Math of aerodynamics in MotoGP & Formula1
Updated: Feb 17, 2021
“It’s not hard to explain how a rocket works but, explaining how a wing works takes a rocket scientist”
Please allow this introduction, intended to be just a way to set the focus of the post.
I found this in a video on YouTube (link below)
Keeping in mind the difference between theory and practice, the basic principles behind a rocket may indeed be a little more intuitive than the ones behind a wing. In rockets, something is accelerated out of the rocket and thrust is generated in the opposite direction through Newton’s 2nd and 3rd law. Moving the discussion on wings … well, one unique and common discussion is usually not easy to find - even actual R&D relies a lot on testing, [maybe] more than other engineering domains.
The actual problem
I ended up thinking about wings, F1 and MotoGP’s aerodynamics while reading through the infos of the Ducati Panigale R (post’s main picture). I was browsing through the Ducati’s website looking for technical nuances and, looking at the Panigale, I noticed the lateral black wings (picture below) already seen in their MotoGP vehicles. Reading along, I found out that the total stabilizing down-force generated by the aerodynamics of the motorbike is about 30kg at 270km/h. While that is the force generated by the entire vehicle (including the front body), I was interested in figuring out the wings' component alone.
Maybe, the best way to hack this question would be to leave any calculation aside and just pretend to be at about 200 km/h in our car - please don’t - extend our arm out of the window and tilt our hand simulating those wings … we would pretty much guess a down-force of about a few kilograms. We could compute that estimate in seconds, ahead of anybody starting going through the conservation of linear momentum or Navier-Stokes equations; maybe we would be real engineers.
Anyway, after that first estimate, we would need a slightly more quantitative or technical formulation of the problem, just to be able to write down some design parameters. We’ll try to do that without any particular knowledge about physics or engineering. We would try to approach it through intuition.
Some notes before diving into the main point:
1) This will be a simplified two-dimensional (2D) discussion. We will not consider important 3D dynamics along the length of the wing. We'll be basically looking at a wing as represented in the image below ("wing section") - slightly dramatic depiction having flux-separation. Moreover, we’ll consider the flux laminar, meaning, we’ll neglect those vortexes depicted in the same picture below.
2) 3D phenomena are mainly vortexes in the reference-plane longitudinal to the wing. For example, the Ducati's wing in the image above has a vertical down-ward wingtip. That detail allows the wing to exploit the vortexes at its tip and pushing air from top to bottom (being the wing designed to push the motorbike down-ward). In planes those wingtips have the opposite orientation being designed for lifting the plane.
3) Even though we are talking about wings, the same thoughts could be applied to pretty much any appendix of a racing vehicle or even to the entire body of the vehicle acting as a wing – usually pushing down rather than up like in flying objects. However, when it comes to Formula1, where aero-forces can reach a couple of times the weight of the car, almost 50% of that could be generated by the under-body of the car alone and the air passing underneath the car - rather than the two main wings. That too could be approached through the following discussion, but Bernoulli approaches more focused on pressure's differentials may be more effective for the under-body of an F1.
We are looking for a force, specifically, the force that the wing exerts on the motorbike pushing it down. That force is the same that the air exerts on the wing itself and it is basically the opposite of the force that the wing exerts on the air by deflecting it upwards (sketch below).
We may recall that a force is equal to a mass times an acceleration or F = m x a. So, we can get an idea of the vertical force by multiplying the mass of the air deflected by the wing times its up-ward acceleration.
We would be basically assuming that, the entire volume of air affected by the wing at a specific instant is shot up-ward simultaneously. This is obviously not what exactly happens in reality where things change and evolve as air proceeds around the wing - rather than happening all at once. This idea would be more appropriate for a rocket, however, it seemed a good intuitive approach to start with - I will also add final notes on more rigorous approaches like linear momentum and circulation.
Proceeding with our estimate, we can guess that the mass of air interested is a small volume of comparable size to the wing itself - as we may think by looking at the picture of the wing's section above. Moreover, we could estimate the air's vertical acceleration as the difference between the exit vertical velocity of air and the entering one, divided by the time taken to cover the length of the wing.
Our final calculation would be:
Force = mass x acceleration =
= mass of air deflected x acceleration of mass of air deflected =
= (density x volume of air deflected) x
x [(air's exit upward velocity – entering one) / time to cover the wing’s length]
We'll call the one above the “main equation”
We’ll need few data but, first, let’ recap the main assumptions that are allowing us to simplify the problem:
Since we are dealing with a velocity of 270km/h (Ducati’s stated velocity), approximately below 0.3 Mach (let’s think about Mach as the supersonic barrier), we’ll consider air to be non-compressible (constant density). We are here comparing air to water
We’ll neglect also friction or losses
We’ll consider fairly well-guided flow. This assumption would be more verified in multiple-wing systems like a turbine or compressor where the flow is forced to stay within subsequent wings placed in parallel around the rotor of the machine.
We need the density of air: 1.225 kg / m^3 (Wikipedia)
We also need an estimate of the dimension of those two wings; from the picture, let’s say 15 cm length x 10 cm width. Actually, we need the volume of air interested by the upward deflection, so, let’s assume the interested height of the volume of air deflected to be about 10 cm (5 above and 5 below the wing; something slightly excused by approximation #3 above). Therefore, a total volume of air per wing of 1,500 cm^3 or 1,500 x 10^-6 m^3 (we are basically saying that the air outside that box around the wing is not affected by the wing itself). Maybe, even considering approximation #3, we may argue we should consider only the air above the wing and a slightly higher height of the interested volume ...
Since we want to calculate the vertical down-force generated, we’d need the orientation of those wings (exit angle); we’ll approximate that number with 25 degrees (tested with the measuring app of the smartphone)
We’ll need everything in [meter], [second], [kg] … in order to be dimension-consistent
As we said at the beginning, we are also assuming a motorbike’s speed of 270km/h or about 75 m/s (dividing by 3.6 in order to move from km/h to m/s; we could find it knowing 1km = 1,000m and 1hr = 3,600sec). From that we can calculate the exit upward velocity through some trigonometry: 75 m/s x sin(25) = 75 x 0.42 = 31.5 m/s. Note, the entering vertical velocity is 0 m/s being the flux horizontal. Also, we assumed an exit velocity at 25 degrees equal to the entering horizontal one in module.
The only thing left to calculate is the time needed to allow for that change in vertical linear momentum, that is the time to steer vertically the air-flux. We may just calculate the time to cover 15cm – wing’s length, equal to 0.15m – but, I think we neet to consider its horizontal projection that is 15cm x cos(25) = 15cm x 0.9 = 13cm or 0.13m (10% error would not be a big deal anyway considering our overall approximation). Finally, the time to cover 0.13m at 75m/s is 0.0017 seconds.
So, we now have everything we need to compute the “main equation” above:
F = (1.225 [kg/m^3] x 1,500 x 10-6 [m^3]) x [(31.5 [m/s] – 0 [m/s]) / 0.0017 [s] = 34 [kgm/s^2] = 34 [N]
Since we obtained a force of 34 newtons, we can get the equivalent force by dividing it by 10 (9.8 m/s^2 being the gravity acceleration) obtaining about 3 and ½ kilograms. Therefore the 2 wings would account for about 7 kg or 20-25% of the total 30kg of aerodynamic force the Panigale R is able to create.
If we repeated a similar calculation for the front part of the motorbike - the one the rider can hide his head behind - we would likely get close to the total down-force. However, we should consider then many other factors going all around the motorbike in order not to make too much of an approximation of something in reality much more complex. As per the Formula1's comment made above, the entire body may have an important aerodynamic role too; the complete calculation should not be reduced to a unique big wing, especially when simplifications have been made and they could scale pretty badly when applied to the overall system.
You may now be wondering: so, why aren’t all wings just flat panels tilted up/down-ward rather than being carefully shaped in such complex and diverse ways? Big part of the answer is linked to those three simplifying assumptions we made above. In real life we do have friction – affecting also drag – and we are not assured to have perfect guided flow – risking to have detached flux and stall behind the wing. Also, we are not even mentioning here the completely different requirements of a subsonic wing vs one for supersonic applications. Therefore, we may say that a proper design of a wing would be about creating the right pressures' profile across and along the wing resulting in the right lift - up/down-ward depending on the application.
If you are expert on these topics you may already have better answers, however, If you just have some knowledge and you want to go deeper, you can try the calculation through a more rigorous conservation of linear momentum - stationary case maybe, therefore deleting the time components and considering only the extremities of a control volume.
Or, you may want to try the circulation-integral on a path enclosing the wing. In this last case, I'd start assuming a perfectly guided flow with symmetry above and below the wing in order to have two symmetric curved paths canceling each-other out in the integral and leaving only two straight lines easy to integrate in front and at the end of the wing.
Or, another approach, you may want to think in terms of Bernoulli ...
Hopefully, more rigorous calculations will get you similar results, otherwise, please feel free to correct or contact me; I'd love to chat on these topics.
Link to the video the initial quote is from