Minimum film thickness Using linear interpolation, = 0.718 = 0.0179mm.
5. Temperature rise of oil Using linear interpolation, = 24.29 Here
kgf/cm2 …………………………………………PSG 7.36) ,
= 20.53 °C
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
6. Oil flow rate Using linear interpolation, = 3.472 = 5.20 cm3/s
7. Heat generated and heat dissipated. Heat generated, Hg = µW.V Hg = µW
…………………..... (PSG 7.34)
Hg = 361
= 361
Watt = 60.16 Watt
Heat dissipated, Hd = Tavg = 58° C Th = Tavg -
...................................................... (PSG 7.34) ............................................................................................ (PSG 7.41)
= 58 -
= 47.73° C
Th – Ta = 47.73 – 30 = 17.73° C Hd =
watt = 12.17 Watt
Hg > Hd Hence, artificial cooling can be provided
8. Mass of lubricated oil for cooling Hg = ms
Where, m= Mass of lubricated oil for cooling, s = specific heat of lubricated oil,
change in temperature (Assume 20°C) s = (1840 – 2100) JKg/°C, et s = 1900 JKg/°C Hg = ms
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Summary: Sommerfeld number =0.5096 Co-efficient of friction = Minimum film thickness = 0.0179mm Temperature rise of oil = 24.29°C Oil flow rate = 5.20 cm3/s Heat generated = 60.16 Watt Heat
dissipated = 12.17 Watt Mass of lubricated oil for cooling ……………………………………………………………………………………………
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
2) For a given data for a 360 degree hydrodynamic bearing, find i.) Dimensions ii. ) Co-efficient of friction iii.) Minimum film thickness, iv.) Viscosity of oil v.) Temperature rise in oil vi.)
Operating temperature of oil vii.) Heat generated and heat dissipated Given Data: W=20kN, N=1000rpm, p=1.3 N/mm2, l/d=1, fit = H8e8
Solution: 1. Dimensions W = 20kN, p = W/LD 1.3 = 20/L2 ,
∴L = D = 125 mm
From PSG 3.7 and 3.9 For shaft diameter: -
, for hole diameter:-
Max. Clearance = 0.211 mm Min. clearance
= 0.085 mm
∴ Avg. clearance = 0.148 mm Now, clearance ratio (D/C) = 125/0.148 = 845, ∴ D/C = 845 For L/D = 1, ϴ = 360 2. Minimum film thickness From PSG 7.36, ϵ=1-
= 0.415, ∴ ho = 0.030695 mm
∴ ϵ = 0.585
From PSG 7.36 for full bearing ϵ
14.2 122
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
S = 0.137 (using interpolation) 3. Co-efficient of friction = 5.79 + (3.22-5.79) (
= 3.41, ∴
= 0.004
4. Viscosity of oil S=
8452, ∴ Z = 14.36 cp
, 0.137 = 5. Temperature rise of oil
= 0.585,
= 24.3 + (14.2-24.3) (
= 14.96,
=14.2 kgf/cm2 ………………………..…………… (PSG 7.36)
= 13.69
6. Operating temperature of oil Ta = To +
, Ta = 30 + 13.69, Ta = 43.69
7. Heat generated Hg = Wv, Hg = 0.004 x 20000 x
, Hg = 523.598 Watt
8. Heat dissipated Hd =
...... (PSG 7.34)
Tavg = 58° C Th = Tavg -
.....(PSG 7.41) = 58 -
= 51.155° C
Th – Ta = 51.155 – 30 = 21.155° C Hd =
Watt = 91.36 Watt
Hg > Hd Hence, artificial cooling is provided.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
9. Mass of lubricated oil for cooling Hg = m.s Where, m= Mass of lubricated oil for cooling, s = specific heat of lubricated oil, change in temperature (Assume 20°C) s = (1840 – 2100) JKg/°C, Let s =
1900 JKg/°C ,
Summary: Dimensions: L= D = 125mm Sommerfeld number =0.137 Co-efficient of friction = Minimum film thickness = 0.0306 mm Temperature rise of oil = 13.69°C Heat generated = 523.598 Watt Heat
dissipated = 91.36 Watt Mass of lubricated oil for cooling
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
3) Design a journal bearing for the following specifications: Load= 12 kN,
Speed = 1440 RPM
Diameter of journal = 75 mm Ambient temperature = 16 Operating oil temperature = 60 Absolute viscosity of oil at 60oc = 0.023 kg/ms
Data: W = 12x103 kN, N = 1440 RPM, D = 75 mm, To = 16
, T = 60 , Z = 0.023 kg/ms
Solution: i) Bearing pressure P=
(Assume L/D =1 ) = 2.133 N/mm2
ii) Sommerfeld number S =
(Assume D/C =1000 )
S = 0.258 From PSG 7.36 for full bearing S 0.121
iii) Minimum oil film thickness By interpolation, =
= 0.591 125
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
, ∴
iv) Co-efficient of friction By interpolation, =
= 5.682
, ∴
v) Oil quantity in circulation By interpolation, =
,∴ ,
= 4.004 m3/s
∴ m3/s = 98864.197 X
q = 98864.197 X
X 60 lit /min
q = 5.931 LPM
vi) Side Leakage By interpolation, 23.833 = = 0.564,
= 0.504, qs = 2.992 litre/min
vii) Power loss in friction Pf
, Pf = 0.385 kW
viii) Temperature rise of oil ϵ=1-
= 1 – 0.591 = 0.409
By interpolation, = ∴
, ∴
= 23.87
, ∴ 126
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
ix) Heat generated Hg = Pf, ∴
= 0.385 kW
x) Heat dissipated Hd =
) =
(Here Hd = 0.407 kW ,
….. (PSG 7.36)) Hg < Hd Hence, artificial cooling is not needed.
Summary: Sommerfeld number = 0.258 Co-efficient of friction = Minimum film thickness = 0.0221mm Temperature rise of oil = 3.58 °C Oil flow rate = 5.931 lpm Heat generated = 385 Watt Heat dissipated =
407 Watt
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
MODULE 4 CAM AND ROLLER FOLLOWER MECHANISMS Introduction: A cam is a rotating machine element which gives reciprocating or oscillating motion to another element known as follower. The cam and the
follower have a line contact and constitute a higher pair. The cams are usually rotated at uniform speed by shaft, but the follower motion is pre-determined and will be according to the shape of the
cam. The cams are widely used for operating the inlet and exhaust valves of internal combustion engines, automatic attachment of machineries, paper cutting machines, spinning and weaving textile
machineries, feed mechanism of automatic lathes etc. In design of cam, it is necessary: 1. To generate the profile of cam for the given type of follower motion; and 2. To calculate the contact stress
between cam and follower so as to ensure the safety of cam and follower against pitting failure
Terminology Used in Cam Follower 1. Base circle: It is the smallest circle that can be drawn to the cam profile. 2. Pressure angle: It is the angle between the direction of the follower motion and
a normal to the pitch curve. This angle is very important in designing a cam profile. If the pressure angle is too large, a reciprocating follower will jam in its bearings. 3. Trace point: It is a
reference point on the follower and is used to generate the pitch curve. In case of knife edge follower, the knife edge represents the trace point and the pitch curve corresponds to the cam profile.
In a roller follower, the centre of the roller represents the trace point. 4. Pitch point: It is a point on the pitch curve having the maximum pressure angle. 5. Pitch circle: It is a circle drawn
from the centre of the cam through the pitch points. 6. Pitch curve: It is the curve generated by the trace point as the follower moves relative to the cam. For a knife edge follower, the pitch curve
and the cam profile are same whereas for a roller follower, they are separated by the radius of the roller. 128
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
7. Prime circle: It is the smallest circle that can be drawn from the centre of the cam and tangent to the pitch curve. For a knife edge and a flat face follower, the prime circle and the base circle
are identical. 8. Lift or stroke: It is the maximum travel of the follower from its lowest position to the topmost position.
Figure: Terminology in cam and follower
Pressure Angle and Its Significance Pressure angle (ø): It is the angle between the axis of the follower stem and common normal (axis of transmission) or the line of force exerted by the cam on the
follower. Pressure angle varies in magnitude during the rotation of the cam. The components of the force along the line of motion that is Ncosø is useful component in overcoming the output load. The
perpendicular component Nsinø should be kept as small as possible to reduce friction between the follower and its guide way. 1. When pressure angle is zero, complete transmitted force goes into the
motion of the follower and there is no side thrust on the guide of the follower. 2. When pressure angle is 90 degree, there is no force acting on the follower and hence no motion of the follower. 3.
The pressure angle should be kept up to 30 degree for translating follower and upto 35 degree if the follower is oscillating on the pivoted arm. 129
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Pressure angle equation: ,
The maximum pressure angle: ,
Where, r – Distance between can centre and roller centre, – Prime circle radius,
y- Instantaneous displacement of follower, Base circle radius of cam, – Radius of the roller, - Angular velocity of cam.
TYPES OF CAM: Cams can be classified according to the shape, according to the follower moment and according to the construction of the follower.
According to the shape, it is further divided into six types they are:
1. Wedge or flat cams A wedge cam has a wedge and also has translational motion. The follower can either have translated or oscillation motion. To maintain a contact between the follower and cam, a
spring is used. 130
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
2. Radial or Disc Cams The follower moves radially along with the centre of rotation, is known as radial cam. Due to the simplicity and compactness the radial cams are very popular. The follower
oscillates about an axis parallel to the axis of rotation of the cam. 3. Spiral cams In the spiral cam, groove is cut in the form of a spiral and is also known as face cam. The grove consists of
teeth and they are meshed with a pin gear follower. From the axis of the cam, the follower velocity is proportional to the groove radial distance. Mainly the spiral cams are used in computer. 4.
Cylindrical cams Cylindrical cams are also known as drum cams or barrel cams. In the cam, the cylinder consists of a circumferential contour cut in the surface and the cylinder rotates about its
axis. 5. Spherical cams In the spherical cam, the follower oscillates about the axis perpendicular to the axis surface of rotation of the cam. In the spherical cam consists of a spherical surface
which transmits motion to the follower.
According to the follower moment:
1. Rise – Return – Rise (RRR) In the Rise- return – Rise, there is an alternate rise and return of the follower with no periodic form of dwells. The follower has a linear or an angular displacement.
2. Dwell – Rise – Return _ Dwell (D-R-R-D) In Dwell – Rise – Return – Dwell after a dwell there are rise and return of the follower. This type is used more normally than the Rise – Return – Rise type
of cam. 3. Dwell – Rise – Dwell – Return – Dwell (D-R-D-R-D) 131
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Dwell – Rise – Dwell – Return – Dwell is the most commonly used cam. Cam dwelling is followed by dwell and it rises and returns successively. In the case of return of the supporter is by a fall. The
motion is known as Dwell – Rise – Dwell. According to the construction of the follower: 1. Preloaded spring cam The preloaded spring is used to make a contact between the follower and the cam 2.
Positive drive cam In the positive drive cam, the contact between the follower and the cam is maintained by a roller follower working in the groove of a cam. During normal working conditions, the
follower does not move out of the groove. By using the conjugate cam, positive drive is obtained. 3. Gravity cam Rising surface of the cam is achieved by rise of a cam, and the cam must be returned
by the force of gravity. This can also return due to the weight of the cam. There are certain limitations due to the uncertain behaviour.
TYPES OF FOLLOWER: Classification of followers is done according to the shape, according to the movement and according to the location of the line of movement. According to the shape: 1. Knife
edge Follower Knife edge follower is known as the end of the contacting follower has a “sharp knife edge”. The sliding motion must be taken between the cam surface and knife edge. The small
contacting surface results in excess wear. The side thrust must exist between the guide and the follower. 2. Roller follower If the follower contacting end is a roller then it is known as the roller
follower. The rolling motion must be takes place in between the cam and the roller. The wear must be critically reduced. The side trust exists between the guide and the follower. One can observe the
roller follower mainly when there is availability of more space in oil engines, stationary gas and aircraft engine. 3. Mushroom Follower or Flat Faced follower If the contacting end of the follower
is flat faced, then it is known as the flat face or mushroom follower. In this condition the side thrust between the guide and follower is reduced very much. Due to limited space in cam the flat
faced followers are used to operate the automobile engine valves are used. 132
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
4. Spherical faced follower The follower connecting end is spherical, and then it is known as spherical faced follower. In order to minimize the stress that is developed by the flat faced follower,
the spherical faced followers are used.
According To The Motion Of The Follower:
1. Translating follower or reciprocating follower The cam rotates uniformly when the follower reciprocates within the guides, it is known as translating follower or reciprocating follower. 2.
Rotating follower or oscillating follower In the cam when the uniform rotary motion is converted into the determined oscillatory motion of the follower, it is known as the rotating follower or
oscillating follower. According to the path of motion of the follower: 1. Radial follower If the line of movement of the follower passes through the centre of rotation of the cam, it is known as
radial follower. 2. Off-set follower If the line of movement of the roller follower is offset from the centre of rotation of the axis then it is known as the off – set follower.
TYPES OF MOTION OF FOLLOWER: 133
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Cam follower systems are designed to achieve a desired oscillatory motion. Appropriate displacement patterns are to be selected for this purpose, before designing the cam surface. The cam is assumed
to rotate at a constant speed and the follower raises, dwells, returns to its original position and dwells again through specified angles of rotation of the cam, during each revolution of the cam.
Some of the standard follower motions are as follows: 1. 2. 3. 4.
Simple harmonic motion Uniform velocity Uniform acceleration and deceleration Cycloidal motion
Simple harmonic motion: The motion executed by point Pl, which is the projection of point P on the vertical diameter is
called simple harmonic motion. Here, P moves with uniform angular velocity ωp, along a circle of radius r (r = s/2). From figure it is seen that velocity of the follower is zero at the beginning and
at the end of each stroke and it is maximum at the mid of each stroke. The acceleration is maximum at the beginning and at the end of each stroke and is zero at the mid of each stroke. The jerk is
zero at beginning and at the end of each stroke and is maximum at mid of each stroke. However there are two infinite jerks at the beginning and end of each stroke because the acceleration is suddenly
brought to zero from finite value. Therefore, this motion is used only for moderate speed. Displacement,
Where, s = Stroke or displacement of the follower. θo = Angular displacement during outstroke. θr = Angular displacement during return stroke 134
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
ω = Angular velocity of cam. Time taken for outstroke
Time taken for return stroke,
Max. Velocity of follower during outstroke = Vomax = r ωp,
Similarly Max. Velocity of follower during return stroke, Max. Acceleration during outstroke = aomax = rω2p = Similarly, Max. Acceleration during return stroke,
Uniform Velocity:
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Figure shows the displacement, velocity and acceleration patterns of a follower having uniform velocity type of motion. Since the follower moves with constant velocity, during rise and fall, the
displacement varies linearly with θ. Also, since the velocity changes from zero to a finite value, within no time, theoretically, the acceleration becomes infinite at the beginning and end of rise
and fall. The line BC and DE are the straight lines representing the rise and return of the stroke. AB, CD and EF represent dwell period.
Motion of follower with uniform acceleration and retardation or parabolic motion: Fig shows the displacement diagram of follower when it moves with uniform acceleration and retardation. During the
first half of the outward motion, the follower moves with uniform acceleration and during the remaining half of the stroke it moves with the uniform retardation. During the first half of the return
stroke, the follower moves with uniform retardation and during the remaining half of the stroke it moves with uniform acceleration. The procedure for drawing the uniform displacement diagram of
follower with uniform acceleration and deceleration is as follows: a. Divide the angular displacement of cam during out stroke and return stroke into equal number of parts (say 6) and draw vertical
lines through these points. b. Draw a vertical line through point 3 and divide it into same equal number of parts say 6. Named A, B, C, D, E and F. c. Join OA, oB, oC, fD, fE and fF etc.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
d. The intersection of these lines with the vertical lines drawn from points 1,2,3,4,5,and 6 give the points o,a,b,c,d,e and f. e. The curve joining these points represent the displacement curve for
outstroke. f. Follow the same procedure for return stroke also. Figure shows the displacement, velocity acceleration and jerk of follower with uniform acceleration and retardation (parabolic motion).
From figure it is seen that velocity increases from 0 to maximum during uniform acceleration and it again reduces to 0 during retardation. The magnitude of acceleration remains throughout the cycle
but in second half the direction is opposite to first half of acceleration. The jerk is infinite throughout the cycle, the infinite jerk will give rise to shock load which causes vibrations and high
stresses .therefore this motion can be used for low and moderate cam motion only. Motion of follower with Cycloidal motion: Cycloid is the path generated by a point on the circumference of a
circle, as the circle rolls without slipping, on a straight/flat surface. The motion executed by the follower here, is similar to that of the projection of a point moving along a cycloidal curve on a
vertical line as shown in figure. The cycloid is the locus of a point on the circle which is rolled on a straight line,
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Figure : displacement diagram of the follower when it moves with cycloidal motion .
The procedure of drawing the displacement diagram with cycloidal motion is as follows: 1. Divide the angular displacement of cam into equal parts (say 6) 2. Draw the diagonal line Of. 3. Draw a
circle at point ‘o’ such the circumference is equal to follower displacement. i.e. 2 r =h .Therefore, r =hm/2 . This circle is also divided into the same no. of parts as cam displacement. 4. Project
the circle point in its vertical diameter and then in a direction parallel to the diagonal ‘of’ to the corresponding vertical line1.2.3.4,……6 etc giving the intersection points a,b,c,…..f etc 5. Draw
the curves through the points a,b,c,d…f etc which gives the cycloidal displacement curves for the follower during outstroke, 6. Project points a,b,c…..d etc corresponding vertical line 7,8,9,10….etc
Which give the intersecting points h,i,j,k….etc which gives the cycloidal displacement curve for the follower during return stroke. From figure it is seen that the velocity of the follower is zero at
the beginning and at the end of stroke and it is maximum at the mid of each stoke. The acceleration is zero at the beginning, mid and at the end of stroke and is maximum at one fourth and three
fourth of each stroke. The jerk is maximum at the beginning, end and mid of each stroke and is zero at one fourth and three fourth of each stroke. It may be noted that jerk is having finite value
throughout the cam angle therefore cycloidal curve is best suited for high speed cams as compared to other motions discussed so far.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
DESIGN OF VALVE SPRING The design of spring involves the determination of following parameters Wire diameter (d), Mean coil diameter (D), Number of active coils (n), Total number of coils (n')
Maximum deflection of spring ( ), Solid length ( ), Free length ( ) ,Pitch of coil (p), Natural frequency of spring ( ) =maximum spring force, N ,
=minimum spring force, N
=maximum compression of spring, mm , d = wire diameter, mm D= mean coil diameter, mm C= spring index . It is taken as 6 to 10 for valve spring. =shear stress correction factor = =Wahl shear stress
factor= =number of active coils =total number of coils G=modulus of capacity for spring material, =mass of spring, kg =solid length of a spring, mm =free length of spring, mm p=pitch of coil, mm K=
stiffness of spring , N/mm. It is generally taken as 10 to 12 N/mm for valve spring. =mean shear stress induced in spring, =shear stress amplitude in spring , =ultimate tensile strength for spring
wire, 139
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Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
=yield strength in shear for spring wire,
=endurance limit in shear for spring wire,
G=modulus of rigidity for spring material, =factor of safety. It is 1.5 to 2.0 for valve spring (i)Wire diameter (d): The spring is subjected to fluctuating forces. It is subjected to hundreds of
millions of stress cycles during its life time. Therefore, it is to be designed for fatigue loading using the modified Soderberg diagram. The mean force ' =
' and force amplitude ' ' on spring are, .... (a)
The mean shear stress ' = Where,
' and shear stress amplitude ‘ ' are given by:
...(c) =
... (d)
The wire diameter can be determined by using relation: ...(1)
(ii) Mean diameter of coil (D): (iii) Number of active coils (
D = Cd ):
The number of active coils is given by, (iv) Total number of coils (
... (2)
... (3)
The square and ground ends are used for valve springs. Therefore, ... (4) (v) Maximum compression of spring (
The maximum compression of the spring is given by,
... (5) 140
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Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
(vi) Solid length ( ): (vii)Free length ( );
... (6) = Solid length +Maximum compression + Total clearance between coils
in compressed condition. The total clearance between the coils in compressed condition is taken as 15% of Maximum compression. Therefore, Or
... (7)
(viii) Pitch of coil (p): The pitch of coil is determined by using relation, ,
... (8)
(ix) Natural frequency of spring ( ): If the natural frequency of the spring coincides with the frequency of operation of cam, a resonance occurs and very large deflections of coil will be produced
with correspondingly high stresses. This phenomenon is as surge in spring. Under this condition, the failure of spring may take place. To avoid the possibility of surge in springs, it is necessary to
ensure that natural frequency of spring be considerably remote from frequency of application of the load. The natural frequency of spring is given by
Design Procedure for Cam and Follower Step 1: Motion Analysis of follower 1. Displacement Analysis Displacement equations for different notions are given on PSG 7.110. Put different values of θ and
plot the graph between h vs θ. 2. Velocity Analysis Similarly get the velocity equation from PSG 7.110 and plot the respective graph. 3. Acceleration Analysis Similarly get the acceleration equation
from PSG 7.110 and plot the respective graph. 141
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Step 2: Calculation of prime circle radius Rp and base circle radius Rb. Also calculate ρkmin and ρkmax. Where, ρkmin – radius of curvature of pitch curve, ρkmax – radius of curvature of cam profile
(Refer PSG 7.113 and PSG 7.114 for above calculations) Step 3: Force Analysis Analyze the forces acting on the cam and follower mechanism by initially excluding the spring force and then including
the spring force. Step 4: Determination of cam width and pin diameter 1. From PSG 7.115 get the formula for contact stress. Select suitable cam and follower material from PSG 7.115 and calculate the
cam width. Select suitable value of cam width for safe design. The safety condition is
2. Select suitable pin material from PSG 1.9and calculate the pin diameter considering double shear is acting on the pin. Select suitable value for safe design.
Roller Pin Design Step 5: Spring and Cam Shaft Design 1. Select required parameters from PSG 7.105 and PSG 7.100. Select spring material from PSG 7.102 and PSG 7.105 and design spring accordingly. 2.
Select suitable material for Cam Shaft from PSG 1.9. Calculate the shaft diameter considering shear stress is acting on the shaft.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Numerical: 1. A rotary disc cam with central translatery roller follower has following motion. Forward stroke of 30mm in 900 of cam rotation with SHM motion, dwell of 500 of cam rotation and return
stroke of 30mm in 1000 of cam rotation with SHM. Remaining dwell to complete the cycle. Mass of the follower is 1kg and cam shaft speed is 500rpm. The maximum pressure angle during forward stroke and
return stroke is limited to 220. The external force during forward stroke is 400N and that of return stroke is 50N. Find cam dimension, roller follower along with pin and spring. Solution: Given
that, h = maximum follower displacement/stroke of follower = 30mm θO = angular displacement of cam in out stroke = 900 θR = angular displacement of cam in return stroke = 1000 Maximum pressure angle,
αmax = 220 during forward stroke External force Fext = 400N (Forward Stroke), Fext = 50N (Return Stroke) Follower motion = SHM (both) ω=
= 52.359 rad/sec
Step 1: Motion analysis of follower A. Displacement analysis For SHM, displacement is given by the equation from PSG 7.110
Where, y – follower displacement instantaneous h – Maximum follower displacement Displacement diagram:
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
B. Velocity analysis From PSG 7.110, Where, θ – Cam angle for displacement y β – Cam angle for rise h h – Maximum rise of follower At θ = 00 and θ = β, v=0 At θ = β/2 for forward stroke Therefore, V
= Vmax =
= 1.57 m/sec
For return stroke, At θ = 00 and θ = β2, V=0 At θ = β2/2, V = Vmax for return stroke Therefore, V = Vmax =
= 1.178 m/sec
Fig: Velocity Diagram
C. Acceleration analysis
From PSG 7.110, At point 2 and 5; a = 0 for forward stroke At θ = 0; a = amax Therefore, amax =
= 164.49 m/sec2
At point 3; θ = β1 amax =
= -164.49 m/sec2
During return stroke At point 4; θ = 00 145
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Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
= 133.24 m/sec2
Therefore, amax = Similarly at point 6; θ = β2
= -133.24 m/sec2
Therefore, amax = Acceleration Diagram:
Step 2: Calculation of prime circle radius Rp and base circle radius Rb. Also calculate ρkmin and ρcmin Based on forward stroke , From PSG 7.113,
Therefore, dy/dθ = (dy/dt)(dt/dθ) = 1/ω (dy/dt)max Therefore, dt/dθ = ω and dy/dt = vmax yθ = displacement of follower at this angle y(θ=β/2) at vmax = 0.015m
Therefore, Rp = 0.0592 m
60 mm
Now based on return stroke Vmax = 1.178 m/sec, αmax = 220 , Therefore, Rp = 0.0406 m
40.6 mm 146
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Selecting higher value of Rp such that the pressure angle remains within permissible limit and also to avoid interference undercutting of cam. Rp = 60 mm Now assuming roller radius Rr = 15 mm Base
circle radius, Rb = Rp –Rr = 60 -15 = 45 mm Pitch circle radius, Ra =Rp + yθp Now for θp i.e. angle corresponding to pitch point is given by. For SHM, , On solving, we get θp = 39.290 i.e. cam angle
at pitch point Therefore, yθp =
= 12.03 mm
Therefore, Pitch circle radius, Ra = Rp + yθp = 72.03 mm For roller follower (convex profile)
From PSG 7.114 ,
For y at point 3, y = 30 mm, Rp + y = 90 mm dy/dt = 0 at point 3;
= 0 and
= -133.24/(52.359)2 = -0.0486
Therefore, ρkmin = 0.0584m = 58.4 mm > Rr (radius of roller) … to avoid undercutting From PSG 7.114 , Radius of curvature of cam profile, ρcmin = ρkmin – Rr = 58.4 – 15 = 43.4 mm > 0
… to avoid undercutting 147
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Step 3: Force Analysis on cam and follower mechanism Initially excluding spring force Fresultant = mg + mẍ + Fext Points
Force in N
1x9.81 + 1x164.49 + 400 = 574.3
1x9.81 + 0 + 400 = 409.81
1x9.81 + 1x(-164.49) + 400 = 245.32
1x9.81 + 1x(-133.24) + 50 = -73.43
1x9.81 + 0 + 50 = 59.81
1x9.81 + 1x133.24 + 50 = 193.05
To maintain a continuous contact between roller follower and cam, let us introduce a spring of stiffness k. Generally the force between cam and follower at any instant should not be less than (20-30
N). Therefore, Fresultant
20 N, F4 + Fspring = 20, Therefore, Fspring = 97.43 N
Stiffness, k = F/d, Therefore, k = 97.43/30 = 3.248 N/mm Initial compression of spring. Fs = k x ϰ Therefore, ϰ = 6.158 mm (When spring is compressed at that time force on spring is Fresultant)
Actual Force Analysis including spring force Point
Deflection in spring
F = mg + mẍ + Fext + k(ϰ + d) [Total force in N]
574.3 + 20 = 594.3
409.81 + 20 + 3.348x15 = 480.03
245.32 + 20 + 3.348x30 = 365.76
-73.43 + 20 + 3.348x30 = 47.01 > 20
59.81 + 20 + 3.348x15 = 130.03
193.05 + 20 = 213.05 148
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Now, Fmax = 594.3 N at point 1, Therefore, Normal force at the point of contact, Pn =
= 594.3/cos22 = 640.97 N
Step 4: Determination of cam width and pin diameter From PSG 7.115, Contact Stress is given by,
(Note: +ve sign for convex profile and –ve sign for concave profile) Considering roller material as hardened steel and cam material as C-20 then for steel E1 = E2 = 2 x 105 N/mm2 From PSG 7.115, [σc]
= 5000 kgf/cm2 = 500 N/mm2 (Assuming 8.3% sliding) Therefore, On solving b = 8.033 mm [Let b = 10 mm] Checking for condition,
Therefore, 10/90 = 0.111
0.3, Hence, safe
Assuming pin material as C-20, From PSG 1.9, σy = 260 N/mm2 and [τ] = 45 N/mm2 Considering double shear of pin, Pn = 2 x Therefore, 640.97 = 2 x
x dp2 x [τ]
x dp2 x 45, Therefore, dp = 3.011 mm
5 mm
Step 5: Spring and Cam Shaft Design Fs max = (h + ϰ) x k = (30 + 6.158) x 3.348 = 121.06 N Assuming design force, [Fs max] = 121.06 x 1.3 = 157.37 N For spring, steel material PSG 7.102 and 7.105 149
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Therefore, design stress, [τ] = 0.263 σu (for severe service) Assume dwire = 5 mm and Grade Ⅰ (PSG 7.105) [τ] = 0.263 x 1210 = 318.23 N/mm2, τinduce =
…….[PSG 7.100]
where, ks – stress factor, for spring index C=6, ks = 1.2525 Therefore, τinduce = 1.2525 x
= 120.46 N/mm2 < [τ] …………. Safe
No of turns,
= 85.34
86 turns
… [PSG 7.100]
D = C x d = 6 x 5 = 30 mm Design of Cam Shaft: Fnet max = 594.3 N [from force table] , Vmax = 1.57 m/s [Comparing points 2 and 5] Therefore, Torque =
= 17.82 N-m Fmax = 594.3 N
L = 3b = 30 mm B.M. max = (Fmax x L)/4 = 4.457 N-m Equivalent Torque, Te =
= 35.92 N-m
Assuming C-30 material for shaft, σy = 300 N/mm2, [τ] = 0.5 σu/4, Let [τ] = 40 N/mm2 150
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Te = Π/16 [τ].ds3, 35920 = Π/16 x 40 x ds3 ds = 16.599mm Let ds =18mm
SUMMARY Parameters
Maximum Velocity in Forward Stroke
1.57 m/sec
Maximum Velocity in Return Stroke
1.178 m/sec
Maximum Acceleration in Forward Stroke
164.49 m/sec2
Maximum Acceleration in Return Stroke
133.24 m/sec2
Prime Circle Radius (Rp)
60 mm
Base Circle Radius (Rb)
45 mm
Pitch Circle Radius (Ra)
72.03 mm
Radius of curvature of pitch curve
58.4 mm
Radius of curvature of cam profile
43.4 mm
Spring Force
97.43 N
Spring Stiffness
3.248 N/mm
Initial Compression of Spring
6.158 mm
Normal force at point of contact
640.97 N
Cam width
10 mm
Pin Diameter
5 mm
No. of turns
86 turns
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Diameter of Spring
30 mm
Diameter of Cam Shaft
18 mm
NUMERICAL 2: A rotary disc cam with central translatery roller follower has following motion. Forward stroke of 30mm in 1000 of cam rotation with parabolic motion, dwell of 400 of cam rotation and
return stroke of 30mm in 800 of cam rotation with cycloidal. Remaining dwell to complete the cycle. Mass of the follower is 1.5 kg and cam shaft speed is 600 RPM. The maximum pressure angle during
forward stroke and return stroke is limited to 250. The external force during forward stroke is 500N and that of return stroke is 100N. Design cam, roller follower, spring, cam shaft. Solution: Given
that, h = maximum follower displacement/stroke of follower = 30mm θO = angular displacement of cam in out stroke = 1000 θR = angular displacement of cam in return stroke = 800 Maximum pressure angle,
αmax = 250 during forward stroke External force Fext = 500N (Forward Stroke),
Fext = 100N (Return Stroke)
Follower motion = Parabolic motion (forward stroke) Follower motion = Cycloidal motion (return stroke) ω=
= 62.359 rad/sec
Forward Stroke – Parabolic motion , Return Stroke – Cycloidal motion
STEPS 1 :- MOTION ANALYSIS OF FOLLOWER A) Displacement analysis :for for For return stroke 152
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Plot h- for different values of
B) VELOCITY ANALYSIS :For forward stroke from PSG 7.110 for parabolic motion for
, For return stroke , at for cycloidal from PSG 7.110 = 2.639 m/s
c) ACCLERATION ANALYSIS:-
For forward stroke from PSG 7.110 for parabolic motion AND
, =155.51 m/
For return-stroke – for cycloidal motion at
= 381.56 m/
at =-381.68 m/
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Step 2: Calculation of prime circle radius Rp and base circle radius Rb. Also calculate ρkmin and ρcmin. Rp - Prime circle radius, Rb - Base circle radius, ρkmin - Radius of curvature of pitch curve,
ρcmin - Radius of curvature of cam profile
Based on outward stroke From PSG 7.113,
, Therefore, dy/dθ = (dy/dt)(dt/dθ) = 1/ω (dy/dt)max
Therefore, dt/dθ = ω and dy/dt = vmax yθ = displacement of follower at this angle y(θ=β/2) at vmax = 0.015m , Therefore, Rp = 0.05872 m
59 mm
Now based on return stroke From PSG 7.113, Therefore, dy/dθ = (dy/dt)(dt/dθ) = 1/ω (dy/dt)max Therefore, dt/dθ = ω and dy/dt = vmax = 2.7/62.83=0.0429731 154
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Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
yθ = displacement of follower at this angle y(θ=β/2) at vmax = 0.015m
Therefore, Rp = 0.077 m
77 mm
Here maximum negative acceleration is -155.51m/s2 At maximum negative acceleration, y=h/2=30/2=15mm=0.015m (Rp +y)=59+15=74mm=0.074m dy/dt=2.16 at point of maximum negative acceleration. =
=0.0343, &
To avoid undercutting (see PSG 7.114) , and radius of curvature of cam profileFrom PSG 7.114 ,
ρcmin = ρkmin – Rr = 50.50 – 15 = 35.5 mm > 0 … to avoid undercutting
Step 3: Force Analysis Initially excluding spring force. The force acting on cam , Fresultant = mg + mẍ + Fext POINTS
FORCE IN NEWTON (N) 1
1.5*9.81 + 1.5*155.51+500 = 747.48
1.5*9.81 + 1.5*(155.51) +500 = 747.48
1.5*9.81 + 1.5*(-155.51) +500 = 281.45
1.5*9.81 + 1.5*(-155.51) +500 = 281.45
1.5*9.81 + 1.5*0 + 100 = 114.715
1.5*9.81 + (-1.5)*381.68 + 100 = -457.8
1.5*9.81 + 1.5*0 + 100 = 114.715
1.5*9.81 + (-1.5)*(-381.68) + 100 = 687.235
1.5*9.81 + (-1.5)(0) + 100 = 114.715 155
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Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
From above table at point 7 there is a chance of separation of cam and follower. Therefore let us introduce spring of Stiffness k. Generally the force between cam and follower at any instant should
not be less than 20-30N. Therefore, we can write, FResultant ≥ 30N, F7 + FSpring = 30 FSpring = 30-(-457.8) = 487.8N, ,
, k = 21.68 N/mm
Initial Compression of the spring Fs = k x, 30 = 21.68 * x , x = 1.3837 mm Force analysis including the spring force
F = mg + mẍ + Fext + k(ϰ + d) [Total force in N]
747.98 + 30 = 777.98 N
747.98 + 30 + (21.68*15) = 1103.18 N
281.45 + 30 + (21.68*15) = 636.65 N
281.45 + 30 +(21.68*30) = 961.85 N
114.715 + 30 + (21.68*30) = 795.115 N
-457.8 + 30 + (21.67*22.5) = 607.30 N
114.715 + 30 + (21.68*15) = 469.915
687.235 +30 + (21.68*7.5) = 879.835
114.715 + 30 + (21.68*0) = 144.715 N
Now, Fmax = 1103.18 N Normal force acting on cam (PSG 7.115) Pn =
= 1217.22 N 156
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Step 4: Determination of cam width and pin diameter From PSG 7.115, Contact Stress is given by,
Considering roller material as hardened steel and cam material as C-20, then [c] = 500 kgf/cm2 = 500 N/mm2 Assuming 8.3% sliding, From PSG 7.115 E1 = E2 =
b = 16.12 mm, Taking b = 18 mm Check for condition b/dbase ≤ 0.3 , Hence safe. Assuming pin material ZSC – 20, y = 260 N/mm2……..PSG 1.9 = 0.5 * y/f.o.s = 0.5 * 250/3 , = 43.033 N/mm2., Take =
45 N/mm2 Consider double shear of pin Pn =
dp = 4.15mm , Take dp = 6 mm
STEP V: - Spring and Cam Shaft Design Fs max = (h + x)k = (30=1.3837) 21.68 Design load , Fs max 1.3 * 680.39 = 884.5N For spring steel material from PSG 7.102 and 7.105 Zper = 0.263 * u (For severe
service) Assume d = 5 33 and Gr.1 157
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Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
u = 1210 N/mm2
Zper 0.263 * 1210 , Zper =318.23n/mm2 Zind =
Ks=Wahl’s stress factor , For spring index 6 = 1.2525 Zind
= 677.05N/mm2 > Zper= 468.14N/mm2
Hence failed , Take Gr. 4 Zper=0.263*1780 =468.14 N/mm2 Zind= 677.05 N/mm2 > Zper=468.14N/mm2, hence failed let take d = 6mm and Gr. 4 , Zind =
=470.18 N/mm2
and Zper=0.263*1700 = 447.1 N/mm2 Till Zind>Zper hence failed, Take d= 7 mm and Gr. 4 , τinduce = 345.43N/mm2, and τper=0.263 1640
τinduce =
i.e. τper=431.32N/mm2, Now, τinduce τper, Spring is safe Number of turns n = n =
from PSG 7.19
=15.69 turns, D = c d=6 7= 42
Design of Cam Shaft Fnetmax=1103.18N Vmax=2.7m/s (Comparing both Vmax) Torque=
Taking design Torque as Td=2Torque=2 47.4=94.8Nm B.M = = = 14892.93 N-m Equivalent torque = Te = (Tf2 + BM2)1/2 = (94.82 + 14892.932) = 14893.23 N Assuming C-30 material of shaft, N/mm2 ……………….. PSG
1.9 158
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Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
= 0.5*
= 0.5
40 N/mm2, Te =
14843.23 *103 =
= 37.5 N/mm2
* 40 * ds3, ds = 123.77 mm
Take ds = 125 mm
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Chapter 5 BELT AND CHAIN DRIVE DESIGN
Introduction Belt, chain and rope drives are called ‘flexible’ drive. There are two types of drives- rigid and flexible. Gear drives are called rigid or non-flexible drives. In gear drives, there
is direct contact between the driving and driven shafts through the gears. In flexible drives, there is an intermediate link such as belt, rope or chain between the driving and driven shafts. Since
this link is flexible, the drives are called ‘flexible’ drives. In gear drives, rotary motion of the driving shaft is directly converted into rotary motion of the driven shaft by means of pinions and
Selection of belt Belt drives are built under the following required conditions: speeds of and power transmitted between drive and driven unit; suitable distance between shafts; and appropriate
operating conditions. The equation for power is: Power (kW) = (Torque in newton-meters) × (rpm) × (2π radians)/ (60 sec × 1000 W) Factors of power adjustment include speed ratio; 1. Shaft distance
(long or short); 2. Type of drive unit (electric motor, internal combustion engine); 3. Service environment (oily, wet, dusty); 4. Driven unit loads (jerky, shock, reversed); 5. pulley-belt
arrangement (open, crossed, turned)
Types of belt drive There are four basic types of belts 1. Flat belts The flat belts are rectangular in cross section. In flat belts, the width of belt is substantially higher than the thickness of
belt. They can be used to transmit a moderate amount of power from one pulley to another, when the two pulleys are not more than 15 metre apart. 160
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Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
2. V-belts The V-belts are trapezoidal in cross section. They can be used to transmit a large amount of power from one pulley to another, when the two pulleys are relatively close to each other. 3.
Circular belts or ropes The circular belts or ropes are circular in cross section. They can be used to transmit a large amount of power from one pulley to another, when the two pulleys are more than
5 metre apart. 4. Timing belts The timing belts or toothed transmit the power by means of teeth rather than friction; hence there is no slip.
Types of flat belt drives: 1. 2. 3. 4. 5. 6. 7. 8.
Open belt drive Cross belt drive Open belt drive with idler pulley Multiple belt drive Quarter turn belt drive Compound belt drive Stepped or cone pulley drive Fast and loose pulley drive
Materials: The desired properties of the materials used for the belts are High coefficient of friction, High flexibility, Durability, High strength The different materials used for the flat belt
are as follows 1. Leather, 2. Fabric, 3. Rubber, 4. Polyester or polyamide Rubber canvas material properties :- [σu] = 40-45 N/mm2, E = 100 N/mm2, µ = 0.3 , ρ = 1000-1200 kg/m3
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Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Belt Speeds: Linear speed of the belt V=
D*N = d*n,
= , Speed Ratio, i =
If the thickness of belt is considered, then Speed Ratio, i =
The belt speed should be between 17.5 m/s < V < 22.5m/s
Slip of the belt: The difference in the linear speeds of the pulley rim and the belt is known as slip and is generally expressed as a percentage. Let, S1 = percentage slip between the driving
pulley and the belt S2 =percentage slip between the belt and the driven pulley V = linear speed of the belt, mm/s Linear speed of the belt, V = Or Again,
… (a)
= V [1-
… (b)
=V–V* Or
Substituting the value of ‘V’ from equation (a) in equation (b) =
] [1-
= [1Speed ratio, i =
], ] =
= [1-
= [1[1-
] [1-
] ],
= [1-
] , Where, s = S1 + S2 = Total percentage slip
If the thickness’’ of the belt is considered, then,
Speed ratio, i =
] 162
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Creep of belt: When the belt moves over the pulley to transmit power, the tension in the tight side is more than the tension in slack side. As the belt material is elastic, it elongates more on
tight side than on slack side, resulting in unequal stretching on two sides of the drive. Therefore, the length of the belt received by the driving pulley is more than the length that moves of the
driving pulley. Hence, belt must creep back slightly relative to driving pulley rim. On the other hand, the length of the belt received by the driven pulley is less than the length that moves of the
driven pulley. Hence, the belt must creep forward slightly relative to driven pulley rim. This motion of belt relative to driving and driven pulley due to unequal stretching of the two sides of the
drive is known as creep. Difference between slip and creep: Creep is due to the elastic property of belt whereas, the conventional slip is due to insufficient frictional grip between the belt and
pulley. However, the effect of the creep as well as slip is to reduce the speed ratio, and hence power transmission.
Power transmission of the belt: Power transmitted between a belt and a pulley is expressed as the product of difference of tension and belt velocity:
P = (T2 – T1) v Where, T1 and T2 are tensions in the tight side and slack side of the belt respectively. They are related as:
Tension ratios: Tension ratios is defined as ratio of tension in the tight side to tension in the slack side
.....For flat Belt,
.....For V-Belt
where, μ is the coefficient of friction between belt material and pulley material (0.2 to 0.3) , α is the angle subtended by contact surface at the centre of the pulley in radian and β is the semi
cone angle of V-cross section (widely β =20 degree).
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Centrifugal tensions: When the belt continuously runs over the pulleys, therefore, some centrifugal force is caused whose effect is to increase the tension on both the tight as well as the slack
sides. The tension caused by centrifugal force is called centrifugal tension. At lower belt speeds (less than 10 m/s), the centrifugal tension is very small, but at higher belt speeds (more than 10 m
/s), its effect is considerable and thus should be taken into account.
Let Tc = Centrifugal tension in the belt, N M = mass of the belt per unit length, Kg/m V = linear speed of the belt, m/s r = radius of the pulley over which the belt runs, m Consider a small element
‘PQ’ of the belt subtending an angle of ‘dϴ’ at the centre of the pulley. Centrifugal force acting on belt element ‘PQ’ = mass * centripetal acceleration = M r dϴ *
= m V2 dϴ
Consider an equilibrium of force on belt element ‘PQ’ in Y-direction M V2 dϴ - 2 Tc sin (dϴ / 2) = 0 M V2 dϴ - 2 Tc(dϴ / 2) = 0 Tc= M V2 Thus the centrifugal tension depends upon 164
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
1. The mass of the belt per unit length ‘M’, 2. The belt speed ‘V’.
Condition for maximum power with centrifugal tension (Tc): When centrifugal tension is neglected , T1 / T2 = eµθ T1 = Tight side tension in Newton (N) T2 = Slack side tension in Newton (N) µ =
Coefficient of friction between the pulley and belt materials θ = Angle of contact or angle of lap on the smaller pulley Initial tension = Ti = (T1 + T2)/2 Power transmitted = (T1 –T2) v Watts (W) v
is the linear velocity of the belt in meters , v =πD N/60 Where N is rpm and D is the diameter of the pulley in meters, When centrifugal tension is considered , Tc = m v2 Where m is mass of belt PER
UNIT LENGTH, kg/m Initial tension = Ti = ((T1 + T2)/2 + Tc) Ttight = T1 + Tc Tslack = T2 + Tc Ttight / Tslack = eµθ Power transmitted = (Ttight — Tslack) v Watt Condition for maximum power
transmission : Tmax = 3 m v2 T1 = (2/3) Tmax T2 = (1/3) Tmax V = (Tmax/3 m) 0.5 Max Power = (2/3) Tmax (1 — eµθ)x( (Tmax/3 m)0.5) Watts
Initial Tension in the belt:
When a belt is wound round the two pulleys (i.e. driver and follower), its two ends are joined together, so that the belt may continuously move over the pulleys, since the motion of the belt (from
the driver) and the follower (from the belt) is governed by a firm grip due to friction between the belt and the pulleys. In order to increase this grip, the belt is tightened up. At this stage, even
when the pulleys are stationary, the belt is subjected to some tension, called initial tension. When the driver starts rotating, it pulls the belt from one side (increasing tension in the belt on
this side) and delivers to the other side (decreasing tension in the belt on that side). The increased 165
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
tension in one side of the belt is called tension in tight side and the decreased tension in the other side of the belt is called tension in the slack side. Let, T0 = Initial tension in the belt, T1
= Tension in the tight side of the belt, T2 = Tension in the slack side of the belt, and α = Coefficient of increase of the belt length per unit force. A little consideration will show that the
increase of tension in the tight side = T1 – T0 and increase in the length of the belt on the tight side = α (T1 – T0)
Similarly, decrease in tension in the slack side = T0 – T2 and decrease in the length of the belt on the slack side = α (T0 – T2) …….(ii) Assuming that the belt material is perfectly elastic such
that the length of the belt remains constant, when it is at rest or in motion, therefore increase in length on the tight side is equal to decrease in the length on the slack side. Thus, equating
equations (i) and (ii), we have α (T1 – T0) = α (T0 – T2) .’. T0 = (T1+ T2)/2
T1 – T0= T0– T2
(Neglecting centrifugal tension)
To = (T1 + T2 + 2Tc)/2
(Considering centrifugal tension)
Note: In actual practice, the belt material is not perfectly elastic. Therefore, the sum of the tensions T1 and T2, when the belt is transmitting power, is always greater than twice the initial
tension. According to C.G. Barth, the relation between T0, T1 and T2 is given by, T1 + T2 = 2 T0
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Design Procedure of Flat Belt STEP 1: Calculate Design Power STEP 2: Calculation of diameter using savarin relations from the data given, take the centre distance, dia. of the pulley & speed ratio.
(Or assuming belt velocity) STEP 3: Calculation of actual belt speed, v =𝜋 d N/60 m/s. STEP 4: From limiting value of speed selecting pulley material STEP 5: Calculate diameter of bigger pulley STEP
6: Calculate center distance STEP 6: Calculate the length of the belt STEP 7: Calculation of arc of contact (PSG 7.54). STEP 8: Determine the belt material STEP 9: Determine expected life of belt
STEP 10: Calculation of allowable tensile stress and Belt dimensions. STEP 11: Design of shaft STEP 12: Pulley construction STEP 13: Arm construction
NUMERICALS Problem 1 : Design a Flat Belt drive for the following specification;
Power = 15KW, Input Speed = 1440rpm, Output Speed = 440rpm , expected life of the belt is 1year. Solution:
STEP 1:- From PSG 7.53, Assuming steady load Design power , [P] = 1.2 x P = 1.2 x 15 = 18 KW STEP 2 :- Let D1 and D2 be the diameters of smaller and bigger pulley respectively, By Severain's
relation, 1100 1100
< Dmin < 1300
< Dmin < 1300
255.18 < Dmin < 301.70 mm , 167
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
From PSG 7.54 Selecting standard diameter of pulley as , D1 = 280 mm STEP 3:-Check for belt speed (PSG 7.54) V=
= 21.11 m/s
STEP 4:- Material for pulley, Cast Iron as V < 30 m/s CI pulley is selected. STEP 5 :- Bigger pulley diameter D2 = D1 x
D2 = 280 x
[Assuming η = 98%, 2% slip]
x 0.98
= 898.036 mm
Selecting standard diameter of the pulley as
D2 = 900 mm.... PSG 7.54,
Checking for output speed, N2 N2 = 21.11 x 60 x 1000/pi x D2 = 447.96 rpm STEP 6:- Centre distance and Belt length Assuming Centre distance , C = 2D2 = 2 x 900 = 1800 mm , Considering Open Drive Belt
Length (L) is given by L = 2C + (D2 + D1) +
....... PSG 7.53
L = 5506.928 mm = 5.506 m STEP 7 :- Angle of lap (θ) For smaller pulley, Angle of lap θ = 180 - 2 α , where, α=
= 9.91 ,let α = 10 degree
Hence , θ = 180 - 2 x 10 = 160 degree STEP 8: Selection of belt material Rubber Canvas σu = 4.5 N/mm2, ρ = 1200 kg / m3, E = 100 N/ mm2, µ = 0.25 168
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
STEP 9 :- Belt tension = eµθ =
T1 = 2T2 T1 - T2 = T1 - T2 = 18/21.11, T2 = 852.67 N , T1 = 1705.35 N STEP 10 :- Selected life in hours Assume 10 hours 1 day - 360 days/year Lhr = 1 year x 360 x
x 10 = 3600 hours
Expected life of belt in number of stress cycles x 2 = 99.57 x 106 cycles
n = Lhr x 3600 x
STEP 11 :- Maximum allowed tensile stress in belt for given life is given by ] ^m =
where σ-1 = 0.2 σut ( based on 10-6 cycles ) σ-1 = 0.2 x 4.5 = 9 N/mm2 ,
m = 6 for flat belt
σtmax = 6.13 N/mm2 σtmax = Assuming
+ ρv2 x 10-6
= 10, as ( = 6
20 )
b = 10t 6.13 =
+ 1200 x 21.112 x 10-6
t = 7.78 mm , t = 8 mm, b = 80 mm STEP 12 :- Pulley dimension & shaft design For driving pulley D1 = 280 mm < 400 mm, hence it is web type construction. width of pulley B = b + 13 = 80 + 13 = 93 mm
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Bearing installation distance B1 = 2B = 186 mm Bearings are installed 186 mm apart T1 + T2 = 852.67 + 1705.352 = 2558.02 N By symmetry RA + RB = 1279.01 N Total bending torque, Tb1 = 1279.01 x 93 =
118.94 x 103 N-mm Consider torque on both pulleys Tt1 =
= 119.36 x 103 N-mm
Tt2 =
= 390.852 x 103 N-mm
Hence the shaft is shifted to both bending & torsion ds = σy = 360 N/mm2 , σb =
= 120 N/mm2, τ = 0.5 x σb = 0.5 x 120 = 60 N/mm2
ds1 = 35.11 mm , from PSG 7.25 selecting 38mm Length of hub = 2 ds1 = 2 x 38 = 76 mm
Step 13: Driven pulley design D2 = 900 mm , Arm type construction. Selecting 6 no, of arms T = (T1- T2)(D2/2) = 456.356 KN-mm Selecting C45 from PSG 1.9, [τ] = 45N/mm2
ds =
= 45 mm
STEP 14: Arm design
Fig: Cast Iron pulley with 4 arm
Selecting GCI-15 as pulley material, from PSG 1.9, Assuming [σb] = 10 to 13 N/mm2 170
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Cross section of arm is elliptical, Assuming b = To accommodate arm as the web, condition is , so, a = 56.54 mm, b = 56/2 = 27 mm PSG 7.120, section modulus,
Bending Moment of arm = M = (
= 8872.332 mm ) *e = 396.375 *
Induced bending stress, (σb)=
= 7.446 < [σb], Design is safe
PROBLEM 2: Design a Flat Belt drive for the following specification Power = 20 KW, Input Speed = 960 rpm, Output Speed = 320 rpm, using KW rating method design Flat Belt and find the life in hours.
Also design shaft and pulleys. Solution: Given data, P = 20KW, N1 = 960 rpm, N = 320 rpm STEP 1:- [P] = 1.3 x (P) Service Factor = 1.3 [PSG 7.53] [P] = 26KW STEP 2:- Assuming velocity = 20 m/s 17.5
, D = 397.887 mm
Selecting standard pulley: - D = 400 mm [PSG 7.54] STEP 3:- Driven Pulley, D=Dx
x η = 400 x
x 0.98 = 1045.33 mm
PSG 7.54 Selecting standard pulley, D = 1120 mm STEP 4:- Center distance 171
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Assume the center distance as, optimal center distance = (1.5 to 2.5) D = 2D C = 2240 mm Belt length, L = 2C + L = 2 x 2.24 +
(D + d) +
(1.12 + 0.4) + (1.12-0.4)
(4* 0.224)
L = 4.48 + 2.3876 + 0.057857 = 6.925 m STEP 5:-Angle of lap (ϴ) For smaller pulley ϴ = 180 - 2α , α = sin-1(
Change in output speed N2 = N1 x
x η = 960 x
x 0.98 = 336 rpm
STEP 6:- Selecting belt using KW rating Selecting FORT949g [PSG 7.34] KW rating of FORT belt 1 mm & width 1 ply at 10 m/s belt speed is given by, For [KW] 10 m/s = 0.0289 KW/per mm per ply [KW] 23.75
m/s =
x 0.0289 = 0.0578 KW/mm per ply
Millimetre plies at belt, I x b = corrective load/ [KW]v Now, corrective load, P x load correction factor =26 x 1.02 = 26.52 KW It is known that, 6 < < 20, I x b =
= 458.823
Length , l
Width, b
Std. Width, b
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Selecting i = 5 and b = 100 Let b/t = 10 σtmax= σt + σb + σc =
+ ρv2 x 10-6
STEP 7:- Material - Rubber canvas ρ = 1200 kg/m2, E = 100 N/mm2, σt = 45 N/mm2, σ-1 = 9 N/mm2 STEP 8:-belt tension T1 – T2 =
T1/T2=eμϴ T1 = 2.329T2 Therefore, T2= 977.88 N, T1 = 2268.68 N STEP 9:- Maximum allowable tensile stress in belt +
+ ρv2 x 10-6 =
+ 1200 x 202x 10-6
σtmax = 5.24 N/mm2 [
]m =
, σ-1 = 9 N/mm2, m = 6 , np = 2, nb = 107
For flat belt, n = 25.67 x 107 = Lhr x 3600 x
x 2 = 22345.9 hours
B = b + 13 = 100 + 13 = 113 mm Bearing span = B1 = 2B = 2 x 133 = 226 mm STEP 10:- Force analysis (i) Bending of shaft T1 + T2 = 3246.56 N, RA = RB = 1623.28 N Total bending torque Tb1 = 1623.28 x
113 = 183.43 x 103 N-mm 173
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Consider torque on both pulleys Tb1 =
= 258.626 x 103 N-mm
Tb2 =
= 775.88 x 103 N-mm
Consider larger value of torsional torque for shaft design, STEP 11:- Pulley and shaft design The shaft is subjected to both bending & torsion, ds = Assuming shaft material C45 & FOS = 3 σy = 360 N/
mm2 σb =
= 120 N/mm2
[τ] = 0.5[σb] = 0.5 x 120 = 60 N/mm2 ds = ds = 44.44 mm
45 mm
Length and diameter of the hub, Dh = Lh= 2 ds1 = 2 x 45 = 90 mm The driving pulley diameter, D1 = 400 mm
STEP 13: Driven Pulley design: Since D2 = 1120mm (i) Number of arms of driving pulley = 6 (ii) Cross-section of arms = Elliptical Selecting GCI-15 as pulley material, from PSG 1.9 Assuming σb =[ 10-
13] N/mm2 Assuming b = 174
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
To accommodate arm as the web, amax so, a = 47.123 mm a= 0.9 * 47.123 = 42.41 b = 42/2 = 21 mm PSG 7.120, section modulus = 3744.35 mm4
BM of arm = M = ( σb =
) *e = (2268.68 - 977.88)* 155
= 8.905 N/mm2 < [σb] , Design is safe
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
V-Belt V belts (V-belts, vee belts, or, less commonly, wedge rope) solved the slippage and alignment problem. It is now the basic belt for power transmission. They provide the best combination of
traction, speed of movement, load of the bearings, and long service life. They are generally endless, and their general cross-section shape is roughly trapezoidal (hence the name "V"). The "V" shape
of the belt tracks in a mating groove in the pulley (or sheave), with the result that the belt cannot slip off. The belt also tends to wedge into the groove as the load increases—the greater the
load, the greater the wedging action—improving torque transmission and making the V-belt an effective solution, needing less width and tension than flat belts. V-belts trump flat belts with their
small center distances and high reduction ratios. The preferred center distance is larger than the largest pulley diameter, but less than three times the sum of both pulleys. Optimal speed range is
1,000–7,000 ft/min (300–2,130 m/min). V-belts need larger pulleys for their thicker cross-section than flat belts. High variety of belt sizes available permits the application of V-belt in a wide
range of drive applications, such as: Machine tools, industrial washing machines, textile machines, Continuous paper machines, high power mills, and stone crushers. For high-power requirements, two
or more V-belts can be joined side-by-side in an arrangement called a multi-V, running on matching multi-groove sheaves. This is known as a multiple-V-belt drive (or sometimes a "classical V-belt
drive"). V-belts may be homogeneously rubber or polymer throughout or there may be fibers embedded in the rubber or polymer for strength and reinforcement. The fibers may be of textile materials such
as cotton, polyamide (such as Nylon) or polyester or, for greatest strength, of steel or aramid. When an endless belt does not fit the need, jointed and link V-belts may be employed. Most models
offer the same power and speed ratings as equivalently-sized endless belts and do not require special pulleys to operate. A link vbelt is a number of polyurethane/polyester composite links held
together, either by themselves, such as Fenner Drives' PowerTwist, or by metal studs, such as Gates' Nu-T-Link. These provide easy installation and superior environmental resistance compared to
rubber belts and are length adjustable by disassembling and removing links when needed. CONSTRUCTION: The main V-belt components are: 1) Belt body made of a special rubber compound which provides,
due to its excellent mechanical characteristics, high transmission efficiency and assures a minimum rubber wear off; 2) Tensile member consisting in high-strength low stretch cords, which grant
length stability over the belt life time; 3) Fabric jacket or cover made of fabric, protecting the tensile member and permitting the use 176
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
of back side idler.
Following are the advantages and disadvantages of the V-belt drive over flat belt drive:Advantages:1. The V-belt drive gives compactness due to the small distance between centers of pulleys. 2. The
drive is positive, because the slip between the belt and the pulley groove is negligible. 3. Since the V-belts are made endless and there is no joint trouble, therefore the drive is smooth. 4. It
provides longer life, 3 to 5 years. 5. It can be easily installed and removed. 6. The operation of the belt and pulley is quiet. 7. The belts have the ability to cushion the shock when machines are
started. 8. The high velocity ratio (maximum 10) may be obtained. 9. The wedging action of the belt in the groove gives high value of limiting ratio of tensions. Therefore the power transmitted by
V-belts is more than flat belts for the same coefficient of friction, arc of contact and allowable tension in the belts. 10. The V-belt may be operated in either direction, with tight side of the
belt at the top or bottom. The center line may be horizontal, vertical or inclined. Disadvantages:1. The V-belt drive cannot be used with large center distances, because of larger weight per unit
length. 2. The V-belts are not so durable as flat belts. 3. The construction of pulleys for V-belts is more complicated than pulleys of flat belts. 4. Since the V-belts are subjected to certain
amount of creep, therefore these are not suitable for constant speed applications such as synchronous machines and timing devices. 5. The belt life is greatly influenced with temperature changes,
improper belt tension and mismatching of belt lengths. 6. The centrifugal tension prevents the use of V-belts at speeds below 5 m/s and above 50 m/s.
V-belt power rating: Each type of belt section has a power rating. The power rating is given for different pitch diameter of the pulley and different pulley speeds for an angle of wrap of 180 degree.
A typical nature of the chart is shown below. Here, for example, for pitch diameter of D1, power rating of the A section belt is kW1, kW2, kW3, kW4 for belt speeds of N1, N2, N3, N4 respectively. 177
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Service Factor: (PSG 7.69) A belt drive is designed based on the design power, which is the modified required power. The modification factor is called the service factor. The service factor depends
on hours of running, type of shock load expected and nature of duty. Design Power = service factor X Required Power (P). Modification of kW rating: (PSG 7.63 to7.67) Power rating of a typical V-belt
section requires modification, since, the ratings are given for the conditions other than operating conditions. The factors are as follows, Equivalent smaller pulley diameter and belt speed. In a
belt drive, both the pulleys are not identical, hence to consider severity of flexing, equivalent smaller pulley diameter is calculated based on speed ratio. The power rating of V-belt is then
estimated based on the equivalent smaller pulley diameter. Equivalent Pitch diameter = pitch diameter x small diameter factor (Fb). Fb can be selected corresponding to velocity ratio from PSG. 7.62.
Angle of wrap correction factor: (PSG 7.68) The power rating of V-belts are based on angle of wrap, α =180 degree. Hence, Angle of wrap correction factor is incorporated when α is not equal to 180
degree. Belt length correction factor: (PSG 7.59, 7.60) There is an optimum belt length for which the power rating of a V-belt is given. Let, the belt length is small then, in a given time it is
stressed more than that for the optimum belt length. Depending upon the amount of flexing in the belt in a given time a belt length correction factor is used in modifying power rating.
PULLEY SYSTEMS A belt and pulley system is characterised by two or more pulleys in common to a belt. A belt drive is analogous to that of a chain drive, however a belt sheave may be smooth (devoid
of discrete interlocking members as would be found on a chain sprocket, spur gear, or timing belt) so that the mechanical advantage is approximately given by the ratio of the pitch diameter of the
sheaves only, not fixed exactly by the ratio of teeth as with gears and sprockets. 178
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
TYPES OF PULLEY These are different types of pulley systems: Fixed: A fixed pulley has an axle mounted in bearings attached to a supporting structure. A fixed pulley changes the direction of the
force on a rope or belt that moves along its circumference. Mechanical advantage is gained by combining a fixed pulley with a movable pulley or another fixed pulley of a different diameter. Movable:
A movable pulley has an axle in a movable block. A single movable pulley is supported by two parts of the same rope and has a mechanical advantage of two. Compound: A combination of fixed and movable
pulley forms a block and tackle. A block and tackle can have several pulleys mounted on the fixed and moving axles, further increasing the mechanical advantage. STANDARD DIEMENSIONS OF THE PULLEY:
Width: It is the width of belt at its pitch zone. Nominal top width: It is the top width of the trapezium outlined on cross sectional area. Nominal height: It is the height of trapezium outlined on
the cross section of the belt. Pitch length: It is the length of pitch line of belt. Angle of belt: It is the included angle obtained by extending the sides of the belt.
Pulley generally has three main parts as below: 1. Rim, 2. Hub, 3. Arm Pulley diameter and width are calculated from design of required standard cross section. Minimum pulley diameter depends on
following factors: 4. Number of belts, 2. Belt speed 179
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Relationship between pulley width and width of rim are shown and based on various manufacturing catalogue, pulley diameter and crown height is also shown in table below: Belt width
Pulley width
Pulley diameter (mm)
Crown height (mm)
up to 125 mm
40-112 125-140 160-180 220-224 250-280 315-355
0.3 0.4 0.5 0.6 0.8 1.0
From 125 mm to 250 25mm From 250 mm to 375 38mm mm From 375 mm to 500 50mm mm
CROWNING OF PULLEY Before V-belts were invented, machinery was usually powered through flat canvas belts running on crowned pulleys. These flat belts stayed centered on pulleys without any guides
or flanges. The key to keeping them tracking centered on the pulleys is the use of "crowned pulleys". A crowned pulley is a pulley that has a slight hump in the middle, tapering off ever so slightly
towards either edge. How a crowned pulley keeps the belt tracking on it is a mystery to most people. An example of a crowned pulley can be found in the drive wheel of a shop band-saw. This
demonstrates that the principle also works with steel belts, which are much more rigid laterally than leather or rubberized canvas belts. Band saws also rely on crowned pulleys to help keep the blade
aligned. The thrust bearing behind the blade is there to help absorb the force of pushing the stock into the blade. But when the saw is just running idle, blade should ideally not touch the thrust
bearing. The crown alone is enough to keep the blade lined up.
Cast Iron Pulleys: a) Number of arms: 1) For pulleys up to 200 mm diameter, web construction is used. 2) For pulleys above 200 mm diameter and up to 450 mm diameter 4 arms are used. 3) For pulleys
above 450 mm diameter 6 arms are used. b) Cross-sections of arms - elliptical. 180
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Thickness of arm b, near rim = taper 4 mm per 100 mm. Radius of the cross-section of arms, R=3/4 b Minimum length of the bore l = 2/3 a; It may be more for loose pulleys, but in no case it exceeds a.
c) Radius: rl = b/2 (near rim), Radius r2 = b/2(near rim).
Minimum length of boss - The length of boss is equal to half the width of face, subject to a minimum of 100 mm in the case of pulleys with 19 mm diameter spokes and minimum of 138 mm for pulleys with
22 mm diameter spokes. The length of the boss is practically greater than the width of the pulleys. Thickness of rims - The thickness of 5 mm for the rim of all the pulleys tabulated in table may be
applicable for mild steel pulleys only.’ For cast iron flat pulleys the thickness may be specified as below: Rim thickness = D/200 + 3 mm for single belt and Rim thickness = D/200 + 6 mm for double
In most of the applications, belts are generally selected by the designer from the manufacturer’s catalogue. This helps in the use of standard available sizes. Following input data is required for
the selection of belt: 1. Power to be transmitted 2. Transmission ratio 3. Centre distance Select suitable V-belt section: Five types of standard V-belt sections are available.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Determine diameters of pulley: Recommended diameter of smaller pulley (d) can be taken from table 20.4 for selected cross section. Diameter of larger pulley (D) can be calculated, for required
velocity ratio i.e. for given input speed (N1) and required output speed (N2), using the following relationship: N1d = N2d Calculate design power For design purpose, maximum power transmitted by the
belt is obtained by multiplying the required power (P) by service factor (Ks). Value of Ks can be taken between 1 and 2, depending upon the service conditions i.e. light, medium, heavy or extra-heavy
duty, type of driver and driven machinery and operational hours. [P] = Ks X P Determine Pitch Length and Centre Distance Calculate length of the belt from its relation with d, D and C. Select the
nearest standard value. Determine corrected power rating. Power transmission capacity / rating (R): for a single V-belt, for different types of cross sections, can be taken from manufacturer’s
catalogue. Corrected power rating is obtained by multiplying the power rating by Pitch Length Correction Factor (Klength) and Angle of Contact Factor, Ka Determine the number of belts required
Required number of belts is determined by dividing the design power with corrected power rating for one belt.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
NUMERICAL 1. Design V belt drive for following specification:P= 30KW, Input speed = 960 rpm, Output speed = 500 rpm, Medium duty, medium shock level. Find the no. of V-belt from 1) manufacturing
catalogue 2) for expected life of 12 months. Also design shafts and pulley Solution: Given data, P= 30KW, N1 =960 RPM, N2 =500 RPM and Medium duty, medium shock level. Calculating velocity ratio, VR
= 𝑁1/ 𝑁2 = 1.92 Service factor for medium duty, for 12 hours a day = 1.2 Design Power [PSG 7.69] [P] = 30 x1.2 = 36 kW , Cross section selection [PSG 7.58] Based on power rating C, D & E cross
sections are available, Selecting E, where, W1 = 38 mm, T = 23 mm Assuming semi-cone angle =20°, W2 =W1 – 2t X tanα = 38 – (2 X 23 X tan 20) = 21.257 mm C/S Area= 0.5 (W1 + W2 )T = 0.5(38 + 21.257)23
= 681.46 mm2 Pulley dimensions: Using Savrain’s relation
368.188 mm < d< 435.131 mm, From psg 7.54 d = 400 mm For driven pulley,
,D= 400 X 1.92 X 0.98 = 752.64 mm, From PSG 7.54, D= 800 mm Centre distance (C):
from PSG7.61
Cmin=0.55(D+d) + T, T=0, Cmin=0.55(800+400) =660 mm Cmax = 2(D+ d) =2400 mm Assuming C =1600 mm,
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Angle of lap = 165.67° Pitch Length from
L =5109.955 mm,
PSG 7.61
PSG 7.60 standard length, L = 5426 mm
Substituting in above equation, C = 1759.153 mm Belt speed V = (𝜋𝑥 𝐷 𝑥 𝑁)/ 60000 = 𝜋 𝑥 960 𝑥 400/ 60000 = 20.104 m/s T1-T2 = [𝑃]/ 𝑉 = 36000/ 20.106 = 1790.49
, 𝑇1 / 𝑇2 = 12.63 , T1 = T2 X 12.63 From above equations, T1 = 1944.44 N, T2 =153.95N Belt material selection: Rubber canvas is selected, σu= 40 N/mm2, E = 100 N/mm2, μ = 0.3, σ-1 = 0.2 X σu = 9 N/
mm2 ρ = 1000 Kg/m3, From manufacturing catalogue, No. of belts = 𝑃∗ 𝐹𝐴/ 𝐾𝑊∗𝐹𝐶∗𝐹𝐷 From PSG 7.70 Where, P x Fa= [P] = 30kw, For V-V belt; PSG 7.68 Fd= 0.97 Fc = 0.94, PSG 7.62 Equivalent pitch diameter
de = d x Fb, For velocity ratio = 1.92, Fb= 1.13 de = d x Fb =400 x 1.13 = 452 mm No of belts = 𝑃∗ 𝐹𝐴/ 𝐾𝑊∗𝐹𝐶∗𝐹𝐷 = 36/21.77∗ 0.94 ∗ 0.97 = 1.92 = 2. Expected life in hours: The expected life of belt
is 12 months Assuming 12 hrs a day and 25 days a month of working time of belt Lhr = 12 X 25 X 12 = 3600hrs Lmr = Lhr X 3600 X Np X (𝑉/𝐿) = Lhr X 3600 X 2 X (20.106/5.109)= 10.2 X 107 cycles 184
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Maximum allowable tensile stress:
nb = 10^7, m = 8 for V belt σ-1= 9 N/mm2, n = 10.2X107 σtmax = 6.732 N/mm2 σtmax= (𝑇1/A∗𝑛) + (𝐸∗𝑡/ 𝑑) + 𝜌xv2, solving n = 1.76, let n = 2 Shaft Design: let the material be C30, [τ] = 30 N/mm [P] =
(2𝜋∗𝑁1∗𝑇1)/ 60 T1 = 36∗60∗1000/ ( 2𝜋∗960) , T1 = 358.09 Nm = 358.09 X10^3Nmm
ds1 = 39.32 mm = 40 mm [P] = (2𝜋∗𝑁2∗𝑇2)/60 , T2 = 36∗60∗1000/ (2𝜋∗500) T2 = 687.54 Nm = 687.54 X103Nmm ds2 = ds1X 𝑖 3 = 40 X 1.92 3 = 49.71 mm = 50 mm From PSG 7.70, Assuming Pulley material as GCI
20 Web Type pulley ds1 = 40 mm ds2 = 50 mm dh1 = 2ds1 = 2 X 40 = 80 mm For cross section E, Ip = 32, e = 1.35 X Ip = 1.35 X 32 = 43.2 mm f = 0.9 X Ip = 0.9 X 32 = 28.8 mm h = 0.7 Ip + 1 = 0.7 X 32 +
1 = 23.4 mm Minimum distance to pitch B = 0.3X Ip = 0.3 X 32 = 9.6 mm Lh1 = 2ds1 = 2 X 40 = 80 mm B = (n-1) X e + 2f = (2-1) X 43.2 + 2 X 28.8 = 100.8 mm
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
CHAIN DRIVES Introduction: Chain drives are means of transmitting power between two mechanical elements. Chain drives consist of an endless series of chain links that mesh with toothed sprockets.
Chain sprockets are locked to the shafts of the driver and driven machinery. Chain drives represent a form of flexible gearing. The chain acts like an endless gear rack, while the sprockets are
similar to pinion gears. Chain drives provide a positive form of power transmission. The links of the chain mesh with the teeth of the sprockets and this action maintains a positive speed ratio
between the driver and driven sprockets. Chains can be used to perform three basic functions: 1. Transmitting power 2. Conveying materials 3. Timing purposes Advantages and Disadvantages over belt
drive: Advantages: i. Chain drives do not slip or creep, therefore constant velocity ratio is obtained. ii. Chain drives are more compact than belt drives. iii. Chain drives give higher transmission
efficiency. iv. The chain drive has ability of transmitting motion to several shafts by one chain only. v. It gives less load on shafts. vi. Chains are easy to replace. vii. Chains can operate
effectively at high temperatures. viii. Chains do not deteriorate due to oil, grease, sunlight, or age. ix. Chains withstand chemicals and abrasive conditions. x. Chains can operate in wet
conditions. Disadvantages: i. The production cost of chain is relatively high. ii. Smooth speed transmission is not possible due to chordal action. iii. It requires accurate mounting and careful
maintenance. iv. Noisy compared to belt drive. v. Life is less compared to belt drive. Components of chain drive : i. Chain ii. Sprocket iii. Shafts 186
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
i. Chain: The chains are made up of rigid links which are hinged together in order to provide necessary flexibility for wrapping around the driving and driven wheel. ii. Sprocket: A sprocket is a
toothed wheel that is designed to engage with something that will be pulled over the wheel as the wheel rotates iii. Shafts: Shaft is used to transmit torque between rotating element. Types of
Chain Drives: The chains, on the basis of their use, are classified into following three groups. I. Hoisting and Hauling (or crane) chains II. Conveyor (or tractive) chains III. Power transmitting
(or driving) chains Hoisting and Hauling (or crane) chains : These chains are used for hoisting and hauling purposes. These are of two types: a) Chains with Ovel links b) Chain with square link
Conveyor Chain: - Used for elevating and conveying within 0.8 to 3m/s. a) Detachable or hook joint b) Closed joint type. Power Transmitting Chain: It is used for short center distance. a) Block or
bush chain b) Bush roller chain c) Silent chain Block or bush chain The chains used in conveyor belts are commonly block chains, which consist of solid or laminated blocks connected by side plates
and pins. The blocks engage with teeth on sprocket wheels. Depending on the material being moved, devices are connected to the blocks.
Bush roller chain A roller chain is a development of the block chain in which the block is replaced by two side plates, a pair of bushings, and rollers. This type of chain is used on bicycle and is
adaptable to many other needs, from small small-strand drives for microfilm projectors to multiple-strand 187
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
chains for heavy-duty service in oil-drilling equipment. Roller chains are assembled from pin links and roller links. A pin link consists of the two side plates connected by two tightly fitted pins.
A roller link consists of two side plates connected by two tightly fitted bushings on which hardened steel rollers are free to rotate. When assembled, the pins are a free fit in the bushings and
rotate slightly, relative to the bushings, when the chain goes onto and leaves a sprocket.
Silent chain: A silent chain is essentially an assemblage of gear racks, each with two teeth, pivotally connected to form a closed chain. The links are pin-connected, flat steel plates with straight
teeth. Silent chains are quieter than roller chains, can operate at higher speeds, and can transmit more load for the same width.
The engagement between the chain and sprocket is shown in above figure. Some of the parameters related to chains are: i. Pitch of chain (P): It is the distance between the hinge centre of one link
and the corresponding hinge centre of the adjacent link. ii. Pitch Angle: It is the angle between the two lines joining the centre of sprocket and the hinge centres of two adjacent links, when the
chain is wrapped around the sprocket. Pitch diameter of sprocket: It is defined as the diameter of an imaginary circle that asses through the centre of link pins as the 188
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
chain is wrapped around the sprocket. Let P = chain pitch, mm Z = number of teeth on sprocket α = pitch angle, degree D = Diameter of sprocket, mm
Average Velocity : The average velocity is given by, v = (zpn /60000) Speed Ratio : It is the ratio of input speed to output speed and is given by,
Where, n1 and n2 = speed of driving and driven shafts Z1, Z2 = number of teeth on driving and driven shafts Number of links and Length of chain : The number of links in a chain is determined by
following relationship,
The length of chain is always expressed in terms of number of links, L = Ln ×P
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Centre Distance: The centre distance is given by,
Polygon effect in chain drive: The chain passes around the sprocket as a series of chordal links. This action is similar to that of a non-slipping belt wrapped around a rotating polygon. Chain forms
a polygon instead of a circle when rapped round a smaller sprocket with less number of teeth. Thus the chain does move along the pitch circle but moves through smaller chordal distance causing
vibrations, noise and impact on the sprocket teeth. It is called the polygonal effect. Thus the radius and hence the velocity is continuously changing in between the two teeths. The magnitude of
speed variation is dependent on the number of teeth on the smaller sprocket. Polygonal effect reduces the chain life too. To reduce the polygonal effect, use minimum number of teeth on smaller
sprocket as 19. With this, speed variation is only 1.6 % which is small. 1. 2. 3.
4. 5. 6.
Characteristics of chain drive: They can be used for short to medium distances. Gears need additional idler gears. They can be used for transmission of higher loads compared with belt drives. The
power transmission efficiency of chain drives may be as high as 99%; higher compared to flat and V belts. A chain drive does not slip and to that extent, it is positive drive. Due to the polygonal
effect and wears in the chain joints, they are not suitable for precise speed control. They have longer life, no creep, and the ability of driving several shafts from a single source of power
compared to belt drives. They can only be used for transmitting motion between parallel shafts. Crossed flat belts, bevel gears, worm gears, and some crossed helical gears can be used to transmit
motion between nonparallel shafts. The location (center distance) and alignment tolerances need not be as precise as with gear drives, but require precise alignment of shafts, compared to belt
drives. The best
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
services can be expected when both the input and output sprockets lie in the same vertical plane. 8. They require proper maintenance, particularly lubrication, when compared to gears. Design
procedure of chain drive: i. Assumptions ii. Find the design load iii. Find the number of teeth iv. Find pitch using following relation, v. Select the proper chain vi. Check the failure against
tensile load vii. Find diameter of input and output sprocket viii. Find diameter of shaft Selection of chain from manufacturing catalogue:
i. Given data from requirement ii. Select number of teeth on smaller sprocket (sprocket pinion) iii. Select service factor or load correction factor for given application iv. Calculate design power
v. Find tooth correction factor (K1)
vi. Find the multiple strand factor ( K2 ) vii. Calculate required power rating of chain viii. Select chain from manufacturer’s catalogue having the required power rating ix. Find dimensions of
selected chain from manufacturer’s catalogue x. Find the number of links in chain (M) xi. Calculate the corrected centre distance xii. Find the maximum theoretical power the chain can transmit based
on tensile strength.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
NUMERICAL Design chain drive for following specifications: Power to be transmitted = 10kW, Input speed=100rpm, Output speed=30rpm, Also design shaft, input sprocket and output sprocket. Solution: 1.
Assumptions: i. Horizontal chain, ii. Drive unit is flexible, iii. Machine unit is rigid iv. Nature of duty: 8-10 hrs/day, v. Nature of load: Heavy 2. Design Power: [P] = P*Ks, where, Ks = K1 * K2 *
K3 * K4 * K5 * K6 K1 = 1.5 (Heavy shock) K2 = 1 (Distance regulation factor) K3 = 1 (Centre distance of sprocket) K4 = 1 (Layout factor) K5 = 1 (Lubrication factor) K6 = 1 (Shift factor/rating) [P] =
10 * 1.5 = 15kW Number of teeth: i = N1/N2 = 100/30 = 3.33 Z1 = 23 ……………………………….. (From PSG 7.74) Z2 = i×Z1 = 3.33 × 21 = 76.67= 77 Pitch Selection:
Let m = 2 (Duplex), Z1 = 23, σ = 3.15 kgf/mm2 = 31.5 N/mm2
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
P ≥ 27.8 mm
w = 7.60 kgf Q = 17700 kgf = 177kN
Checking of Tensile Failure:
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Tensile failure is checked indirectly by comparing the actual FOS (Factor of safety) and Minimum FOS specified by manufacturer, Minimum FOS = 7.0 ………………………………... (PSG 7.77)
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Input Sprocket diameter:
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Summary: Chain selection Selecting Chain DR100 Number of Links np = 132 mm Length of Chain L = 4191 mm Exact centre Distance a = 1273 mm Input Sprocket Diameter d1 = 233.17 mm Output Sprocket
Diameter d2= 778.404 mm Shaft Design: Material C30 Input Shaft Design ds1 = 62.417 mm Output Shaft Design ds2 = 88.56mm
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
MODULE 6 CLUTCH
A clutch is a mechanical device that is used to connect or disconnect the driving shaft to the driven shaft at the convenience of the operator. It is used whenever the transmission of power or motion
must be controlled either in amount or over time. In the simplest application, clutches connect and disconnect two rotating shafts (drive shafts or line shafts). In these devices one shaft is
typically attached to an engine or other power unit (the driving member) while the other shaft (the driven member) provides output power for work. In Automobiles the flow of mechanical power from the
prime mover to driven machine is controlled by the means of clutch. In order to change the gears or to temporarily stop the vehicle, the requirement is that the driven shaft should stop but the
engine should continue to run. This is achieved by the means of a clutch, mounted between the engine and the gear box, which is operated by a lever.
Types of Clutches:
Generally the clutches are classified into two types – Positive Clutches Friction Clutches Positive clutches: These clutches are generally used when positive drive is required. The jaw clutch,
which is the simplest type of positive clutch, permits one shaft to drive another shaft through direct contact.
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Fig: Square jaw clutch coupling
Fig - Spiral jaw clutch coupling
The jaw clutch consists of two halves, one of which is permanently fastened to the driving shaft by the key while the other half of the clutch is free to slide axially on the splined driven shaft,
thus permitting it to engaged or disengaged by sliding. Square shaped jaws are used where the engagement or disengagement of the teeth takes places in motion and under load. Such teeth can transmit
power in either direction. While the spiral shaped jaws can transmit power in one direction only. Friction clutches This plate consists of two plates .One plate is rigidly keyed to the driving shaft,
while the other is free to slide axially on splined driven shaft. Friction lining is normally provided on the driven plate. Two plates are held together due to axial force applied by the compression
spring. The frictional force between the two contacting surfaces is responsible for transmitting the torque or power.
Fig -Single plate friction clutch
Fig - Multiple plate friction clutch
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Types of Friction clutches:
Uniform Pressure Criterion-
In this theory, it is assumed that the normal intensity of the pressure is uniform over the entire contacting surface i.e. p=constant, In general if there are ‘N’ pair of contacting surfaces then the
equations for uniform pressure theory is given asW= π.P. (r02-ri2). Where, W is the axial thrust which the friction surfaces are withheld. Uniform Wear CriterionIn this theory, it is assumed that
the rate of wear is uniform over the entire contacting surface i.e. Pr = constant. Here, the wear is considered to be uniform over the entire surface.
Design Considerations in a Friction Clutch
The following considerations must be kept in mind while designing a friction clutch. 1. The suitable material forming the contact surfaces should be selected. 2. The moving parts of the clutch should
have low weight in order to minimise the inertia load, especially in high speed service. 3. The clutch should not require any external force to maintain contact of the friction surfaces. 4. The
provision for taking up wear of the contact surfaces must be provided. 5. The clutch should have provision for facilitating repairs. 6. The clutch should have provision for carrying away the heat
generated at the contact surfaces. 206
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
7. The projecting parts of the clutch should be covered by guard.
Requirements of a Good Clutch
1. Gradual Engagement 2. Good Heat Dissipation 3. Compact Size 4. Sufficient Clutch Pedal Free Play 5. Ease of Operation Design of single plate clutch Consider two friction surfaces maintained in
contact by an axial thrust (W) as shown in fig. (a)
T = Torque transmitted by the clutch, p = Intensity of axial pressure with which the contact surfaces are held together, r1 and r2 = External and internal radii of friction faces, r = Mean radius of
the friction face, and μ = Coefficient of friction. Consider an elementary ring of radius r and thickness dr as shown in figure. It is known that area of the contact surface or friction surface= 2π
r.dr ∴ Normal or axial force on the ring, δW = Pressure × Area = p × 2π r.dr And the frictional force on the ring acting tangentially at radius r, Fr = μ × δW = μ.p × 2π r.dr ∴ Frictional torque
acting on the ring, Tr = Fr × r = μ.p × 2π r.dr × r = 2 π μ p. r2.dr Following two cases can be considered: 1. When there is a uniform pressure, and 2. When there is a uniform axial wear. 1.
Considering uniform pressure. When the pressure is uniformly distributed over the entire area of the friction face as shown in Fig.(a), then the intensity of pressure,
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Where W = Axial thrust with which the friction surfaces are held together. The frictional torque on the elementary ring of radius r and thickness dr is Tr = 2π μ.p.r2.dr Integrating this equation
within the limits from r2 to r1 for the total friction torque. ∴ Total frictional torque acting on the friction surface or on the clutch,
2. Considering uniform axial wear. The basic principle in designing machine parts that are subjected to wear due to sliding friction is that the normal wear is proportional to the work of friction.
The work of friction is proportional to the product of normal pressure ( p) and the sliding velocity (V). Therefore, Normal wear ∝ Work of friction ∝ p.V Or p.V = K (a constant) or p = K/V Let p be
the normal intensity of pressure at a distance r from the axis of the clutch. Since the intensity of pressure varies inversely with the distance, therefore p.r = C (a constant) or p = C / r and the
normal force on the ring,
∴ Total force acing on the friction surface,
We know that the frictional torque acting on the ring, 208
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
∴ Total frictional torque acting on the friction surface (or on the clutch),
Design of multi plate clutch:
Construction: A multiple plate clutch has more number of clutch plates. A typical clutch consist of the following components: Clutch basket or cover, clutch hub, drive (friction) plates, driven
(steel) plates, pressure plate and the clutch springs. The clutch housing is attached to the engine crank shaft flywheel. The pressure plate is fixed on the flywheel through the clutch springs. The
engine flywheel turns the clutch housing. The inner circumference of the clutch basket is splined to carry the thin metal plates. The clutch basket splines engage the tabs on the friction drive
plates. This sources the clutch housing and the drive plates to rotate together. Additionally they are free to slide axially within the clutch basket. Interleaved with the drive plates, there are
many number of driven plates. These driven friction plates have inner splines. These splines engage with the outer splines on the clutch hub. As such, the driven friction plates can slide on the
clutch hub. The clutch hub is linked to the input shaft of the transmission gear box. The drive plates and the driven plates are firmly pressed together by the pressure plate due to the clutch
springs. The drive plates, driven plates and the strong clutch coil springs are assembled within the clutch basket. 209
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Working: During clutch engagement, spring pressure forces the pressure plate towards engine flywheel. This causes the friction plates and the steel driven plates to be held together. Friction locks
them together tightly. Then the clutch basket, drive plates, driven plates, clutch hub and the gearbox input shaft all spin together as one unit. Now power flows from the clutch basket through the
plates to the inner clutch hub and into the main shaft of the transmission. The clutch gets released or disengaged when the clutch pedal is pressed. This causes the clutch pressure plate to be moved
away from the drive and driven plates, overcoming the clutch spring force. This movement of the pressure plate, relieves the spring pressure holding the drive and driven plates together. Then the
plates float away from each other and slip axially. Thus, the clutch shaft speed reduces slowly. Finally, the clutch shaft stops rotating. Power is no longer transferred into the transmission
gearbox. Depending upon the power output of the engine, and the weight of the vehicle, four to eight sets of plates (four to eight drive plates and four to eight driven plates) may be housed in the
clutch basket. The multiple plate clutch may be of dry type or wet type. When the clutch functions in atmosphere, it is called a dry clutch. When the clutch operates in an oil bath, it is named as a
wet clutch. Let n1 = Number of discs on the driving shaft, and n2 = Number of discs on the driven shaft. ∴ Number of pairs of contact surfaces, n = n1 + n2 – 1 , and total frictional torque acting on
the friction surfaces or on the clutch, T = n.μ.W.R Where R = Mean radius of friction surfaces given by
Design of cone clutch:
In a cone clutch, the driver is keyed to the driving shaft by a sunk key and has an inside conical surface or face which exactly fit into the outside conical surface of the driven. The driven member
resting on the feather key in the driven shaft, may be shifted along the shaft by a forked lever. In order to engage the clutch by bringing the two conical surfaces in contact. Due to the frictional
resistance set up at the contact surface, the torque is transmitted from one shaft to another shaft. In some cases, a spring is placed around the driven shaft in contact with the hub of the driven.
This spring holds the clutch faces in contact and maintains the pressure between them, and the forked lever is used to disengagement of the clutch. 210
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Applications: 1) Conveyers: Speed Compensation 2) Mixers: Motor Protection 3) Motorcycles: Gear Box or Transmission 4) Packaging Machine: Spins Down Freely 5) Printing Press: Ink Roller Drive
Advantages: 1) Small axial force is required to keep the clutch engaged. 2) The design is simple. 3) For a given dimension, the torque transmitted by cone clutch is higher than that of a single plate
clutch. Disadvantages: 1) One pair of friction surface only. 2) The small cone angle causes some reluctance in disengagement. Notations and Formulae of Cone Clutch: 𝛼 = Semi-Cone Angle (Generally
12.50) R = Maximum radius of friction lining R = Minimum radius of friction lining F = Axial Force Mt = Torque transmitting capacity P = Pressure intensity on friction lining According to Uniform
Pressure Theory (P is constant), Axial Force and Torque transmitting capacity is given by, and 211
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
According to Uniform Pressure Theory (P.r is constant), Axial Force and Torque transmitting capacity is given by, and Thermal analysis of clutch:
The design of the clutch or the selection of the clutch that is commercially available depends not only upon the torque transmitting capacity but also upon its ability to dissipate heat. The later
consideration is important because the coefficient of friction of the friction lining generally decreases with increasing temperature. If the temperature exceeds the limiting value, it may lead to
the destruction of friction lining. During the engagement of the clutch, there is relative angular motion between the driving surfaces and the driven surfaces, and hence part of the power is lost
in overcoming the friction between these surfaces. The power lost in overcoming the friction is converted into heat, and is absorbed by the clutch assembly. This results in rise in temperature of
the clutch assembly The rate of heat generation (i.e. power lost )during the engagement of the clutch is given by Hg = T θR Where, Hg = rate of heat generation, T = torque transmitted, θR =
instantaneous relative angular velocity between the driving and driven surfaces = θ1 - θ2 = ω1 – ω2 Where, ω1 = angular velocity of driving surface and ω2 = angular velocity of driven surface The
total heat generated (i.e. total energy lost) during the single engagement of the clutch is given by, E=
Or E= T θR
Where, E= total heat generated (or total energy lost) θR= total relative angular displacement between the driving and the driven surfaces during the engagement of clutch = θ1- θ2 θ1 = total angular
displacement of driving surface during the engagement θ2 = total angular displacement of the driven surface the engagement This heat generated is absorbed by the clutch assembly E=m C ΔT The
temperature rise of the clutch assembly is given by, ΔT = 212
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Where , ΔT = temperature rise of the clutch assembly, m = mass of clutch assembly and C = specific heat of clutch assembly
Temperature Time curve:
Initially, the temperature is constant. Then there is a sudden increase in temperature to temperature T1. Afterward, there is a gradual decrease in temperature and it reaches point D. Then, there is
a sudden increase in temperature to T2. Then, there is again gradual decrease in temperature with time. Design Parameters: 1) Single Plate Clutch: 1) Power, 2) Speed, 3) Friction Plate dimensions
with material (Radius and width) 4) Shaft dimensions and material 5) Thermal Analysis (Temperature of clutch) 2) Multi Plate Clutch: 1) Power, 2) Speed, 3) No. of Frictional Surfaces, 4) No. of
Driving and Driven Plates 5) Friction Plate dimension with material (Radius and width), 6) Shaft dimensions and material 7) Thermal Analysis (Temperature of Clutch) 3) Cone Clutch: 1) Power, 2)
Speed, 3) Frictional Plate dimensions (Radius and width) 4) Semi-cone Angle, 5) Shaft dimensions, 6) Thermal Analysis Friction Materials: 213
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
The desirable properties of a good friction material are as follows:. 1) High coefficient of friction. 2) Good thermal conductivity. 3) Remains unaffected by environmental conditions like moisture or
dirt particles. 4) High resistance to abrasive and adhesive wear. 5) Good resilience to provide good distribution of pressure. There are 2 types of friction materials in common use1) Asbestos based:
There are 2 types of asbestos friction materials- woven and moulded. A woven asbestos friction disk consists of; asbestos fibre woven around brass, copper or zinc wires and impregnated with rubber or
asphalt. Moulded asbestos friction disk are prepared from the wet mixture of brass chips and asbestos, which is poured into the mould and given the shape of disk. 2) Sintered Metals: There are 2
varieties of friction disks made from sintered metals- bronze base and iron –base, depending upon the major constituent. The advantages of sintered-metal friction metal friction disks are as follows:
1) Higher wear resistance 2) Can be used at high temperature. 3) Coefficient of friction is constant over a wide range of temperature and pressure. 4) Unaffected by environmental conditions. The
values of coefficient of friction for different combinations are given in following table: Contacting Surfaces Woven asbestos-cast iron Moulded asbestos-cast iron Bronze-base sintered metal-cast iron
Wet 0.1-0.2 0.08-0.12 0.05-0.1
Bronze-base sintered metal-steel
Coefficient of Friction Dry 0.3-0.6 0.2-0.5 0.1-0.4 0.1-0.3
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
NUMERICAL: For the following specifications, power =15 kW , speed =2800 RPM, Engagement per hour = 40 1. Design a single plate clutch (Dry running) 2. Design a multi plate clutch (Wet running)
Solution: Given data : Power P = 15KW , Speed = 2800 rpm , Engagement per hour = 40
1. Single Plate Clutch 1. Design power [P] = (P) *Kw, where Kw = k1 + k2 + k3 + k4 Where k1 = Driver dynamic characteristic factor K2 = Driven dynamic characteristic factor K3 = Wear factor k4=
Frequency of operation factor Kw = Factor based on working conditions.. Let kw = 2 [P] = 30KW Layout of clutch design Single plate clutch parts are: 1. Input shaft 2. Flywheel 3. Pressure plate 4.
Output shaft (spline) 5. Friction plate Torque transmission capacity: T = 𝑃∗60/ 2𝜋𝑁 = 30∗1000∗60/2𝜋∗2800 = 102.31 N-m = 102.31*103 N-m Input shaft and output shaft Material = C-35, [𝜏] = 30N/mm2 Ds1
= 1.1(16∗𝑀𝑡/𝜋∗𝜏3) ^ (1/3) = 30 mm Ds2 = 30 mm 215
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Friction plate Material = C-30 Friction material = feredo lining 𝜇 = 0.3 Allowable pressure Pa = K* Pb …………..PSG 7.89 Pb = 0.25 to 0.3N/mm2 (dry clutch)……….PSG 7.89 Vmax = 𝜋∗30∗2800/60∗1000 = 4.389 m
/s K = 0.84 …………………………… (Graph)PSG 7.90 P a = 0.84*0.25 = 0.21 N/mm2 Principle dimensions: Rmin =2 Ds = 60 mm C = Plate Thickness + 2 * Lining Thickness Let C = 5 mm No of Friction Surfaces: i = 2
(Dry Clutch) i = (𝑀)/ (2𝜋∗𝑃∗𝑏∗𝜇∗𝑅2)……………PSG 7.90 2= (102.31∗1000∗4)/ (2𝜋∗0.21∗ (𝑅𝑚𝑎𝑥−60) ∗0.3∗ (𝑅𝑚𝑎𝑥+60) ∗ (𝑅𝑚𝑎𝑥+60)) Rmax = 84.69 mm, let Rmax = 86 mm b = Rmax - Rmin = 86-60 =16 mm Thermal Check:
Assumptions: 1) Heat transfer under steady state condition.
2) Heat transfer under convection and radiation. 3) Thermal Check under static condition. Heat Generated: Hg = ½ * [Mt] * 𝜃𝑠 𝑛 Let 𝜃𝑠 = 4 π radians Slip time = 0.08 sec, Let ts = 0.08 sec 60 sec :
2800rpm , 0.08 sec : 3.733 rev , 𝜃𝑠 = 8𝜋 radian No. of operations per minute 𝑛= 40/60 = 0.667 Hg = ½ [102.31] * 8 π * 0.667 = 857.54 Watt Heat rejected (dissipation) Hd = Hc + Hr ……………………………………..
(Convection and Radiation) Hc = C A (Th – Tc) Let Ta = 303 K , We have, C = 7.1 (Vm)^0.78 216
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Vm = mean velocity = 𝜋∗𝐷𝑚𝑒𝑎𝑛∗𝑁/60 = 𝜋∗146∗2800/60∗1000 = 21.405 m/s C = 7.1 (21.405) ^0.78 = 77.46 Watt/m2 0C
To find Area, A = 2*π/4 *Dmax2 + π*Dmax*B B = 20+5+10 = 35 mm A = 65.38*1000 mm2 Hr = C1 A (Th4 – Ta4) C1 = 5*10-8 Watt/m2K4 Heat generated = Heat dissipated, Hg = Hd 857.54 = 77.4*65.38*1000 (Th -
303)*10-6 +5*10-8*65.38*103(Th4 - 3034)*10-6 Th = 451.05 K = 178.05 0C…………………..… (Temperature of Clutch) 2. MUTIPLE PLATE CLUTCH Layout of Multi Plate Clutch is as shown in figure: Multiple plate
clutch parts: 1. Input shaft 2. Housing 3. Output shaft 4. Driving plates 5. Driven plates
Material for plate C-30 , μ = 0.3 Pallowable = K*Pb Dmin = 2*Ds = 60 mm, Rmax = (1.25 to 1.8) Rmin Let Rmax = 1.25*60 = 75 mm, Dmax = 150 mm Vmax = 𝜋∗150∗2800/60 = 21.99 m/s, K =0.5 ……………………PSG 7.90
Pallowable = 0.5*0.25 = 0.125 N/mm2 No. of Friction Surfaces: i = [𝑀𝑡]/2∗𝜋∗𝜇∗𝑅𝑚2∗𝑏∗𝑃𝑎 …………………………………….......PSG7.89 Rb = Rmax – Rmax = 15 mm Rm = 67.5 mm i = 102.31∗1000/2𝜋∗0.3∗0.015∗0.125∗0.06752 i=
5.29, let i= 6 217
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Plates: No. of driven plates: M2 = i/2 -1 = 6/2 -1 = 4 No. of driving plates M1 = i/2 = 3
……………….PSG 7.90 ……………… PSG 7.89
Friction plate:
Rmin = 60 mm, Rmax = 75 mm b = 15 mm, Let t = 6 mm Dmax = D + 2t = 150 + 12 = 162 mm B = C1m1 +C2m2 + t + margin = 1*4 + 5*3 + 6 + 5 = 30 mm Thermal check: Assumptions: 1) Heat transfer under steady
state condition. 2) Heat transfer under convection and radiation. 3) Thermal Check under static condition. Heat Generated: Hg = ½ * [Mt] * 𝜃𝑠 𝑛 Let 𝜃𝑠 = 4 π radians Slip time = 0.08 sec, Let ts =
0.08 sec 60 sec : 2800rpm 0.08 sec: 3.733 rev 𝜃𝑠 = 8𝜋 radian No. of operations per minute 𝑛=4060 = 0.667 Hg = ½ [102.31] * 8 π * 0.667 = 857.54 Watt Heat rejected (dissipation) 218
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
Hd = Hc + Hr ……………………………………. (Convection and Radiation) Hc = C A (Th – Tc) Let Ta = 303 K We have, C = 7.1 (Vm)^0.78 Vm = mean velocity = 𝜋∗𝐷𝑚𝑒𝑎𝑛∗𝑁/60 = 𝜋∗150∗2800/60∗1000 = 21.99 m/s C = 7.1 (21.99)
^0.78 = 79.10 Watt/m2 0C To find Area, A = 6*π/4 *Dmax2 + π*Dmax*B, B = 30 mm A = 6*π/4*1502 +π*150*30 = 204.988*103 mm2 Heat generated = Heat dissipated Hg = Hd 857.54= 79.10*204.988*1000(Th - 303)
*10-6 +5*10-8*204.988*103(Th4 - 3034)*10-6 Th = 351.559 K = 78 0C………………………..… (Temperature of Clutch) …………………………………………………………………………………………….......... Numerical 2) Show that for maximum torque
transmission, RI/R0 = 0.577 Solution:, Let R0 = outer radii of friction surface. Ri = inner radii of friction surface. Permissible intensity of pressure = Pmax According to uniform wear theory, the
torque transmitting capacity of plate clutch is given by,
Substituting equation 2 in equation 1,
………… (3) Substituting
in equation (3) ,
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
For maximum torque transmitting capacity,
Therefore, ,
= 0.577 ,
The variation of torque characteristics against d/D is shown in below figure:
……………………………………………………………………………………………………… Numerical 3) Determine the dimension of cone clutch and the axial force required to disengage and engage the clutch for the following specification. I) Power
to be transmitted = 25kW, (ii) speed = 1440 rpm, (iii) maximum pressure = 0.28 N/mm2, (iv) semi - cone angle = 12.5 degree, (v) outer diameter = 250 mm, (vi) Co – efficient of friction = 0.25.
Solution: Given: P = 25kw, N = 1440 rpms, Pmax = 0.28 N/mm2, α = 12.50 D0 = 250 mm , Therefore, r0 = 125 mm and µ = 0.25 To Find: (1) Dimensions,
(2) W
(3) Wd
Force required to engage the clutch, W = 2πPmax*ri*(ro – ri) = 2π*0.28*ri*(125 - ri) [Mt] =
………. (1)
= 165.78 Nm = 165.78 x103 Nmm
Design Torque: Take factor of safety (fos) = 1.25, Tα = 1.25*[Mt] = 1.25*165.78*103 = 207.225*103 220
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai
Machine Design - II / MEC701 / Semester VII / Mechanical Engineering / Mumbai University.
For Cone clutch, Tα =
, Where, R =
207.225*103 =
ri*(125 – ri)*(125 + ri) = 203953.417 (ri)3 * - (125)2 * ri = -203953.4172 ,
ri = 117.87 mm ≈ 118 mm
From (1), W = 2π*0.28*118*(125 – 118)= 1453.175 N =
, b = 32.34 mm
Force required to disengage , Wd = (1 + Wd = (1 +
= 3.09 kN
Answers: (1) Dimensions ,
ri = 118 mm, ro = 125 mm, b = 32.34 mm
(2) Force required to engage, W = 1.453 kN (3) Force required to disengage, Wd = 3.09 kN
Prepared by: Prof. Sanjay W. Rukhande /FCRIT, Vashi, Navi Mumbai | {"url":"https://doku.pub/documents/md2pdf-yl4z4jm5j7qr","timestamp":"2024-11-12T09:05:15Z","content_type":"text/html","content_length":"317483","record_id":"<urn:uuid:78798292-174e-4cf0-9147-2394548014a0>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00673.warc.gz"} |