3
\$\begingroup\$

Im developing simple car racing game. To turn a car, I need to calculate slip angles for each tire and then use those angles to find lateral force (using Hans B. Pacejka magic formula). One thing I don't understand is relation between car`s speed and lateral forces.

Everywhere is written, that lateral force depends only on slip angle. But it means that lateral force will be the same for slow and very fast velocity. Is that true? How to use lateral forces to move a car? In my opinion lateral force should be small if velocity is small.

\$\endgroup\$
3
\$\begingroup\$

Well, first of all I am by no means a physics expert.

Regarding your question, Your observation is correct, faster cars should have higher lateral forces, well but that's not the whole story.

I think your problem is that the question that needs to be asked, where does the lateral force come from in the first place?

In normal circumstances the car's engine generates force that makes the car moves forward. This force along with other forces that act on the car movement direction (or opposite) is called longitudinal forces.

On the other hand, in high speed cornering scenarios, the tires develop lateral forces also known as the cornering or side force. Which means this force is only generated when the tires have an angle between the tire's heading and its direction of travel.

Now the important point is, the angle between the tire's heading and its direction of travel is called alpha. When the movement direction is the same as the tires heading direction then Alpha is Zero. This results in a lateral force that equals zero.

When there is an angle between heading direction and the tire's traveling direction the velocity vector of the car is now split into two components:

  • The longitudinal vector has magnitude cos(alpha) * velocity.
  • The lateral vector has magnitude sin(alpha) * velocity and causes a resistance force in the opposite direction: the cornering force.

Which means that lateral force depends on the As the slip angle grows, so does the cornering force. Now the point is, at low slip angles, the relationship between slip angle and cornering force is linear.

Flateral = Ca * alpha

Now, since this equation is only true for small angles, I believe this equation is using small angle approximation where:

enter image description here

Hence, dropping the sin and the velocity from the equation. Keep in mind that the linear relation is only true for small angles, otherwise the linear relation doesn't apply anymore. But I might be wrong someone with more knowledge might correct me.

\$\endgroup\$
  • \$\begingroup\$ Thanks for your reply! I still don't understand relation between lateral vector and lateral force. Another thing is that I'm not sure if F_lateral = Ca * alpha is correct, because Ca and alpha is unitless and F has unit newtons. And still, if alpha is big and car has very low speed, then lateral force is very big. Which is wrong in my opinion. \$\endgroup\$ – majak-the-coder Dec 31 '13 at 17:19
  • 1
    \$\begingroup\$ @majak-the-coder Constants are not unit less, they are calculated once under some circumstances and their units are derived so they can keep the equation sides equal. But you need to check the constant's unit. \$\endgroup\$ – concept3d Dec 31 '13 at 17:29
  • \$\begingroup\$ remember that force is a vector. \$\endgroup\$ – concept3d Dec 31 '13 at 17:30
  • \$\begingroup\$ @majak-the-coder "if alpha is big and car has very low speed, then lateral force is very big." Then the equation just doesn't apply, and the relation is not more linear. \$\endgroup\$ – concept3d Dec 31 '13 at 17:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.