How can I connect F=ma to Drag=1/2*Cd*(rho)*V^2*A?

I previously posted a different question that was very vague. I have kept working on the problem, and now I have a different question. Please let me know if I should have just added it to the old one.

I am trying to find the drag that a go-kart sees using a coast-down test, in which you accelerate then allow the cart to decelerate on its own (without using the brakes.) I have data on the speed of the cart vs. the time.

For a cart that is decelerating due to drag, my data is saying that the velocity is proportional to the square of time (a parabolic curve of velocity vs. time). I have looked at the data of other people's coast-down tests (I found them online), and their velocity appears to follow the same trend that mine does.

The above would mean that the derivative of velocity, the acceleration, would be linear if plotted against time.

Given these two relationships:

$$v \propto t^2$$ $$a \propto t$$

$$v \propto a^2$$

My understanding is that you can calculate drag using $$F=ma$$, where is the drag force $$F=D$$.

This would mean that $$D \propto \sqrt{v} \quad (?)$$

However, I have read that drag is proportional to $$v^2$$. The data that I have looked at from other people's tests also shows the relationship that drag is proportional to $$v^2$$, but I can't figure out how they got that relationship when their velocity curves are parabolic with respect to time.

What am I doing wrong? I can't figure out how these two equations can both be true.

I guess my main question at this point is: should the velocity vs. time plot for a decelerating vehicle have a curve fit of $$t \propto v^2$$ or $$t \propto \sqrt{v}$$?

Thank you for any input.

I've attached an image of my data to show what I'm seeing.

• "velocity is proportional to the square of time" -- I'm having trouble believing this. Even if you're saying that $v(t) = v_0 - A t^2$, that's saying that your deceleration is minimal when you first release the cart, and at it's maximum right as the cart stops. It also implies that the cart stops, then starts accelerating in reverse until it's eventually going infinitely fast -- I do suspect that's not happening! Could you edit your question to include your check plot of velocity vs. time, or perhaps distance vs. time? – TimWescott Jul 7 '19 at 20:13
• Thanks for the answer! I've attached my plot – Kit Jul 7 '19 at 20:48
• That velocity vs. time plot is so barely convex I'm not sure that the $t^2$ term is significant. Given that you have a model (constant rolling resistance + drag that's proportional to $v^2$) it would be better if you start by seeing how good of a fit you get to that model. I think if it were me, and I had the ability to do more experiments, I'd try to get another set of data with a higher starting velocity, to enhance the expected aerodynamic drag component of the overall drag. – TimWescott Jul 7 '19 at 21:49
• Thanks for the suggestions. I plotted sqrt(v) vs. time and v^2 vs time, and sqrt(v) vs. time came out linear (meaning that v is proportional to t^2 for my data), while v^2 vs. time came out with some curvature. If the model held for my data, I would expect the velocity vs. time graph to show a square root function, I believe (where v^2 is proportional to t). I'm seeing more that sqrt(v) is proportional to t, so at the moment, it seems like I'm either doing something wrong, or my data is bad. I can't test again for a few more weeks, but I'll definitely get more data when I'm able. Thanks! – Kit Jul 7 '19 at 22:22
• The model is for the drag being the sum of a constant, and a $v^2$ term. Did you try to fit to that? – TimWescott Jul 7 '19 at 23:05

There are two parts to drag for a vehicle, aerodynamic and rolling friction.

The aerodynamic drag is the larger part above about 40mph, while that due to rolling friction is larger below 40mph.

The total drag is both combined, obviously rolling friction is there at any speed, so the curve you plot is representing both, unless, your results are below 40mph and are mostly due to rolling friction...

Drag is usually calculated as a function of several things: Drag coefficient, density, velocity squared and area.

D = Cd * rho * (V^2 / 2) * A


rho is density, Cd is drag coefficient, V is velocity and A is area.

So, as you put in your question the drag is basically proportional to the velocity squared plus some other bits.

Here is a link to the drag formula, but inclination will be relevant to an aeroplane, but not really relevant for a vehicle...

Drag formula

Another link you may find interesting is this one as it expands on those things you are looking at:

overview of drag on a vehicle

• Thanks for answering! My data looks similar to other data I found in thesis papers online, though (like this one: dspace.lboro.ac.uk/dspace-jspui/bitstream/2134/10797/1/…). To me, this would indicate that the shape of the velocity curve is correct to find aerodynamic drag. (In addition to rolling drag, but it would seem that that didn't affect the shape of my curve too much). Are you saying that my initial velocity vs. time curve is shaped incorrectly? – Kit Jul 7 '19 at 18:47
• Thank you for adding more to your answer! I'm trying to find the coefficient of drag using this test, though, so I can't calculate it using the formula you gave. I'm trying to figure out how my data connects to that formula. The coast-down test is a test a number of people do, and drag is calculated using F=ma from data obtained from it. So I'm trying to figure out if a) my data is bad (it should be shaped differently), or b) I'm doing something incorrectly in analyzing it. – Kit Jul 7 '19 at 19:35
• "The aerodynamic drag is the larger part above about 40mph, while that due to rolling friction is larger below 40mph." Please put the word "typically" in front of that. Because I have counterexamples just boiling in my head right now (starting with a "cart" on a table with air bearings, another one with a really draggy rolling assembly, one with a giant parachute, and then all sorts of bizarre possibilities in between). – TimWescott Jul 7 '19 at 20:15
• @TimWescott "vehicles" as covered in texts such as this : amazon.co.uk/Fundamentals-Motor-Vehicle-Technology-Hillier/dp/… not carts on tables or other weird contraptions... – Solar Mike Jul 7 '19 at 21:15

Assuming that drag is the only force (this is not accurate - see friction).

Then if you solve the differential equation:

$$m \ddot{x} = -\frac{1}{2}C_d\cdot A\cdot \rho\cdot (\dot{x})^2$$

with $$x(t=0)=0, \dot{x}(t=0) = v_0 [m/s]$$

then the solution is :

• $$x(t) = - \frac{m \log{\left(\frac{m}{v_0} \right)}}{A C_d \rho} + \frac{m \log{\left(A C_d \rho t + \frac{m}{v_0} \right)}}{A C_d \rho}$$
• $$\dot{x}(t) = \frac{m}{A C_d \rho t + \frac{m}{v_0}}$$

you can use the following python/sympy code at repl.it to get the result, or modify it for your needs

import sympy as sp

# %% Define symbols
t = sp.symbols('t' )
x = sp.Function ('x')(t)
dx = sp.diff(x,t)
dx2 = sp.diff(x,t,2)

m ,Cd, rho, A = sp.symbols('m, C_d, rho, A')
v0 = sp.symbols('v0')
#%% Define and solve ode
eq1= sp.Eq(m*dx2, -Cd*rho*A*(dx)**2 )
res = sp.dsolve(eq1,x, ics = {x.subs(t,0):0, dx.subs(t,0):v0})
# %% print x(t)
print(res)
# %% print velocity v(t) in latex format
print(sp.latex(sp.diff((res.rhs),t)))