Thread: Dimmensional Analysis Of Drag On A Sphere Equation Problem

Making Assumptions:



Drag force is equal to a function of radius, velocity, air density, and the viscosity of air.



Solves to:

L : a + b − 3c − d = 1,
T : −b − d = −2,
M : c + d = 1.

From the T equation we have b = 2−d, and from the M equation c = 1−d.
The L equation then gives us a = 2 − d



The book forms an equation using as an arbitrary number. I'm fine with that but, I can't figure out how they create the rest of the equation.
I conclude that a,b,c and d are substituted back into the equation but, I dont see how they get Rho on its own.

Can someone explain that to me?

Thanks.

This is saying that "R" is a length so has units "L" (meters, feet, parsecs, etc.), v is a speed so has units "L/T" (T is seconds, minutes, hours, years, etc.), $\displaystyle \rho$ is a density so have units of "mass over volume" so "M/L^3" (M is grams, kg, slugs, etc. and, of course, volume is "length cubed"). $\displaystyle \mu$ is viscosity and so (as we all know!) has units of "M/LT". We don't know what function of each of these is involved but whatever function is involved, it can only change the "number" of each unit involved, and whether it is in the numerator or denominator so it must have some exponent (negative if in the denominator).

That is, whatever the function f is, we must have the result $\displaystyle (L)^a(L/T)^b(M/L^3)^c(M/LT)^d= L^aL^bT^{-b}M^cL^{-3c}M^dL^{-d}T^{-d}$ and, combining like bases, $\displaystyle L^{a+b-3c- d}T^{-b-d}M^{c+d}$

Because drag has units of "$\displaystyle ML/T^2= MLT^{-2}$, in order that those have the same units we must have a+ b- 3c- d= 1, -b-d= -2, and c+ d= -2.

Again, f can be any function. If it were functions like [itex]e^x[/itex] or sin(x) which have no units for the arguments, we must have constants involved that will cancel any units on x, and then constants multiplying the result of the functions having the correct units for the result. But if we expect simply multiplying or dividing (for some physical reason) then the factors must involve lengths, times, masses as shown.

I don't see how they get Rho on its own.

I've concluded it's a type-o by the publisher. If I solve on my own, I get the same equation the book uses later in the example.



They seem to have double counted the exponent of Rho.