Motorhome question, sorry.

I'm looking at replacing the antisway bars on my motorhome. Stock, it has 1 1/2" bars front and rear.

Hellwig makes a 1 5/8" front bar ($381) Expressed as a percentage, how much stiffer is that?

Roadmaster makes a 1 3/4" (They want a king's ranson for it too :( ) how much stiffer?

On the back, Hellwig sells a 1 5/8" bar ($582) which installs WITH the existing Factory 1 1/2" bar. (I'd guess more than twice as stiff, depending on the length of the arms?

Roadmaster goes with a 1 1/2" on the back, also an addon not a replacment bar.

So... are either a reasonably effective change in front, both cost / benefit?
The Roadmaster bar just flat costs too much regardless of the benefit.
The Hellwig bar is more reasonably priced, but at only 1/8" larger than stock, how much benefit would there be?

I don't know why I'd even think about "balance" with a 36 foot long, ten ton shoebox on wheels, but it's interesting that roadmaster incrfeases the front roll stiffness proportionally much more than hellwig does?

and for those who will spend hours searching the web for a dirt cheap set of perfect swaybars, this is a 1999 Ford F53 chassis, 20,500 GVRW, large rear axle, V10 engine.

Thanks.

Interesting. I just found the roadmaster front bar for $549! Hmmm.

Its expressed as a percentage due to the larger area of the rod itself. If a tube, leverage of the larger diameter.

If I remember correctly, materials ability to withstand torsion (for a round bar) is

I = Pi/64 * d^4

1 1/2" = 1 >>> 1 5/8" == 1.37 so around 40% more stiffer
1 1/2" = 1 >>> 1 3/4" == 1.84 = 85% stiffer if the material is the same




BTW, these could be way off, this was just something I think I remembered from my college days material engineering lectures.

If I remember correctly, materials ability to withstand torsion (for a round bar) is

I = Pi/64 * d^4

1 1/2" = 1 >>> 1 5/8" == 1.37 so around 40% more stiffer
1 1/2" = 1 >>> 1 3/4" == 1.84 = 85% stiffer if the material is the same




BTW, these could be way off, this was just something I think I remembered from my college days material engineering lectures.

That's what my calculator says, provided that the lever arms are the same length and the construction is the same ( not hollow, etc ). The stiffness is related to the 4th power of the diameter.

Here is an online calculator: http://buildafastercar.com/tech/Sway-Bar-Rate-Calculator
It's best to refresh the page after a calculation or two or it may start giving out wrong numbers.

I Have visited Hellwig, and can attest that they only use crome molly material and properly heat treat them in-house. They buy cro-mo steel by the box car load!
A small increase in diameter is worth a lot in stiffness, if the arms are the same length. Mounting stiffness is important too. However on such a huge vehicle I don't know if it would be noticed.

Air bags can help too.
David

Thanks guys. 40% and 85% it is.
Now I jus gotta save some peso's and upgrade this behemoth before our next long trip.

It's completely scary what crosswinds do to this thing, or passing trucks, or even "body roll induced self steering", eeeek.

An extra set of shocks would also help. The springs on motor coaches are softer than for the same GVW used on a HD truck. In case you have not noticed the maximum GVW and the tare or empty weight is not that far apart on a motorhome.

Larger Dave

If I remember correctly, materials ability to withstand torsion (for a round bar) is

I = Pi/64 * d^4

1 1/2" = 1 >>> 1 5/8" == 1.37 so around 40% more stiffer
1 1/2" = 1 >>> 1 3/4" == 1.84 = 85% stiffer if the material is the same




BTW, these could be way off, this was just something I think I remembered from my college days material engineering lectures.


I agree w/the above as long as it is a solid, uniform cross section. I just went about it slightly different, using the polar moment of inertia, J which is

(pi*d^4)/32

1.5" bar, J= .497"^4
1.625" bar, J= .726"^4
1.75" bar, J = .921"^4

1 5/8 bar is 37.7% stiffer, (.726/.497)= 1.377
1 3/4 bar is 85.3% stiffer, (.921/.726)= 1.853

now if you wanted to take it a step further to see how much each one would actually deflect under a load....


deformation = (P*L)/(A*E)

where P is the load on the bar, L is its length, A is the cross sectional area and E is the modulus of elasticity. E for structural A36 steel is 29*10^3 ksi.