Calculate Root Mean Square Velocity of Gas Particles
This example problem demonstrates how to calculate the root mean square (RMS) velocity of particles in an ideal gas. This value is the square root of the average velocitysquared of molecules in a gas. While the value is an approximation, especially for real gases, it offers useful information when studying kinetic theory.
Root Mean Square Velocity Problem
What is the average velocity or root mean square velocity of a molecule in a sample of oxygen at 0 degrees Celsius?
Solution
Gases consist of atoms or molecules that move at different speeds in random directions. The root mean square velocity (RMS velocity) is a way to find a single velocity value for the particles. The average velocity of gas particles is found using the root mean square velocity formula:
μ_{rms} = (3RT/M)^{½}
μ_{rms} = root mean square velocity in m/sec
R = ideal gas constant = 8.3145 (kg·m^{2}/sec^{2})/K·mol
T = absolute temperature in Kelvin
M = mass of a mole of the gas in kilograms.
Really, the RMS calculation gives you root mean squarespeed, not velocity. This is because velocity is a vector quantity that has magnitude and direction. The RMS calculation only gives the magnitude or speed. The temperature must be converted to Kelvin and the molar mass must be found in kg to complete this problem.
Step 1
Find the absolute temperature using the Celsius to Kelvin conversion formula:
 T = °C + 273
 T = 0 + 273
 T = 273 K
Step 2
Find molar mass in kg:
From the periodic table, the molar mass of oxygen = 16 g/mol.
Oxygen gas (O_{2}) is comprised of two oxygen atoms bonded together. Therefore:
 molar mass of O_{2} = 2 x 16
 molar mass of O_{2} = 32 g/mol
 Convert this to kg/mol:
 molar mass of O_{2} = 32 g/mol x 1 kg/1000 g
 molar mass of O_{2} = 3.2 x 10^{2} kg/mol
Step 3
Find μ_{rms}:
 μ_{rms} = (3RT/M)^{½}
 μ_{rms} = [3(8.3145 (kg·m^{2}/sec^{2})/K·mol)(273 K)/3.2 x 10^{2} kg/mol]^{½}
 μ_{rms} = (2.128 x 10^{5} m^{2}/sec^{2})^{½}
 μ_{rms} = 461 m/sec
Answer
The average velocity or root mean square velocity of a molecule in a sample of oxygen at 0 degrees Celcius is 461 m/sec.
9.15: Kinetic Theory of Gases Molecular Speeds
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Other sections state that increasing the temperature increases the speeds at which molecules move. We are now in a position to find just how large that increase is for a gaseous substance. Combining the ideal gas law with Eq. (1) from The Total Molecular Kinetic Energy, we obtain
\(\begin{align} & PV=nRT=\tfrac{\text{1}}{\text{3}}Nm\text{(}u^{\text{2}}\text{)}_{\text{ave}}\\ & \text{or } \text{3}RT=\frac{Nm}{n}\text{(}u^{\text{2}}\text{)}_{\text{ave}} \label{1}\end{align}\)
Since N is the number of molecules and m is the mass of each molecule, Nm is the total mass of gas. Dividing total mass by amount of substance gives molar mass M: \[M=\frac{Nm}{n}\] Substituting in Eq. \(\ref{1}\), we have \(\begin{align} & \text{ 3}RT=M(u^{\text{2}})_{\text{ave}} \\ & \text{or }(u^{\text{2}})_{\text{ave}}=\frac{\text{3}RT}{M} \\ & \text{so that }u_{rms}=\sqrt{\text{(}u^{\text{2}}\text{)}_{\text{ave}}}=\sqrt{\frac{\text{3}RT}{M}}\text{ (2)} \end{align}\) The quantity u_{rms} is called the rootmeansquare (rms) velocity because it is the square root of the mean square velocity.The rms velocity is directly proportional to the square root of temperature and inversely proportional to the square root of molar mass. Thus quadrupling the temperature of a given gas doubles the rms velocity of the molecules. Doubling this average velocity doubles the number of collisions between gas molecules and the walls of a container. It also doubles the impulse of each collision. Thus the pressure quadruples. This is indicated graphically in Figure \(\PageIndex{1}\). Pressure is thus directly proportional to temperature, as required by GayLussac’s law.
The inverse proportionality between rootmeansquare velocity and the square root of molar mass means that the heavier a molecule is, the slower it moves, which is verified by the examples below
We can compare the rates of effusion or diffusion of a known gas with that of an unknown gas to determine the molar mass of the unknown gas. A convenient equation can be derived easily by considering the kinetic energy of individual molecules rather than moles of gas:
Knowing that kinetic energy is proportional to temperature, if the two gases are at the same temperature,
 \(\text{K} \text{E}_{1} = \text{K} \text{E}_{2} \) where 1 and 2 denote the two gases. Since \(KE= \frac{1}{2} m v^{2}\),
 \( \frac{1}{2} m_{1} ( u_{\text{rms, 1}} )^{2} = \frac{1}{2} m_{2} ( u_{\text{rms, 2}} )^{2}\) where m is the atomic weight in amu/average molecule, and u_{rms} is the velocity.
Dividing,
\[\frac{m_{1}}{m_{2}} = \frac{ ( u_{rms,2} )^{2} }{u_{rms,1} )^{2} } \]
Example \(\PageIndex{1}\) : Molar Mass
What is the molar mass of an unknown gas if the gas effuses through a pinhole into a vacuum at a rate of 2 mL/min, and H_{2} effuses at 11 mL/min. Assume that the rate of effusion is proportional to the gas molecule velocities.
Solution
\[\frac{m_{1}}{m_{2}} = \frac{ ( u_{rms,2} )^{2} }{u_{rms,1} )^{2} } \\ \frac{4}{m_{2}} = \frac{2^{2}}{11^{2}} \\ m_{2} = 121 \]
Example \(\PageIndex{2}\) : RMS Velocity
Find the rms velocity for (a) H_{2} and (b) O_{2} molecules at 27°C.
Solution This problem is much easier to solve if we use SI units. Thus we choose
R = 8.314 J mol^{–1} K^{–1} = 8.314 kg m^{2}s^{–2} mol^{–1} K^{–1}
a)For H_{2}\(\begin{align}u_{\text{rms}}=\sqrt{\frac{\text{3}RT}{M}} & =\sqrt{\frac{\text{3 }\times \text{ 8}\text{.314 J mol}^{\text{1}}\text{ K}^{\text{1}}\text{ }\times \text{ 300 K}}{\text{2}\text{.016 g mol}^{\text{1}}}}\\ & =\sqrt{\text{3}\text{.712 }\times \text{ 10}^{\text{3}}\text{ }\frac{\text{kg m}^{\text{2}}\text{s}^{\text{2}}}{\text{g}}}\\ & =\sqrt{\text{3}\text{.712 }\times \text{ 10}^{\text{3}}\text{ }\times \text{ 10}^{\text{3}}\text{ }\frac{\text{g m}^{\text{2}}\text{s}^{\text{2}}}{\text{g}}}\\ & =\sqrt{\text{3}\text{.712 }}\times \text{ 10}^{\text{3}}\text{ m s}^{\text{1}}=\text{1}\text{.927 }\times \text{ 10}^{\text{3}}\text{ m s}^{\text{1}}\end{align}\) b)For O_{2}\[u_{\text{rms}}=\sqrt{\frac{\text{3 }\times \text{ 8}\text{.314 J mol}^{\text{1}}\text{ K}^{\text{1}}\text{ }\times \text{ 300 K}}{\text{32}\text{.00 g mol}^{\text{1}}}}=\text{4}\text{.836 }\times \text{ 10}^{\text{2}}\text{ m s}^{\text{1}}\]Now we can see the microscopic basis for Avogadro’s law. Most of the volume in H_{2}, O_{2} or any gas is empty space, and that empty space is the same for a given amount of any gas at the same temperature and pressure. This happens because the total kinetic energy of the molecules is the same for H_{2} or O_{2} or any other gas. The more energy they have, the more room the molecules can make for themselves by expanding against a constant pressure. This is illustrated in Figure \(\PageIndex{2}\), where equal numbers of H_{2} and O_{2} molecules occupy separate containers at the same temperature and pressure.
The volumes are seen to be the same. Because O_{2} molecules are 16 times heavier than H_{2} molecules, the average speed of H_{2} molecules is 4 times faster. H_{2} molecules therefore make 4 times as many collisions with walls. Based on mass, each collision of an H_{2} molecule with the wall has onesixteenth the effect of an O_{2} collision, but an H_{2} collision has 4 times the effect of an O_{2} collision when molecular velocity is considered. The net result is that each H_{2} collision is only onefourth as effective as an O_{2} collision. But since there are four times as many collisions, each onefourth as effective, the same pressure results. Thus the same number of O_{2} molecules as H_{2} molecules is required to occupy the same volume at the same temperature and pressure.
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Root mean square speed
Root mean square speed (v_{rms})
Root mean square speed (v_{rms}) is defined as the square root of the mean of the square of speeds of all molecules. It is denoted by v_{rms} = √v^{2}
Equation (9.8) can be rewritten as,
From the equation (9.18) we infer the following
(i) rms speed is directly proportional to square root of the temperature and inversely proportional to square root of mass of the molecule. At a given temperature the molecules of lighter mass move faster on an average than the molecules with heavier masses.
Example: Lighter molecules like hydrogen and helium have high ‘v_{rms}’ than heavier molecules such as oxygen and nitrogen at the same temperature.
(ii) Increasing the temperature will increase the r.m.s speed of molecules
We can also write the v_{rms} in terms of gas constant R. Equation (9.18) can be rewritten as follows
Where N_{A} is Avogadro number.
Since N_{A}k = R and N_{A}m = M (molar mass)
The root mean square speed or r.m.s speed
The equation (9.6) can also be written in terms of rms speed
Impact of v_{rms} in nature:
1. Moon has no atmosphere.
The escape speed of gases on the surface of Moon is much less than the root mean square speeds of gases due to low gravity. Due to this all the gases escape from the surface of the Moon.
2. No hydrogen in Earth’s atmosphere.
As the root mean square speed of hydrogen is much less than that of nitrogen, it easily escapes from the earth’s atmosphere.
In fact, the presence of nonreactive nitrogen instead of highly combustible hydrogen deters many disastrous consequences.
EXAMPLE 9.2
A room contains oxygen and hydrogen molecules in the ratio 3:1. The temperature of the room is 27°C. The molar mass of 0_{2} is 32 g mol1 and for H_{2} 2 g mol^{1}. The value of gas constant R is 8.32 J mol^{1}K^{1}
Calculate
(a) rms speed of oxygen and hydrogen molecule
(b) Average kinetic energy per oxygen molecule and per hydrogen molecule
(c) Ratio of average kinetic energy of oxygen molecules and hydrogen molecules
Solution
(a) Absolute Temperature
T=27°C =27+273=300 K.
Gas constant R=8.32 J mol^{1}k^{1}
For Oxygen molecule: Molar mass
M=32 gm=32 x 10^{3} kg mol^{1}
Note that the rms speed is inversely proportional to √M and the molar mass of oxygen is 16 times higher than molar mass of hydrogen. It implies that the rms speed of hydrogen is 4 times greater than rms speed of oxygen at the same temperature.
1934/484 ≈ 4 .
(b) The average kinetic energy per molecule is 3/2 kT. It depends only on absolute temperature of the gas and is independent of the nature of molecules. Since both the gas molecules are at the same temperature, they have the same average kinetic energy per molecule. k is Boltzmaan constant.
(c) Average kinetic energy of total oxygen molecules = 3/2 N_{0}kT where N_{0}  number of oxygen molecules in the room
Average kinetic energy of total hydrogen molecules = 3/2 N_{H}kT where N_{H}  number of hydrogen molecules in the room.
It is given that the number of oxygen molecules is 3 times more than number of hydrogen molecules in the room. So the ratio of average kinetic energy of oxygen molecules with average kinetic energy of hydrogen molecules is 3:1
Learning Objective
 Recall the mathematical formulation of the rootmeansquare velocity for a gas.
Key Points
 All gas particles move with random speed and direction.
 Solving for the average velocity of gas particles gives us the average velocity of zero, assuming that all particles are moving equally in different directions.
 You can acquire the average speed of gaseous particles by taking the root of the square of the average velocities.
 The rootmeansquare speed takes into account both molecular weight and temperature, two factors that directly affect a material’s kinetic energy.
Term
 velocitya vector quantity that denotes the rate of change of position, with respect to time or a speed with a directional component
Kinetic Molecular Theory and RootMeanSquare Speed
According to Kinetic Molecular Theory, gaseous particles are in a state of constant random motion; individual particles move at different speeds, constantly colliding and changing directions. We use velocity to describe the movement of gas particles, thereby taking into account both speed and direction.
Although the velocity of gaseous particles is constantly changing, the distribution of velocities does not change. We cannot gauge the velocity of each individual particle, so we often reason in terms of the particles’ average behavior. Particles moving in opposite directions have velocities of opposite signs. Since a gas’ particles are in random motion, it is plausible that there will be about as many moving in one direction as in the opposite direction, meaning that the average velocity for a collection of gas particles equals zero; as this value is unhelpful, the average of velocities can be determined using an alternative method.
By squaring the velocities and taking the square root, we overcome the “directional” component of velocity and simultaneously acquire the particles’ average velocity. Since the value excludes the particles’ direction, we now refer to the value as the average speed. The rootmeansquare speed is the measure of the speed of particles in a gas, defined as the square root of the average velocitysquared of the molecules in a gas.
It is represented by the equation: [latex]v_{rms}=\sqrt{\frac{3RT}{M}}[/latex], where v_{rms} is the rootmeansquare of the velocity, M_{m} is the molar mass of the gas in kilograms per mole, R is the molar gas constant, and T is the temperature in Kelvin.
The rootmeansquare speed takes into account both molecular weight and temperature, two factors that directly affect the kinetic energy of a material.
Example
 What is the rootmeansquare speed for a sample of oxygen gas at 298 K?
[latex]v_{rms}=\sqrt{\frac{3RT}{M_m}}=\sqrt{\frac{3(8.3145\frac{J}{K*mol})(298\;K)}{32\times10^{3}\frac{kg}{mol}}}=482\;m/s[/latex]
Speed root mean square
Root Square Mean Velocity Example Problem
Gases are made up of individual atoms or molecules freely moving in random directions with a wide variety of speeds. Kinetic molecular theory tries to explain the properties of gases by investigating the behavior of individual atoms or molecules making up the gas. This example problem shows how to find the average or root mean square velocity (rms) of particles in a gas sample for a given temperature.
Root Mean Square Problem
What is the root mean square velocity of the molecules in a sample of oxygen gas at 0 °C and 100 °C?
Solution:
Root mean square velocity is the average velocity of the molecules that make up a gas. This value can be found using the formula:
v_{rms} = [3RT/M]^{1/2}
where
v_{rms} = average velocity or root mean square velocity
R = ideal gas constant
T = absolute temperature
M = molar mass
The first step is to convert the temperatures to absolute temperatures. In other words, convert to the Kelvin temperature scale:
K = 273 + °C
T_{1} = 273 + 0 °C = 273 K
T_{2} = 273 + 100 °C = 373 K
The second step is to find the molecular mass of the gas molecules.
Use the gas constant 8.3145 J/mol·K to get the units we need. Remember 1 J = 1 kg·m^{2}/s^{2}. Substitute these units into the gas constant:
R = 8.3145 kg·m^{2}/s^{2}/K·mol
Oxygen gas is made up of two oxygen atoms bonded together. The molecular mass of a single oxygen atom is 16 g/mol. The molecular mass of O_{2} is 32 g/mol.
The units on R use kg, so the molar mass must also use kg.
32 g/mol x 1 kg/1000 g = 0.032 kg/mol
Use these values to find the v_{rms}.
0 °C:
v_{rms} = [3RT/M]^{1/2}
v_{rms} = [3(8.3145 kg·m^{2}/s^{2}/K·mol)(273 K)/(0.032 kg/mol)]^{1/2}
v_{rms} = [212799 m^{2}/s^{2}]^{1/2}
v_{rms} = 461.3 m/s
100 °C
v_{rms} = [3RT/M]^{1/2}
v_{rms} = [3(8.3145 kg·m^{2}/s^{2}/K·mol)(373 K)/(0.032 kg/mol)]^{1/2}
v_{rms} = [290748 m^{2}/s^{2}]^{1/2}
v_{rms} = 539.2 m/s
Answer:
The average or root mean square velocity of the oxygen gas molecules at 0 °C is 461.3 m/s and 539.2 m/s at 100 °C.
Watch Now: How to Calculate Velocity
Root mean square
Square root of the mean square
In mathematics and its applications, the root mean square (RMS or RMS or rms) is defined as the square root of the mean square (the arithmetic mean of the squares of a set of numbers).^{[1]} The RMS is also known as the quadratic mean^{[2]}^{[3]} and is a particular case of the generalized mean with exponent 2. RMS can also be defined for a continuously varying function in terms of an integral of the squares of the instantaneous values during a cycle.
For alternating electric current, RMS is equal to the value of the constant direct current that would produce the same power dissipation in a resistive load.^{[1]}
In estimation theory, the rootmeansquare deviation of an estimator is a measure of the imperfection of the fit of the estimator to the data.
Definition[edit]
The RMS value of a set of values (or a continuoustimewaveform) is the square root of the arithmetic mean of the squares of the values, or the square of the function that defines the continuous waveform. In physics, the RMS current value can also be defined as the "value of the direct current that dissipates the same power in a resistor."
In the case of a set of n values , the RMS is
The corresponding formula for a continuous function (or waveform) f(t) defined over the interval is
and the RMS for a function over all time is
The RMS over all time of a periodic function is equal to the RMS of one period of the function. The RMS value of a continuous function or signal can be approximated by taking the RMS of a sample consisting of equally spaced observations. Additionally, the RMS value of various waveforms can also be determined without calculus, as shown by Cartwright.^{[4]}
In the case of the RMS statistic of a random process, the expected value is used instead of the mean.
In common waveforms[edit]
If the waveform is a pure sine wave, the relationships between amplitudes (peaktopeak, peak) and RMS are fixed and known, as they are for any continuous periodic wave. However, this is not true for an arbitrary waveform, which may not be periodic or continuous. For a zeromean sine wave, the relationship between RMS and peaktopeak amplitude is:
 Peaktopeak
For other waveforms, the relationships are not the same as they are for sine waves. For example, for either a triangular or sawtooth wave
 Peaktopeak
Waveform  Variables and operators  RMS 

DC  
Sine wave  
Square wave  
DCshifted square wave  
Modified sine wave  
Triangle wave  
Sawtooth wave  
Pulse wave  
Phasetophase voltage  
where:

In waveform combinations[edit]
Waveforms made by summing known simple waveforms have an RMS value that is the root of the sum of squares of the component RMS values, if the component waveforms are orthogonal (that is, if the average of the product of one simple waveform with another is zero for all pairs other than a waveform times itself).^{[5]}
Alternatively, for waveforms that are perfectly positively correlated, or "in phase" with each other, their RMS values sum directly.
Uses[edit]
In electrical engineering[edit]
Voltage[edit]
Further information: Root mean square AC voltage
A special case of RMS of waveform combinations is:^{[6]}
where refers to the direct current (or average) component of the signal, and is the alternating current component of the signal.
Average electrical power[edit]
Further information: AC power
Electrical engineers often need to know the power, P, dissipated by an electrical resistance, R. It is easy to do the calculation when there is a constant current, I, through the resistance. For a load of R ohms, power is defined simply as:
However, if the current is a timevarying function, I(t), this formula must be extended to reflect the fact that the current (and thus the instantaneous power) is varying over time. If the function is periodic (such as household AC power), it is still meaningful to discuss the average power dissipated over time, which is calculated by taking the average power dissipation:
So, the RMS value, I_{RMS}, of the function I(t) is the constant current that yields the same power dissipation as the timeaveraged power dissipation of the current I(t).
Average power can also be found using the same method that in the case of a timevarying voltage, V(t), with RMS value V_{RMS},
This equation can be used for any periodic waveform, such as a sinusoidal or sawtooth waveform, allowing us to calculate the mean power delivered into a specified load.
By taking the square root of both these equations and multiplying them together, the power is found to be:
Both derivations depend on voltage and current being proportional (that is, the load, R, is purely resistive). Reactive loads (that is, loads capable of not just dissipating energy but also storing it) are discussed under the topic of AC power.
In the common case of alternating current when I(t) is a sinusoidal current, as is approximately true for mains power, the RMS value is easy to calculate from the continuous case equation above. If I_{p} is defined to be the peak current, then:
where t is time and ω is the angular frequency (ω = 2π/T, where T is the period of the wave).
Since I_{p} is a positive constant:
Using a trigonometric identity to eliminate squaring of trig function:
but since the interval is a whole number of complete cycles (per definition of RMS), the sine terms will cancel out, leaving:
A similar analysis leads to the analogous equation for sinusoidal voltage:
where I_{P} represents the peak current and V_{P} represents the peak voltage.
Because of their usefulness in carrying out power calculations, listed voltages for power outlets (for example, 120 V in the USA, or 230 V in Europe) are almost always quoted in RMS values, and not peak values. Peak values can be calculated from RMS values from the above formula, which implies V_{P} = V_{RMS} × √2, assuming the source is a pure sine wave. Thus the peak value of the mains voltage in the USA is about 120 × √2, or about 170 volts. The peaktopeak voltage, being double this, is about 340 volts. A similar calculation indicates that the peak mains voltage in Europe is about 325 volts, and the peaktopeak mains voltage, about 650 volts.
RMS quantities such as electric current are usually calculated over one cycle. However, for some purposes the RMS current over a longer period is required when calculating transmission power losses. The same principle applies, and (for example) a current of 10 amps used for 12 hours each 24hour day represents an average current of 5 amps, but an RMS current of 7.07 amps, in the long term.
The term RMS power is sometimes erroneously used in the audio industry as a synonym for mean power or average power (it is proportional to the square of the RMS voltage or RMS current in a resistive load). For a discussion of audio power measurements and their shortcomings, see Audio power.
Speed[edit]
Main article: Rootmeansquare speed
In the physics of gas molecules, the rootmeansquare speed is defined as the square root of the average squaredspeed. The RMS speed of an ideal gas is calculated using the following equation:
where R represents the gas constant, 8.314 J/(mol·K), T is the temperature of the gas in kelvins, and M is the molar mass of the gas in kilograms per mole. In physics, speed is defined as the scalar magnitude of velocity. For a stationary gas, the average speed of its molecules can be in the order of thousands of km/hr, even though the average velocity of its molecules is zero.
Error[edit]
Main article: Rootmeansquare deviation
When two data sets — one set from theoretical prediction and the other from actual measurement of some physical variable, for instance — are compared, the RMS of the pairwise differences of the two data sets can serve as a measure how far on average the error is from 0. The mean of the absolute values of the pairwise differences could be a useful measure of the variability of the differences. However, the RMS of the differences is usually the preferred measure, probably due to mathematical convention and compatibility with other formulae.
In frequency domain[edit]
The RMS can be computed in the frequency domain, using Parseval's theorem. For a sampled signal , where is the sampling period,
where and N is the sample size, that is, the number of observations in the sample and FFT coefficients.
In this case, the RMS computed in the time domain is the same as in the frequency domain:
Relationship to other statistics[edit]
If is the arithmetic mean and is the standard deviation of a population or a waveform, then:^{[8]}
From this it is clear that the RMS value is always greater than or equal to the average, in that the RMS includes the "error" / square deviation as well.
Physical scientists often use the term root mean square as a synonym for standard deviation when it can be assumed the input signal has zero mean, that is, referring to the square root of the mean squared deviation of a signal from a given baseline or fit.^{[9]}^{[10]} This is useful for electrical engineers in calculating the "AC only" RMS of a signal. Standard deviation being the RMS of a signal's variation about the mean, rather than about 0, the DC component is removed (that is, RMS(signal) = stdev(signal) if the mean signal is 0).
See also[edit]
References[edit]
 ^ ^{a}^{b}"Rootmeansquare value". A Dictionary of Physics (6 ed.). Oxford University Press. 2009. ISBN .
 ^Thompson, Sylvanus P. (1965). Calculus Made Easy. Macmillan International Higher Education. p. 185. ISBN . Retrieved 5 July 2020.
 ^Jones, Alan R. (2018). Probability, Statistics and Other Frightening Stuff. Routledge. p. 48. ISBN . Retrieved 5 July 2020.
 ^Cartwright, Kenneth V (Fall 2007). "Determining the Effective or RMS Voltage of Various Waveforms without Calculus"(PDF). Technology Interface. 8 (1): 20 pages.
 ^Nastase, Adrian S. "How to Derive the RMS Value of Pulse and Square Waveforms". MasteringElectronicsDesign.com. Retrieved 21 January 2015.
 ^"Make Better AC RMS Measurements with your Digital Multimeter"(PDF). Keysight. Keysight. Retrieved 15 January 2019.
 ^If AC = a and BC = b. OC = AM of a and b, and radius r = QO = OG.
Using Pythagoras' theorem, QC² = QO² + OC² ∴ QC = √QO² + OC² = QM.
Using Pythagoras' theorem, OC² = OG² + GC² ∴ GC = √OC² − OG² = GM.
Using similar triangles, HC/GC = GC/OC ∴ HC = GC²/OC = HM.  ^Chris C. Bissell; David A. Chapman (1992). Digital signal transmission (2nd ed.). Cambridge University Press. p. 64. ISBN .
 ^Weisstein, Eric W. "RootMeanSquare". MathWorld.
 ^"ROOT, TH1:GetRMS".
External links[edit]
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