Math help please?

[911st]

When an A arm is parallel to the ground it is at its longest position of measured from right angles to the ground.

If it is 10.75" long and moves to be at three, four, or five degrees in angle to the ground, how much shorter in length is it if measured at right angles from the ground.

Thank you in advance if you can help.

[tt_targa]

A_armlength* (1-cos(angle)) is the amout of shortening

At 0 degrees the answer should be 0. At 90 degrees, the answer should be equal to A_armlength.

Make sure you calculate the cosine with the proper unit, degrees or radians.

You might describe why you need to know for all than might be interested.

[911st]

Thanks but this is beyond me. Can you tell me the reduced arm length at say 3 or 4 deg?

From that I can get close to how much camber loss I migh see and if it is significant.

[Poor-sche Lover]

Assuming tt_targa's formula is correct (I'm a mathematician, but not a suspension expert), and assuming a 10.75" A-arm, here is what you get:

3 degrees --> .015" shorter
4 degrees --> .026" shorter
5 degrees --> .041" shorter

[911st]

Alan,

Thank you much! That really helps me.

The most neg camber we can get is when the A arm is parallel to the track as this pushes the bottom of the strut the furthest from the center of the car as possible.

I believe about .3" movement in the strut top or bottom equals about 1 deg of camber change.

In a full on turn a stock suspension car leans about 4 deg including tire compression.

Thus, setting up the front to low can cost us up to 1 deg of neg camber in a full on turn it seems.

I suspect that we need the front A arms set at about 3-5 deg positive inclination to get the most dynamic camber available in a full on turn on a stock suspension car.

I defer to the experts.

[911st]

Boy did I misread those numbers.

The camber curve is very flat near parallel.

Camber should only reduce about 15/100's of a deg if the a arm is set parallel to the ground for a stock suspension 911 in a full on corner.

[petevb]

I think you're asking the wrong question.

a) If the car rolls 4 degrees the A arm does not move 4 degrees- the A arm will move through a larger angle.
b) Maximum camber is not achieved when the A arm is parallel to ground (with the chassis static). This would be the case if the top of the strut was directly above the end of the A arm (90 degrees to the A-arm). Since the strut top is inboard, max camber is achieved later, when the A-arm is 90 degrees to the strut.
c) Unless you are very camber-limited you're not shooting for maximum camber in a turn but rather camber gain.

What are you trying to figure out?

[911st]

Correct, the distance between the center of the car and the arm is 13", the arm is 10.75.

The inner hinge of the A arm lowers about .9" with a 4 deg tilt.

If the ball joint stayed at a fixed height the A arm it would tilt about 20% more than the car. However, the ball joint will also lower due to tire deformation on the load side lessening the angle some.

I suspect there is something to your point about the angle of the arm to the strut and that was my thought at one time. The end of the a arm is the furthest it can be to the strut plane at 90 deg. However, I suspect the point of the most neg camber with the car leaning is somewhere between where the strut is parallel to the ground and when it is 90 deg to the strut. After 90deg I bet the camber starts to fall off faster.

Thank you for helping me understand this better. I still have a lot to learn but I am not as concerned with my arms being close to level or a bit less.