Front Anti-Roll Bar Rates and Torsion Bar Sizes to Wheel Rates.

[Duk]

I'm a bit bored, so I thought I'd quickly crunch the numbers for some of the popular front anti-roll bars sizes. There is no unit of measurement, just a numerical comparison. The wheel rates of popular torsion bars are also here.

Solid Bars:
22mm Bar =    234256
24mm Bar =    331776
27mm Bar =    531441
29mm Bar =    707281
30mm Bar =    810000

Tubular Bars:
1.125" x 0.120" =   380400 So 14.6% stiffer than a 24mm AR bar.
1.375" x 0.058" =  442030 So 33% stiffer than a 24mm AR bar.
1.250" x 0.095" =  490707 So 48% stiffer than a 24mm AR bar.
1.250" x 0.120" =  532896 So 61% stiffer than a 24mm AR bar.
1.375" x 0.095" =  667059 So 101% stiffer than a 24mm AR bar.

As can be seen in with these numbers, the numbers go up very quickly with diameter. A 30mm bar is 2.44 time stiffer than the 24mm bar.

Tubular bars can achieve decent increases in stiffness, but without the weight penalty. You'd be flat out getting a 1.250" diameter bar inside the original saddle clamps. With a 0.120" wall thickness, you would have a bar with a rate very similar to a 27mm solid bar, but with significantly less weight.

Popular Torsion Bar Sizes:

22.8mm TB has a WR of 93lb/in.
25.4mm TB has a WR of 143lb/in.
27.3mm TB has a WR of 191lb/in.
28.7mm TB has a WR of 233lb/in.
30.0mm TB has a WR of 279lb/in.
32.0mm TB has a WR of 360lb/in.


*** TB = Torsion Bar ***
*** WR = Wheel Rate ***

[Duk]

While I'm at it:

The motion ratio of the front damper is 38.67%. So if you add a spring over the front damper, you get 38.67% of its force transferred to the wheel

The standard wheel rate would equal a 242lb/in coil spring over the damper.

Standard TB + 253lb/in spring = WR of 191lb/in (equivalent to a 27.3mm TB)
Standard TB + 360lb/in spring = WR of 233lb/in (equivalent to a 28.7mm TB)
Standard TB + 481lb/in spring = WR of 279lb/in (equivalent to a 30.0mm TB)
Standard TB + 690lb/in spring = WR of 360lb/in (equivalent to a 32.0mm TB)


*** TB = Torsion Bar ***
*** WR = Wheel Rate ***

[djm411]

Hi Duk,
With respect to the torsion bar rates, have these come from Alfa, or measured values? I only ask as i assume that the different machine shops could use different property steels and heat treatment resulting in different spring rates?

Thanks

[Duk]

Hi Duk,
With respect to the torsion bar rates, have these come from Alfa, or measured values? I only ask as i assume that the different machine shops could use different property steels and heat treatment resulting in different spring rates?

Thanks

That's a fair point about different steel properties giving different spring/wheel rates. Ultimately there are only a few grades of steel that are suitable to be used as a torsion bar for long periods of time and their behavior should be be very similar.
Same said for heat treatment. Treat it to hard and it snaps. To soft and it maybe to weak or even take a permanent set

Some of the results were pinched from the AlfaBB and I calculated a few more in the list. I recon the non Alfa sizes are just calculated off of measurements from the the standard TB.

[djm411]

Just wanted to confirm the definitions. WR is wheel rate, calculated at the contact patch of the tyre. This essentially means the upwards force on the tyre required to move the wheel up a certain distance inch, hence lb/inch.

Motion ratio is necessary as the force exerted by the spring does not act directly through the tyre but a short distance from the tyre. Hence why a ratio is necessary

Are these definitions correct??

Reason for my last question was that I tested a set of 30mm bars and they came out with a rate of 500lb/in, measured at the bottom ball joint.

[Jekyll and Hyde]

Reason for my last question was that I tested a set of 30mm bars and they came out with a rate of 500lb/in, measured at the bottom ball joint.

I assume the swaybar was disconnected for that test?  500lb/in sounds spine shatteringly high, in my experience 30mm torsion bars certainly aren't that hard (cross referenced to ride I've experienced in other makes of car with known spring rates, I'd say Duk's figures are reasonably close...)

[Duk]

Just wanted to confirm the definitions. WR is wheel rate, calculated at the contact patch of the tyre. This essentially means the upwards force on the tyre required to move the wheel up a certain distance inch, hence lb/inch.

Motion ratio is necessary as the force exerted by the spring does not act directly through the tyre but a short distance from the tyre. Hence why a ratio is necessary

Are these definitions correct??

Reason for my last question was that I tested a set of 30mm bars and they came out with a rate of 500lb/in, measured at the bottom ball joint.

Yes, that is correct.

500 lb/in for a 30mm TB is very high!

The formula for calculating wheel rate for a torsion bar size is based on the measured (by more than 1 person) standard torsion bar wheel rate:

93(lb/in)/(22.8(TB diameter)^4) or 93/(22.8 x 22.8 x 22.8 x 22. = 0.000344146
                                                                                     ^ (LOL!  )
New WR: 0.000344146 x New TB Diameter^4                 

Or: 0.000344146 x 30^4
= 0.000344146 x (30 x 30 x 30 x 30)
= 0.000344146 x 810,000
= 278.76 lb/in

[djm411]

Agree that number does seem very high, and yes the front sway bar was disconnected. Rear spring rate it 205 lb/inch so doesn't really match. Will do another check. By the way, this was done using a jack and a load cell with the car secured to the hoist so it didn't go up when the car was jacked the ball joint.

Why is the diameter to the power 4? Assume your looking at cross sectional area

[Darryl]

Why is the diameter to the power 4? Assume your looking at cross sectional area

It is not cross sectional area. To understand this intuitively, consider that if you make a tube (rather than solid rod) out of the same amount of material you make the diameter bigger but keep cross sectional area the same - and of course the result is stiffer.

Taking that thought experiment further, imagine a tube of radius R that is very thin walled compared to the radius - and you are trying to twist it by applying force to the tube wall (ie at a radius R also). that will then mean you need to apply R^2 * 2pi * R * thickness of force to twist it a given angle (force in mystery units - multiply by material dependent constant to get it in something useful).

Now consider that you have a huge number of very thin such tubes stacked inside each other. The tubes are stuck together to form a solid cylinder. Each tube applies force to the next one in so....

Call the thickness of each tube dr. and add up (integrate) for all those tubes, all of thickness dr but of different radius (R, R-dr, R - 2dr etc) to get the polar moment of inertia of a rod J = pi/2 * R^4.

So if you want to calc for a hollow tube of inner diameter d1 and outer d2, use the equation Duk gave for each diameter and subtract inner from outer to get the result so eg a 30mm OD, 24mm ID bar (ie 3mm wall thickness) would give in the same mystery units 810000 - 331776 = 478224 or about the same as a 25mm bar but with a mass ration of (30^2-24^2)/25^2 or about 1/2 the weight.

[djm411]

The reason for my question was based on the fact that part of the discussion was on radius and part was on diameter. I then realized that pi/2 (radius) or pi/32 (diameter) can be removed when figuring a ratio out.

Thanks Darryl for the explanation. Should the last formula be (od^4-id^4)/od or (30^4-24^4)/30 or were you simply determining the weight ratio comparison?

[Darryl]

Just the weight ratio (not "ration" - don't know where that n came from!).

With the rest of the steps left in and replacing rough in-head attempts at 4th roots the exact maths for the hypothetical 30mm OS 3mm wall tube is just....

OD^4 - ID^4 = 30^4 - 24^4 = 478224

and taking the 4th root of that gives us the equivalent solid bar which is actually 26.3mm when I bother to use a calculator...

So really the weight ratio is  (30^2-24^2)/26^2 = 0.48. Not bad...

Of course, you can also make the bars stiffer and lighter if you can figure out a way to make them shorter (and attach them)...