Calculating Vehicle Speed based on Downforce (Aerodynamics)

[giulioooooh]

TL;DR Summary

Basically, I'd like to create a function that displays speed or acceleration based on aerodynamic downforce which in turn is determined by speed. I'm not really getting any further with my math skills so I'm desperate for someone who actually knows what they're doing to tell me how easy this actually is.

First off: I'm currently working on an RC car heavily focused on aerodynamics for a Project I'd like to add to my Uni application. I've created a model, ran it through CFD and came up with some numbers. Now I'd love to plot out an equation that displays either speed or acceleration in dependence of time so I could tweak my Aero design. Turns out, I've got no clue of what I'm doing.

Here we go:

I'm trying to Calculate Downforce (aerodynamics and gravity) with this Formula:
F_downforce=C_{downforce=lift}*rho*A*v²+g*m
And drag with this one:
F_drag=0.5*C_drag*rho*A*v²
(C being the coefficients of drag or downforce=lift, rho the air density, m being 50% of the total vehicle mass when accounting for 50:50 weight distribution)
These are the Formulas I found on the NASA website and I adjusted them to solve for Force.

Putting in my values I got out of CFD and some other rough calculations I came up with, I get these formulas:
F_downforce(x)=0.073*(1.225kg/m³)*6245cm²*(x*m/s)²+(9.81N/kg)*400g
F_drag(x)=0.5*0.052*(1.225kg/m³)*6245cm²*(x*m/s)²
(Yes, I know this is a lot of area, I just went with what CFD came up with since it doesn't really matter as long as both reference areas are the same)

or, simplified:
F_downforce(x)=0.056N*x²+3.924N
F_drag(x)=0.02N*x²
These equations seem to work fine and match up with CFD results for given speeds.

If I want to calculate the maximum possible acceleration, I've got to account for mass and tire grip. I decided to go for a traction coefficient μ _t=0.9 and a total mass m_total=800g. (Quick question: the traction coefficient also accounts for tire contact patch right?)

This gives me following equation that I'm kinda proud of, though I can't shake off the feeling that I've got something terribly wrong:
a_max(x)=(F_downforce(x)*0.9-F_drag(x))/800g

My current problem is that "(x)". The formula for acceleration is dependent on speed, which is very much self-referential and I've really got no idea on how to turn that into an equation that tells me speed or acceleration in relation to time.
Normally, you'd just have e.g. a(t)=s/t², the integral of that being velocity v(t)=a*t (I feel like this line is absolute bs but I'm too tired to really realize what exactly is wrong with it, though I think you get the idea.)

Furthermore, I've come up with the max acceleration the motors could output, calculated like this:
Torque=(9550*255W)/22500rpm=0.108Nm
F_1motor=0.108Nm/2.5cm*sin(90)=4.32N => F_2motors=2*F_1motor=8.64N

a_max=8.64N/800g=10.8m/s² (not accounting for drag and with infinitely high downforce that doesn't crush the car)

So, can someone please help me out of my misery and give me some tips on how to create that function I crave which should tell me the acceleration in relation to time up until the point in time when max acceleration (limited by the motors) is achieved?

Thanks so much in advance guys and gals<3333

P.S: I'm in Europe and about to go to bed, so I'm sorry if I can't reply for the next like 10 hours, I'll try to get back to everyone bothering to put up with my (hopefully way too easy) questions

 

[Baluncore]

Welcome to PF.

Good morning.
The motor drive and vehicle inertia are related to ground speed.
The downforce is related to air speed.
Ground speed and air speed will only be the same when there is no wind.
What assumptions are you making with regard to wind ?

 

[jack action]

The equation you are looking for is equation (1b) on this page:
$$a= \frac{1}{\lambda_m}\left(\frac{F_t}{m}-f_r g - \frac{0.5}{m}\left(C_D A - f_r C_L A\right)v^2\right)$$
Where:

• $\lambda_m$ is the mass factor that takes into account the effect of the inertia of the rotating parts;
• $f_r$ is the rolling resistance coefficient;
• The tractive force $F_t$ is the smallest between the one based on motor power ($F_t = \frac{P}{v}$) and the traction limit $\left(F_t = \mu mg\left(K_1 -K_2\frac{0.5\rho C_L A}{mg}v^2\right)\right)$. The value of $K_1$ and $K_2$ depend on the vehicle layout (see link for more info).
• Note that $C_L$ is the aerodynamic lift coefficient, based on $F_{lift} = 0.5C_L Av^2$. Thus, for a downforce, $C_L$ is negative.

And then to find the time and distance with respect to speed, you have to do calculations in increments of speed:
$$t_f = t_i + \frac{\Delta v}{a_{avg}}$$
$$d_f = d_i + 0.5(v_i+v_f)(t_f - t_i)$$
Finally, this acceleration simulator does all these calculations and gives something like this for your vehicle:

which should tell me the acceleration in relation to time up until the point in time when max acceleration (limited by the motors) is achieved?

So you would reach a maximum acceleration of 0.45 g at around 14 km/h, which should happen after about 0.9 s.

The maximum speed would be 73 km/h reached after 13.64 s.

the traction coefficient also accounts for tire contact patch right?

what do you mean when you say "tire contact patch"?

 

[giulioooooh]

Welcome to PF.

Good morning.
The motor drive and vehicle inertia are related to ground speed.
The downforce is related to air speed.
Ground speed and air speed will only be the same when there is no wind.
What assumptions are you making with regard to wind ?

Hey Baluncore, thanks for taking the time to have a look at my problem. Yes, I‘m assuming that there is no wind on the straight.

 

[giulioooooh]

The equation you are looking for is equation (1b) on this page:
$$a= \frac{1}{\lambda_m}\left(\frac{F_t}{m}-f_r g - \frac{0.5}{m}\left(C_D A - f_r C_L A\right)v^2\right)$$
Where:

• $\lambda_m$ is the mass factor that takes into account the effect of the inertia of the rotating parts;
• $f_r$ is the rolling resistance coefficient;
• The tractive force $F_t$ is the smallest between the one based on motor power ($F_t = \frac{P}{v}$) and the traction limit $\left(F_t = \mu mg\left(K_1 -K_2\frac{0.5\rho C_L A}{mg}v^2\right)\right)$. The value of $K_1$ and $K_2$ depend on the vehicle layout (see link for more info).
• Note that $C_L$ is the aerodynamic lift coefficient, based on $F_{lift} = 0.5C_L Av^2$. Thus, for a downforce, $C_L$ is negative.

And then to find the time and distance with respect to speed, you have to do calculations in increments of speed:
$$t_f = t_i + \frac{\Delta v}{a_{avg}}$$
$$d_f = d_i + 0.5(v_i+v_f)(t_f - t_i)$$
Finally, this acceleration simulator does all these calculations and gives something like this for your vehicle:

View attachment 299210
View attachment 299211​

So you would reach a maximum acceleration of 0.45 g at around 14 km/h, which should happen after about 0.9 s.

The maximum speed would be 73 km/h reached after 13.64 s.what do you mean when you say "tire contact patch"?

Hey jack, what you posted is looking really promising. Thanks so much! I‘ll have another look or two at this and get back to you later when I got some more time.

 

[giulioooooh]

The equation you are looking for is equation (1b) on this page:
$$a= \frac{1}{\lambda_m}\left(\frac{F_t}{m}-f_r g - \frac{0.5}{m}\left(C_D A - f_r C_L A\right)v^2\right)$$
Where:

• $\lambda_m$ is the mass factor that takes into account the effect of the inertia of the rotating parts;
• $f_r$ is the rolling resistance coefficient;
• The tractive force $F_t$ is the smallest between the one based on motor power ($F_t = \frac{P}{v}$) and the traction limit $\left(F_t = \mu mg\left(K_1 -K_2\frac{0.5\rho C_L A}{mg}v^2\right)\right)$. The value of $K_1$ and $K_2$ depend on the vehicle layout (see link for more info).
• Note that $C_L$ is the aerodynamic lift coefficient, based on $F_{lift} = 0.5C_L Av^2$. Thus, for a downforce, $C_L$ is negative.

And then to find the time and distance with respect to speed, you have to do calculations in increments of speed:
$$t_f = t_i + \frac{\Delta v}{a_{avg}}$$
$$d_f = d_i + 0.5(v_i+v_f)(t_f - t_i)$$
Finally, this acceleration simulator does all these calculations and gives something like this for your vehicle:

View attachment 299210
View attachment 299211​

So you would reach a maximum acceleration of 0.45 g at around 14 km/h, which should happen after about 0.9 s.

The maximum speed would be 73 km/h reached after 13.64 s.what do you mean when you say "tire contact patch"?

Hey Jack, I'm back again. That Calculator you linked and that website in general is very useful, thanks so much for linking it. I still got a bunch of questions though:

1. You set power to 255W although I have two motors with 255W each, so I think that I didn't explain myself completely, right? Or am I overlooking something?

2. I don't really understand how you came up with $F_{lift} = 0.5C_L Av^2$. When looking at http://hpwizard.com/aerodynamics.html, C_L,D*A=C_L,DA, so I don't quite understand why you have that 0.5 in your lift equation. Is that just to cancel out the 0.5 in the Drag equation? Do you even need those "0.5"s in the first place or can you just leave those out if you do it for both?

3. (Still kind of part of question 2 and not really a question:) Regarding the settings you put into the acceleration sim, it doesn't really make sense to me why the drag factor should be bigger than the absolute value of the lift factor if (at least in CFD) downforce is bigger than drag. Assuming that I leave out that 0.5 factor I just talked about on both, I get a C_LA=457.14cm² and C_DA=161.45cm².

This results in the following inputs into the acceleration sim:

(I also added stuff like aero balance, weight distribution etc., though those are approximations)

This gives me the following results:

What's interesting (but maybe totally irrelevant) is that for this set of values I get a very smooth quarter mile curve which isn't a given for all values, eg. those you picked iirc.

Finally, about the tire contact patch:
I mean the amount of area of the tire that actually touches the ground. Like, a tire that is slightly flat will have higher grip than an overinflated one as the latter will tend to spin when accelerating hard. I thought that that coefficient also takes into account how much of the tire actually touches the ground, but I think that the optimal tire pressure is to be determined experimentally.

Anyway, thanks for your answers and sorry for the confusion. Hoping that I don't sound too confused :)

 

[Baluncore]

Finally, about the tire contact patch:

In theory, the friction coefficient between the rubber and the road, multiplied by the wheel loading, decides the shear force before a wheel spins or slides. It is not dependent on tire contact patch area.

The total wheel load is the product of contact patch area by internal tire pressure. If the tire pressure is too low, the contact patch will be large and the tire side wall will flex more and overheat, which wastes energy. If the tire pressure is high, the patch will be small, with less flex and less heat. If the road surface is rough it is better to have a bigger patch to average more road surface.

Very low pressure is really only needed off road. If the surface material can be sheared, then a greater contact patch will spread the shear over a greater area and allow more traction before the wheel slips. If the ground is soft, to prevent the vehicle from becoming bogged, the tire pressure must be reduced to a pressure the ground can support.

 

[jack action]

You set power to 255W although I have two motors with 255W each, so I think that I didn't explain myself completely, right? Or am I overlooking something?

No, I thought you had only one.

2. I don't really understand how you came up with $F_{lift} = 0.5C_L Av^2$. When looking at http://hpwizard.com/aerodynamics.html, C_L,D*A=C_L,DA, so I don't quite understand why you have that 0.5 in your lift equation. Is that just to cancel out the 0.5 in the Drag equation? Do you even need those "0.5"s in the first place or can you just leave those out if you do it for both?

The equation used throughout the website^[1] is $F_{lift} = 0.5\rho C_L Av^2$. (It is defined here)

You have to be careful with aerodynamic coefficients as they can be based on very subjective data. The important relation is:
$$C_L \propto \frac{F_{lift}}{A_{ref}\rho v^2}$$
Where $A_{ref}$ is some reference area that represents the size of the object. The frontal area is commonly used in the automobile industry, but sometimes, some people use the bottom view area of the car, as they find it more representative. Since both areas are usually proportional, it doesn't change anything in the end, but the value of $C_L$ will not be the same in both cases. This is why it is better to know $C_L A$ as a whole rather than just $C_L$ and try to guess which area was used as a reference.

Similarly, it is usual to use the $0.5$ factor, because it comes from the dynamic pressure definition $\frac{F}{A} = 0.5\rho v^2$. But if you don't use it to define $C_L$, it is OK as long as everybody knows when using it for calculations. Otherwise, it will be off by a factor of 2. You don't seem to use it in your calculations: Are you sure how your $C_L$ was defined?

As for the numbers I used for your aerodynamic coefficients, I based them on this:

Fdownforce(x)=0.073*(1.225kg/m³)*6245cm²*(x*m/s)²+(9.81N/kg)*400g
Fdrag(x)=0.5*0.052*(1.225kg/m³)*6245cm²*(x*m/s)²

If I understood your equations correctly, for $C_D A$, I used $0.052 \times 6245 \text{ cm²}$ and I use $-0.5 \times 0.073 \times 6245 \text{ cm²}$ for $C_L A$. I wanted to compensate for the lack of 0.5 in your equation, but I realize that I made a mistake and I should have used $-2 \times 0.073 \times 6245 \text{ cm²}$. But maybe your equation is wrong and your $C_L$ value is correct to be used as is in the program. You'll have to check this out.

---

[1] Full disclosure: I wrote the website.

 

[giulioooooh]

Similarly, it is usual to use the $0.5$ factor, because it comes from the dynamic pressure definition $\frac{F}{A} = 0.5\rho v^2$. But if you don't use it to define $C_L$, it is OK as long as everybody knows when using it for calculations. Otherwise, it will be off by a factor of 2. You don't seem to use it in your calculations: Are you sure how your $C_L$ was defined?

I used the equations that I uploaded some shots of which I calculated w my CAS. This is the right side of the equations I uploaded, unfortunately I don't have enough screen real estate to show them completely:

So yeah, I took both equations without the 0.5 factor as I thought they would cancel out either way. Multiplied by 6245cm² I got the values I put into vehicle characteristics earlier (which seem to be incorrect).

If I add that .5 factor into both equations as is shown on the Acceleration page I get the following values (entered into Vehicle characteristics):

This seems logical and all, the only thing bothering me right now is that weird jagged section in the quarter mile graph I didn't get with the earlier values.

Finally, this acceleration simulator does all these calculations and gives something like this for your vehicle:

Though, when quadrupled, the value you entered up here for C_LA lines up with the new values I got, which in turn lines up with what you said here:

f I understood your equations correctly, for CDA, I used ²0.052×6245 cm² and I use ²−0.5×0.073×6245 cm² for CLA. I wanted to compensate for the lack of 0.5 in your equation, but I realize that I made a mistake and I should have used ²−2×0.073×6245 cm².

So yeah, I think the values I got now are pretty much correct.Anyways, I'm going to go to bed now and think about this for a while. Thanks so much for helping me, and I got to say that your website is a great resource for aero amateurs. The only thing about it I'd change is how hard it is to find. The best website of the world isn't going to help anyone if you can only find it on the 8th page of google, yknow? Nevertheless, great work and keep going strong!