Physics Quickie: Consistency of Measure

Sometimes maintaining consistent measure in a physical equation is the most important feature determining the final result. Many people get confused as you start to throw mass, length, and time together to create what are called derived units. For instance, area is a function of length and you achieve area by multiplying two lengths together:

10m * 10m = 100m^2

The unit measures on each side of the equation must remain equal. We have two length units on the left side and so we need two on the right side. How confusing can that be when your term is multiplied, but your unit remains? Well, look at the unit as if it were a variable in algebra. What would happen if you multiplied:

5a * 4a = 20a^2

a = 4
(5)(4) * (4*4) = 20 * 16 = 320
(20)(4^2) = 20 * 16 = 320

You see how our scalar values (called coefficients) can be combined, while the variable can't? It has to maintain consistency until the point where it is replaced by an actual value. In physical equations things like meters and feet have no universal meaning. You don't replace meters with some fixed value, nor do you replace seconds. You can operate a conversion, but then the unit simply changes form, and is still a unit.

With that in mind you can start to explain the distance equation a bit better. The confusing term for most people is in squaring the time within the equation. I was surprised how often this is confused after reading a few quick help guides on the Internet that were simply wrong in the way they discussed the workings of the time squaring term. The final result of the distance equation allows for the squaring the the time term to cancel out the square seconds of the acceleration term. The only unit of measure left after this cancellation is the length unit of the acceleration.

d = 0.5at^2

d = m
t = s
a = m/s^2

??m = (0.5)(9.8m)(??s^2) / s^2
??m = (0.5)(9.8m)(??^2)

When working a physical equation you can disregard the units entirely. They aren't needed for the computation to get a resulting value. What they are important for is consistency checking. By making sure the equation we are using is balanced in terms of units we can ensure the final result is going to carry the unit we are looking for. This not only finds situations where you've grossly mis-matched units that measure different properties, but also when you've mismatched units of different measure. Take the following:

??ft = (0.5)(9.8m)(??s^2) / s^2

When the equation is checked for balance we have feet on one side and meters on the other. When we mix the result in meters, but give it the unit feet, we are coming up with an inaccurate measure. In fact, we'd be over 300% off target. Imagine how badly that could effect your next bungie jumping excursion (and yes there have been gross miscalculations of this type before where units weren't properly balanced and people have died). Didn't we almost lose or actually lose one of our planet exploring satellites to a unit error? I'll have to look that up.

The great thing about consistency of measure is that as long as we keep the units consistent we can mix and match. In fact, I could easily replace the gravitational pull of 9.8m/s^2 with the same value in feet 32.17ft/s^2 (acc?) and get a valid result in feet. With that in mind, if you head back to the bouncing ball sample, you can find yet another short-cut that allows you to get rid of the computations in meters, and allows you to turn them into computations in pixels.

a = 30px/s^2

Pixels are a unit of length and measure, and so you can replace meters or feet with pixels in the acceleration constant. Now all of your resulting calculations will be in screen pixels. Because pixels represent a virtual unit, your conversion between real units and pixels are always going to be variable. The conversion factor can be whatever you'd like. This is unlike meters and feet where the conversion factors between them are fixed values.

Bonus: Can anyone draw a relationship between the distance formula of acceleration and the geometric formulas of a triangle?

Published Thursday, September 02, 2004 12:57 PM by Justin Rogers
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Comments

Thursday, September 02, 2004 4:32 PM by Addy Santo

# re: Physics Quickie: Consistency of Measure

None of your posts from the last few months are even remotely .Net related. Do us all a favor and set the Posts->Advanced Options->Display on Homepage to 'NO' for future posts of this type... Thanks!
Thursday, September 02, 2004 6:59 PM by Justin Rogers

# re: Physics Quickie: Consistency of Measure

Not only is your statement highly inflammatory, but it is desperately wrong. Looking at my 20 posts for the month of August, I see that about half of them are .NET related... I'd definitely say that isn't bad.

If you don't like the math, then don't read my posts. I'm a game developer. I develop games. It takes a lot of math and physics to do that and so that is what I'm going to blog about.
Friday, September 03, 2004 4:28 AM by Barry Kelly

# re: Physics Quickie: Consistency of Measure

> Can anyone draw a relationship between the
> distance formula of acceleration and the
> geometric formulas of a triangle?

If you graph velocity (Y-axis) against time (X-axis), acceleration looks like an upward slope, constant velocity looks like a flat line, and deceleration looks like a downward slope.

The funny thing about a graph drawn in this way is that the area under the curve (i.e. the line indicating velocity at that moment) is equal to the distance travelled (or, if you consider area under the X-axis to be "negative area", then the total area is equal to the displacement.)

The reason the area under the curve is equal to the distance travelled has to do with calculus. The area under any curve is the integral of the equation of the curve. When you consider the relationship between velocity and displacement, it's obvious that the rate of change of displacement is velocity. Mathematically speaking:

s: displacement
v: velocity

ds/dt = v

Since the differential of the displacement with respect to time is the velocity, the integral of the velocity is going to be the reverse - the displacement.

Considering your equation d = 0.5at^2 (although when I learned that in school, we called it 's = ut + 1/2 at^2'), you're starting from the speed of zero (i.e. you have no 'ut' term since initial velocity, u, is zero).

This means that if you draw a simple constant acceleration from 0, it looks like a triangle, and the total distance travelled is the area under the triangle. The time taken is t (measured along the X-axis) and the final velocity is v (but is equal to a*t since the acceleration is constant). This means the total area under the triangle is 1/2 a*t*t, or, in other words, d = 0.5at^2.
Friday, September 03, 2004 4:46 AM by Justin Rogers

# re: Physics Quickie: Consistency of Measure

Very thorough post. To draw direct correlations the area of a triangle is:

b = base
h = height
area = 0.5bh

b = t
h = a*t
area = 0.5at^2

By using basic substitution we go from the classical equation for area of a triangle of 0.5bh to the equation of distance by acceleration of 0.5at^2.

Glad to see someone took the time to solve the bonus problem. Thanks Barry!
Friday, September 03, 2004 11:28 AM by brianbec

# re: Physics Quickie: Consistency of Measure

It would be very interesting to create "dimension-sensitive" math libraries for physics that would use types and coersions to make units errors impossible. In pseudocode, instead of writing "42.6", for instance, you'd have to write something like "Cm(42.6)", and, if a function expected feet, such as

Seconds timeRequired (Feet distance, Feet/Second velocity)
{
if (velocity != Feet(0.0)/Second))
return distance / velocity;
}

and you called it like this

timeRequired (Cm(42.6), Meters(60.2)/Second);

the appropriate transformations would be applied.

The tricky part, using C# or C++ types, would be getting the compound units like Feet/Second right in a generic way. I would know how to do something like this in Scheme or Haskell, which latter has a very elaborate and interesting type system. Nice inspiriration. http://weblogs.asp.net/brianbec
Friday, September 03, 2004 5:02 PM by Justin Rogers

# re: Physics Quickie: Consistency of Measure

Interesting way to write your velocity I have to say. Meters is given the property of being a dimensional unit, while seconds is given the property of being dimensionless... I think a rational approach might be the very verbose:

Term(Coeffficient, Units[])

timeRequired( new Term(42.6, new Units[] { Cm }), new Term(60.2, new Units[] { Meters, Neg1Seconds })

Of course you start to be very verbose in this system... The units themselves carry the unit type, and the unit power. Division is units with negative power. Examining the units would have to be more functionally driven as well.

distance.EnsureUnits( Feet );
velocity.EnsureUnits( new Unit[] { Feet, Neg1Seconds });

Seems like a dead ringer for a dynamic language though. If you make any progress on making this work in C#, let me know. I'm very curious about the performance characteristics and would love to look at some optimization schemes.
Friday, September 03, 2004 5:25 PM by brianbec

# re: Physics Quickie: Consistency of Measure

looks like Andrew Kennedy of Microsoft Research is a specialist in this field. I'm looking into it ...

http://research.microsoft.com/~akenn/units/

Monday, September 06, 2004 3:57 PM by George V. Reilly

# re: Physics Quickie: Consistency of Measure

There's a bunch of work going on in the C++ Boost world to handle dimensional analysis: see http://www.artima.com/cppsource/metafunctions.html for an introduction.
Thursday, September 08, 2011 8:43 PM by yaco

# re: Physics Quickie: Consistency of Measure

Pienso que no sois derecho. Soy seguro. Puedo demostrarlo.      

http://www.shampes.com/      

darek

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