why does it matter dude.
just adjust it to where it feels good. u shouldnt go off preset values.
I’m trying to work out anti-rollbar stiffness for a project toy (non Toyota). It’s 1 of those projects that I don’t really want to spend much money on, so there is going to be plenty of improvise and DIY
.
A diagram would probably help, but I’m shite on the compoota when it comes to that.
Bar diameter is 24mm and the distance from the centre of the chassis mount to the centre of the lower control arm link is 298mm.
If I raise the diameter value to the 4th power I end up with a value of 331776 (no units).
If I then divide that number by the anti-rollbar arm length of 298mm, I get a value of 1113.34 (no units).
Bar Diameter (24mm) raised to the 4th power/Anti-rollbar arm length (298mm) = 1113.34 (no units).
Does this sound/look/smell the correct way to calculate anti-rollbar stiffness values?
"Don't worry what people think, they don't do it very often."Originally Posted by oldcorollas
Daily: Glorified Taxi (F6 Typhoon). Out Of Action: Twin-charged Adub. Ongoing Nightmare: Over re-engineered (not) Alfa Romeo 75.
why does it matter dude.
just adjust it to where it feels good. u shouldnt go off preset values.
Originally Posted by RAd28
That's the thing, it's not adjustable, it's stock. I want to try and figure out what percentage increase shortening the bar arm length will have. If that made any sence![]()
"Don't worry what people think, they don't do it very often."Originally Posted by oldcorollas
Daily: Glorified Taxi (F6 Typhoon). Out Of Action: Twin-charged Adub. Ongoing Nightmare: Over re-engineered (not) Alfa Romeo 75.
Moment lb/inch = (Pi x Dia^4 x lever arm) / (16 x Dia )
All dimensions in inches
Sure helps to have an understanding, beats 'suck it and see'.
Thanks for some info, but are you sure it's correct?
Moment lb/inch = (Pi x Dia^4 x lever arm) / (16 x Dia )
All dimensions in inches
According to your formula, shortening the lever arm results in a lower spring rate![]()
3.1415 x 0.95 x 0.95 x 0.95 x 0.95 x 11.73(arm length) / (16 x 0.95) = 1.975lb/inch
3.1415 x 0.95 x 0.95 x 0.95 x 0.95 x 10.95(shorter arm length) / (16 x 0.95) = 1.843lb/inch
or am I missing something![]()
"Don't worry what people think, they don't do it very often."Originally Posted by oldcorollas
Daily: Glorified Taxi (F6 Typhoon). Out Of Action: Twin-charged Adub. Ongoing Nightmare: Over re-engineered (not) Alfa Romeo 75.
Sorry for that Plonka, the difference between r for radius and r for the greek rho is difficult to see. I will have my new glasses next week !
The lever arm should have been shear stress and that formula gives Moment at shear stress which wont help you.
The easiest way to get a relationship to stiffness is to take a set angular deflection of the lever arm, say 10 degrees, then the tangent of the angle * lever length.
The result is the % difference.
Lever arm = 10", angle = 10 deg, tangent of 10 deg = .17633. Result = 1.7633
Lever arm = 5", angle = 10 deg, tangent of 10 deg = .17633. Result = .088165
.088165/1.7633 = .5 or 50% stiffer at the same angular deflection
If you change bar Dia then D^4 is the relativity.
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