Torque coordinate transformation

[Jan Grossmann]

Hi all,

I am currently working on a creating a mathematical model of a longboard and am in need of advice. The pictures describe the sitaution.

Side view

Top view

Back view

The pictures depict a simplified longboard - the brown line is the deck and the black lines represent the trucks - at an angle because the longboard is turning. Point C,D,E,F are where the wheels are. The center of gravity and rotation lies at the coordinantes 0,0, e.i. in the middle of the deck on line A,B. Now my question regards the vectors u,v,w,a - these are force vectors generated by the wheels in a turn. These forces generate a torque about the COR in the axis perpendicular to line CD. I need to know the torque in the axis on line AB. Can I calculate the torque simply by transfroming the force vectors into the axis perpendicular to AB and using these transformed forces to calculate, or is this approach not correct? Thank you in adavance.

John

 

[haruspex]

These forces generate a torque about the COR in the axis perpendicular to line CD

I'm not sure that's what you mean. If the cornering is at constant speed and radius there is no torque. The rotation rate is constant. Rather, there is a radial (centripetal) acceleration.
If there is angular acceleration, what is driving it?

 

[A.T.]

I need to know the torque in the axis on line AB.

Do you mean the torque around the axis AB? Compute the total torque vector around some point on AB, and ignore its component perpendicular to the axis AB.

 

[Jan Grossmann]

I decided to redraw the problem, as I clearly did not explain it well enough.

In this picture you can see how the longboard is travelling, not all forces are drawn, as they are not necessary for this problem. The force F1 always act in the x-axis and represents the pushing force. The green vector v represents the speed vector and it is tangent to the circular path the longboard is travelling. Forces T1 to T4 in pink represent the tractive froce generated by the wheels and S1 to S4 represent the side forces acting on the wheel, making the turn possible. Forces T are a function of longitudinal slip and forces S are a function of side slip. The deck is one body, the front and back truck (wheel axle) are individual bodies and each wheel is an individual body. That is to understand the problem.

This picture now shows only the front two wheels. The left diagram shows a view from the side of the wheel (the viewpoint is shown in the right side of the image). The orange force Fz is the calculated load for the front truck (axle, not important, just to clarify) .The tractive forces T1 and T2 generate a torque in the axis as show by the purple cross on the left and arrow on the right Torque = T x h. Now, looking back to the first image of the whole longboard, I need to find out what the effect of the torque is in the y axis.

To calculate the effects of T forces and S forces in the x and y coordinates, I simply project the forces using the known angle Beta. My question is, can I calculate the torque effect in the y-axis using the projected forces or is this not the correct approach. I hope I managed to explain my question better. Thank you.

John

 

[haruspex]

The wheels do not generate a tractive force. The tractive force is F. The T forces would be the (small) frictional forces necessary to overcome axial friction in the wheels and keep them turning.
Ordinarily this will be rolling contact, not sliding, so I do not understand why you refer to T as being a function of longitudinal slip. Perhaps you do not mean sliding.
Similarly, there should be no side slip; it should be successfully opposed by the S forces.

To the extent that these small T forces produce a torque about the centre of the board, the torques on opposite sides oppose each other and pretty much cancel.
The only time there will be a net torque on the board is when speeding up or slowing down while cornering. Otherwise there is no angular acceleration.
If you apply F to accelerate in a corner, the S forces at the front of the board will exceed those at the rear and lead to a net torque.

 

[Jan Grossmann]

The wheels do not generate a tractive force. The tractive force is F. The T forces would be the (small) frictional forces necessary to overcome axial friction in the wheels and keep them turning.
Ordinarily this will be rolling contact, not sliding, so I do not understand why you refer to T as being a function of longitudinal slip. Perhaps you do not mean sliding.
Similarly, there should be no side slip; it should be successfully opposed by the S forces.

To the extent that these small T forces produce a torque about the centre of the board, the torques on opposite sides oppose each other and pretty much cancel.
The only time there will be a net torque on the board is when speeding up or slowing down while cornering. Otherwise there is no angular acceleration.
If you apply F to accelerate in a corner, the S forces at the front of the board will exceed those at the rear and lead to a net torque.

Well, that is not true. I am modeling the wheels according to the Pacejka tire model (see https://en.wikipedia.org/wiki/Hans_B._Pacejka#The_Pacejka_"Magic_Formula"_tire_models) and using the magic formula to calculate the longitudinal and lateral forces and self aligning torque of each wheel. So in this case T is a function of longitudinal slip and there actually needs to be a side slip in order for turning to happen, as no S force is generated if there is no side slip. The force F1 always acts in line with x-axis and without the S forces, it would never turn.

The answer I am looking for is irelevent of this anyway, I only gave it for context. I need to know if I can calculate the torque effect in the y-axis using the projected forces, as mentioned in my previous post.

Thank you for the quick reply, haruspex, I hope you can help me find the answer.

John

 

[haruspex]

self aligning torque of each wheel

Is that the torque you are interested in? If so, I misunderstood; I had thought you were interested in torque on the board as a whole.
Self-aligning torque acts to restore alignment. When the wheels are perfectly aligned there is no torque.
Even when not perfectly aligned, if the bogey is rotating steadily (about a vertical axis) to restore alignment, still no torque. That torque acts to change the rotation rate of the bogey. Is that really what you are asking about?

T is a function of longitudinal slip

needs to be a side slip in order for turning to happen

Thanks for clarifying what slip means here. But whether such slips are needed to create these forces or they are merely side effects of these forces while rolling is moot. E.g. if we replace a rubber tire on road with a steel tire on a pinion track we still get these forces without slip.
Saying the slip causes the forces is like saying that when I use a lever it is the bending strain in the lever that produces the force on the object. It is a reasonable statement but too detailed a view for most purposes.

without the S forces, it would never turn.

Yes, and I did not disagree with that. I wrote that the S forces do not need to produce a torque when maintaining a constant speed around a curve. A torque on the board is needed to change speed during a turn, or to change radius of turn.

So, please clarify, which torque is of interest to you?