Mcpherson 1

[modificater]

What geometries effect the bump camber curve of mcpherson struts?


It looks like control arm length (the radius it travels) in relationship to the top strut mount?

 

[modificater]

"The A-arm to leg angle both set the rate and the + or - camber effect."

Relative to each other? At some point as stated above they change signs, if you're at dead center of the sign change, is that the point in the A-arm swing radius that for a given leg/arm length ratio and angle that the camber change rate is the most accelerated?

 

[GregLocock]

Look, you need to draw it out. It's either that or we draw it.

Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[modificater]

Ouch...

I did, I was trying to verify that I did it correctly. I'll take it elsewhere.

"On a paper model it looks like a horizontal (with respect to the ground) control arm is one that will gain and lose camber at the fastest rate. Am I correct?"

 

[hemipanter]

When the A-arm is at 90 dgr angle to the leg you will have the slowest rate of camber change, as the balljoint is moving in the direction of tangent (same as leg direction). In order to create chamber change the lower leg balljoint must move sideways.
So, the longer the leg, the greater the Tw change vs camber effect. Tw=wheel sideway movement.
This also means that at 90 dgr you will have the max -camber. So, if the starting point is paralell to ground A-arm and 10 dgr leg tilt, the camber will go - (during bump) until 90 dgr leg to A-arm angle and then go +.
Goran

 

[modificater]

Thank you Goran