Torque steer...
Dec. 11th, 2006 10:47 am![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
growler_southI was looking over some old backups and came across this wee gem, a piece I wrote for a magazine 6 years ago explaining front-wheel-drive torque steer as it relates to unequal length driveshafts. (Non car nuts: that's when you stomp on the accelerator and the steering wheel wrenches to the left. Or right. Depends on the car).
I just read it again and it still makes sense. Might as well post it here for the entertainment and enlightenment of those of you who like this sort of thing...
Here goes.. try to follow.
Example 1;
Imagine, if you will, a normal wheel connected to a normal driveshaft. The driveshaft is exactly in line with the wheels rotational axis (parallel with the ground). A torque is now applied to the driveshaft- you can see how all of the torque goes into rotating the wheel about rotational axis. Even if the wheel were free to rotate about another axis (like the steering axis or kinpin) it wouldnt have any reason to do so, as there is no torque along the steering axis.
That was the easy bit.
Example 2;
Next we will imagine a situation in which the driveshaft is not at right angles to the plane of the wheel. To make it easier, we will imagine a completely vertical driveshaft aligned with the wheel's steering axis, the torque being transmitted to the wheel, from above, via a bevel gear. Apply a torque to your imaginary driveshaft. NOtice how, due to the bevel gear, half of the torque goes into turning the wheel about its rotational axis, the other half goes into turning the wheel about its steering axis.
NOw this bit's fun.
Example 3;
Your imaginary driveshaft is currently aligned with the wheel's steering axis. Lower the driveshaft until the driveshaft is halfway between the steering axis and the rotational axis (IE 45 degrees). In this position, 3/4 of your torque is going into rotating the wheel about its rotational axis, the other 1/4 is still going into rotating the wheel about its steering axis.
Would it help if I pointed out that a normal CV joint acts like a bevel gear when on an angle?
Here we go;
Since the driveshafts on most FWD cars are on an angle, there is always some steering torque being aplied to the wheels along with the driving torque.
On a car with equal length driveshafts, both CV joints are acting like bevel gears to equal degrees, and so the steering torques on both sides of the car are equal and opposite. They thus cancel each other out.
But unequal length driveshafts (and an off-centre diff) means unequal driveshaft angles- which means one CV joint is acting more like a bevel gear than the other, and the opposing steering torques dont cancel each other out any more.
Got it?
Now why does the car steer towards the shorter driveshaft?
If you think back to the bevel-gear driven example 2 above, you'll see that the steering torque tries to push the front of the wheel (viewed from above) away from the car, and draw the back of the wheel into the car. So the side with the shorter driveshaft, being on a more acute angle and thus imparting more steering torque to the wheel, tries to steer towards itself.
So there it is. Easy really.
I just read it again and it still makes sense. Might as well post it here for the entertainment and enlightenment of those of you who like this sort of thing...
Here goes.. try to follow.
Example 1;
Imagine, if you will, a normal wheel connected to a normal driveshaft. The driveshaft is exactly in line with the wheels rotational axis (parallel with the ground). A torque is now applied to the driveshaft- you can see how all of the torque goes into rotating the wheel about rotational axis. Even if the wheel were free to rotate about another axis (like the steering axis or kinpin) it wouldnt have any reason to do so, as there is no torque along the steering axis.
That was the easy bit.
Example 2;
Next we will imagine a situation in which the driveshaft is not at right angles to the plane of the wheel. To make it easier, we will imagine a completely vertical driveshaft aligned with the wheel's steering axis, the torque being transmitted to the wheel, from above, via a bevel gear. Apply a torque to your imaginary driveshaft. NOtice how, due to the bevel gear, half of the torque goes into turning the wheel about its rotational axis, the other half goes into turning the wheel about its steering axis.
NOw this bit's fun.
Example 3;
Your imaginary driveshaft is currently aligned with the wheel's steering axis. Lower the driveshaft until the driveshaft is halfway between the steering axis and the rotational axis (IE 45 degrees). In this position, 3/4 of your torque is going into rotating the wheel about its rotational axis, the other 1/4 is still going into rotating the wheel about its steering axis.
Would it help if I pointed out that a normal CV joint acts like a bevel gear when on an angle?
Here we go;
Since the driveshafts on most FWD cars are on an angle, there is always some steering torque being aplied to the wheels along with the driving torque.
On a car with equal length driveshafts, both CV joints are acting like bevel gears to equal degrees, and so the steering torques on both sides of the car are equal and opposite. They thus cancel each other out.
But unequal length driveshafts (and an off-centre diff) means unequal driveshaft angles- which means one CV joint is acting more like a bevel gear than the other, and the opposing steering torques dont cancel each other out any more.
Got it?
Now why does the car steer towards the shorter driveshaft?
If you think back to the bevel-gear driven example 2 above, you'll see that the steering torque tries to push the front of the wheel (viewed from above) away from the car, and draw the back of the wheel into the car. So the side with the shorter driveshaft, being on a more acute angle and thus imparting more steering torque to the wheel, tries to steer towards itself.
So there it is. Easy really.
