# Dimensional Analysis - Comparison of exponents

1. Jan 5, 2009

### ajassat

I have been reading a paper which introduced classical mechanics concerning only one dimension. Before we get into anything about displacement and velocity, I read about dimensional analysis.

There is a section of the paper which isn't very well explained and I hope someone can explain it to me, so that I can continue reading.

"A film company wants to shoot the leaning tower of pisa falling. They have built a model tower. They cannot just let it drop otherwise it would look too unrealistic. Therefore they need to slow the video down so it looks more realistic by a factor t. This factor depends on mass of tower,m, the height of the tower,h, and acceleration due to gravity, g. In other words...

t = C m^x h^y g^z
Where C is a dimensionless constant.

The dimensions =

[T] = [M]^x [L]^y ([L]/[T-squared])^z

We can now compare the exponents of [L], [M] and [T].

0 = y + z
0 = x
1 = -2z

While I understand what the author is attempting to do, I do not understand how we gain the above equations, and hence the following.

x = 0
y = 0.5
z = -0.5

t = C (sqrt) h/g

ps - I want to understand fully how we compare the exponents, get three equations and the formula.

2. Jan 5, 2009

### Haths

I think what it's explaining is that the factor t is multiplied by the speed of fall, to slow it down.

M=mass
L=lenght
T=time

Therefore your end units of t*v (where v is you speed) should be units of v. Speed is distance divided by time, so it's dimensional units are L1 T-1. I assume that you understand that this is the same as writing Lenght / Time ;).

Hence:

t*v=v

so;

C Mx Ly gz * L1 T-1 = L1 T-1

Now g is not in fundamental units it's Newtons per Kilogram, but Newtons are Kilograms-Meters per second per second. So;

g => M1 L1 T-2 / M1

g => M1 L1 T-2 M-1

So

g = L1 T-2

Ok knowing that;

C Mx Ly (L1 T-2)z * L1 T-1 = L1 T-1

We can ignore C because it's not got a dimension and consider this as a set of equations. So basically we are going to count up the powers for each of the dimensions on each side of the equation. So for M, then L, then T; we have;

C Mx Ly (L1 T-2)z * L1 T-1 = L1 T-1

C Mx Ly (L1 T-2)z * L1 T-1 = L1 T-1

C Mx Ly (L1 T-2)z * L1 T-1 = L1 T-1

(1) x = 0
(2) y+z+1 = 1
(3) -2z+-1=-1

Equation 1, tells us that this factor t doesn't depend on the mass at all. Which should make sense as a tonne of feathers and a ton of iron fall at the same rate. 2 and 3 can be rewritten;

(2) y+z=0
(3) -2z=0

Errm different to what the last one is, but let's continue anyhow. These are a set of simultaneous equations, these we can solve as -2z=0, implies that z needs to be 0, but that would imply that y also equals zero, so somewhere in my adding up of the powers I've gone wrong. Ok using;

(3) -2z=1

We can see that z must be -1/2 because -2*-1/2 = 1 As we know y+z=0 and y+-1/2=0 so y=1/2

Going back to;

C Mx Ly (L1 T-2)z

We put our values of x,y and z into the powers to find out what our constant is. Like;

C M0 L1/2 (L1 T-2)-1/2

Hence;

t = C sqrt{h/g}

As the power of 1/2 is the same as a square root.

As for the problem with equation 3, it might be because t is not a scale factor for the speed, but rather the acceleration of the tower. So it would actually be;

t*a=v

C Mx Ly (L1 T-2)z * L1 T-2 = L1 T-1

Which would give; -2z-2=-1 which is -2z=1 so it's a scale factor for the acceleration of the tower. So the formula for the velocity of the tower would be;

C sqrt{h/g}*a = v

I think...
Haths

3. Jan 5, 2009

### daschaich

Here's a different take on the problem statement, which I think could be worded better.

To my reading, it sounds like we want to make the tower take time t to fall (note that it is given dimensions of time (T) later on). That is, t itself is not the scale factor, but the desired scaled time.

Per the problem, we suppose it can depend in as-yet-unknown ways on the mass of the tower m, its height h, g, and an arbitrary dimensionless constant C. Turning that last sentence into an equation just gives
t = C mx hy gz,​
where x, y and z are constant exponents we want to determine.

Dimensional analysis makes this easy: t has dimensions of time (T), so we demand that C mx hy gz have dimensions of time as well. Since m has dimensions of mass (M), h has dimensions of length (L), g has dimensions of length per time-squared (L/T2 = LT-2), and C has no dimensions at all, we have
T = Mx Ly (LT-2)z = Mx L(y + z)T(-2z).​

Now all we have to do is match the exponent of each dimension on each side of the equation. Writing everything out explicitly,
M0 L0 T1 = Mx L(y + z)T(-2z).​
Separately matching the exponents of M, L, and T on each side of the equation immediately gives the three equations
0 = x
0 = y + z
1 = -2z,​
which have the solution
x = 0
z = -1 / 2
y = 1 / 2.​
Plugging x, y and z back into the original formula, we have
t = C m0 h1/2 g-1/2 = C(h / g)1/2,
$$t = C\sqrt{h / g}.$$​
And that's all there is to it.

4. Jan 6, 2009

### ajassat

Excellent. Both of your replies helped me to understand things better. Any confusions/doubts were gotten rid off when I read daschaich's reply.

Thank you.