# Units And Equations

Discussion in 'Physics & Math' started by Saith, Jan 28, 2005.

1. ### SaithRegistered Senior Member

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149
How do you figure out what units to use in an equation? I come across tons of equations, but rarely do they ever tell you want units to use.

Does anyone know of any websites with a comprehensive list of equations and the units they use? I want to make a calculator that deals with equations, but Im having a hard time finding information.

3. ### James RJust this guy, you know?Staff Member

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37,747
Most equations are unit-free. For example, take the equation

w = mg

for the weight w of a body of mass m in gravitational field of strength g.

If you choose to use SI units, then putting m in kilograms and g in metres per second squared gives you and answer for w in Newton. If, on the other hand, you use m in slugs, g in feet per second squared, then you get an answer in pounds (if I've got the imperial units right!)

More important in physics is that equations are dimensionally correct. For example, the quantity w in the above equation has the dimensions of force, and so does the combination of m multipled by g, regardless of the system of units used. By dimension, in this context, we mean that whatever units you use, w will always reduce to a mass multiplied by a length squared, divided by a time squared. This is often written as [M][L]<sup>2</sup>[T]<sup>-2</sup>.

The important point is that for any physical equation to be correct, at the very least the dimensions on the left of the equals sign must be the same as the dimensions on the right. This is one way to quickly check if a given equation is at least plausible.

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Huh? That equation example you gave us (w = mg) has units. Since you're using the SI system, the units for w is the newton. The unit of mass is the kilogram, and gravity invokes the meter as a unit.

7. ### QuarkHeadRemedial Math StudentValued Senior Member

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1,735
Which is precisely what he said.
But its a long time since we used slugs as a unit of measurement. Ah, the good old days.....

8. ### SaithRegistered Senior Member

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149
So if I was going to use the equation E=mc² to find out how many kilojoules are in 165 pounds of mass, how would I figure out what kind of units to use for m?

Last edited: Mar 2, 2005
9. ### Janus58Valued Senior Member

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2,318
For joules you use kilograms for mass and meters/sec for c

So you would need to convert pounds to kg and divide your answer by 1000 to convert joules to kilojoules.

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http://www.ex.ac.uk/cimt/dictunit/dictunit.htm
To use the equation, you need to know what the units represent. Since we're talking about the SI system, the E represents energy in joules, the m is mass in kilograms, and the letter c is a constant for the speed of light in metres per second. That means that one kilogram of mass converted into energy gives you 90,000,000,000,000,000 ([300 million] squared or 90 quadrillion or 90 x 10<Sup>15</Sup>) joules.

11. ### SaithRegistered Senior Member

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149
This is a little off topic of my original post, but while I was looking for information on the MKS/SI, CGS and FPS unit systems I came across this this website: http://www.blazelabs.com/f-u-suconv.asp

The author of the article says that the 7 base units in the SI system aren't independent. How I understand the article, he says that a unit system can be based off of only 2 units. There is more than one combination, but he chose space and time. There is a table that shows the space/time equivalent of the equations/units. I don't understand it. I understand that S3 is volume, but how can you get mass from T3*S-3?

If anyone would like to explain it to me, or point me in the right direction, I would also like to know how the candela, mole, Amp and Kelvin can be derived from the other base units. This is what the article says about them:

"If for instance, one had to change the definition of the Kg unit, we see that the fundamental units candela, mole, Amp and Kelvin would change as well."

CANDELA
"The candela is an old photometric unit which can easily be derived from radiometric units (radiated power in Watts) by multiplying it by a function to describe the optical response of the human eye. The candela unit, together with its derived units as lux and foot-candelas serve no purpose that is not served equally well by watt per steradian and its derivatives." *The steradian of what and what is its derivatives that are the same as the lux, etc...?

KELVIN
"Hence the Kelvin could also be defined as a derived unit, equivalent to 1.3806E-23Joule per degree of freedom, having the same dimensions of energy." *I don't really understand "degree of freedom", but Im guessing that you have to divide energy by that number.

I don't think it mentions the ampere again and all it says about the mole (along with the candela) is that it's redundant and unneeded.