I tried to search this out first

https://honda-tech.com/zerothread?id=1174419&page=2

And while it was helpful, it didn't answer my question...

I am going to be *experimenting* with the bigger Si front bar on my new 2006 Civic LX (H-Stock).

While this is something I would never normally consider on a FWD car, Nate Whipple made some compelling statements regarding his "theories" success on Chang's H-Stock Civic being consistently quicker with a bigger front bar.

So... All non-Si Coupes come with 25.4 x 3.5mm hollow front sway bars.

I will be moving to a solid 28mm front bar from the Si.

These formulas really don't take hollow vs. solid into consideration. My OEM hollow bar would equate to what diameter if it were solid? I am guessing 21mm - 22mm.

 

Most of the torsional resistance or bar rate comes from the steel at the outer diameter of a solid bar anyway. The biggest difference between a solid and a hollow bar of the same diameter is the weight. I'm sure there's an engineer here somewhere who can show us the math, but that's the layman's explanation.

Thawley -- Mr. Layman

 

You'd need to know the hollow bar's wall thickness to calculate the equivalent diameter of a solid bar.

 

to find the difference, calculate for a solid bar using all the parameters, etc. and get your result in pounds, etc.

then calculate for a solid bar the size of the hollow section and get a rate in pounds.

then subtract the hollow section from the complete section and that is your effective rate for the hollow bar.

then play some trial and error or algebra to back calculate the diameter of an equivalent solid bar.

thawley is right, the vast majority of the rate is due to the outer section of the bar.

i really don't think that a couple of pounds is too much of difference though. a good rule of thumb is 500 pounds per cubic foot for steel.

so if you a 1/4" "hole" in the center of your bar and it is 6 feet long total, that is only about one pound.

tom

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by WRXRacer111 &raquo;</TD></TR><TR><TD CLASS="quote">You'd need to know the hollow bar's wall thickness to calculate the equivalent diameter of a solid bar. </TD></TR></TABLE>


<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by honda93 &raquo;</TD></TR><TR><TD CLASS="quote">25.4 x 3.5mm</TD></TR></TABLE>

 

you would actually need to calculate moments of inertia to find the exact "equivalent"

 

http://www.whiteline.com.au/do...r.pdf

Hollow versus Solid Swaybars
Comparison of the 2 separate design strategies and how they differ in
practical use and manufacture.

front swaybars are hard to calculate exactly because of all the bends, but you can get an approximation if you assume it to be straight and measure between the bushing and endlink geometry.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by honda93 &raquo;</TD></TR><TR><TD CLASS="quote">I tried to search this out first

https://honda-tech.com/zerothread?id=1174419&page=2

And while it was helpful, it didn't answer my question...

I am going to be *experimenting* with the bigger Si front bar on my new 2006 Civic LX (H-Stock).

While this is something I would never normally consider on a FWD car, Nate Whipple made some compelling statements regarding his "theories" success on Chang's H-Stock Civic being consistently quicker with a bigger front bar.

So... All non-Si Coupes come with 25.4 x 3.5mm hollow front sway bars.

I will be moving to a solid 28mm front bar from the Si.

These formulas really don't take hollow vs. solid into consideration. My OEM hollow bar would equate to what diameter if it were solid? I am guessing 21mm - 22mm. </TD></TR></TABLE> Your hollow bar is equivalent to a 23.4mm solid bar. A 28mm solid bar will be about twice as stiff. To compare bars just use (Do^4 - Di^4)

 

that is not correct.

you need to account for moments of inertia like i said.
a larger hollow bar is equivalent to some smaller solid bar.

you can see that that paper the other guy posted references I (moments of inertia) . units(mm^4)

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by tjbizzo &raquo;</TD></TR><TR><TD CLASS="quote"> Your hollow bar is equivalent to a 23.4mm solid bar. A 28mm solid bar will be about twice as stiff. To compare bars just use (Do^4 - Di^4)

</TD></TR></TABLE>

That is correct by my reckoning also about the 23.4 mm bar being equivalent to the 25.4 mm hollow with a 3.5 mm wall as both have a moment of inertia of 0.071 in^4 using J=pi*(Do^4-Di^4)/32 for the moment of inertia and D in inches (you can just use the same formula with D in mm and get the answer in mm^4).

The 28 mm bar will have a moment of inertia of 0.145 in^4, so it will be just slightly more than twice as stiff as the first two. We have to assume the arm stiffness is the same, but in reality it is not, as the arms flex, and act like a spring in series with the center section. It will also unfortunately be heavier, and of course move the car more towards understeer than before. You may not like that in the end.

 

i just dont see these formulae allowing room for the mass of each small section of area.

you guys are doing area moments and not mass moments which are what is called for here.

also you should be dividing by 4 not 32 if doing an area moment


Modified by dfoxengr at 6:28 PM 5/2/2006

 

http://en.wikipedia.org/wiki/L...ertia
use the mass ones (with an m in it)

here. wiki is awesome.
-derek

 

Thank you for all the replies. I was merely looking for approximations and didn't want anyone to get upset over the math involved!

I am fully aware of the chance that the car may understeer, but I can compensate with toe settings, and when available, shock settings.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by dfoxengr &raquo;</TD></TR><TR><TD CLASS="quote">i just dont see these formulae allowing room for the mass of each small section of area.

you guys are doing area moments and not mass moments which are what is called for here.

also you should be dividing by 4 not 32 if doing an area moment


Modified by dfoxengr at 6:28 PM 5/2/2006</TD></TR></TABLE>

Strictly speaking, yes the car's suspension is a dynamical system where inertial effects would need to be included for proper analysis. However, the questions asked in this thread have to do with comparative torsional stiffness of either solid or hollow anti-roll bars and thus, the polar area moment of inertia suffices for this discussion. The reason why they can be compared is simply a matter of looking at the equations for both I (polar area moment of inertial) and J (moment of inertia)

J = pi/32(Do^4 - Di^4)*rho*length
I = pi/32(Do^4 - Di^4)

rho = material mass density
length = length of shaft
Do = outside diameter
Di = inside diameter

Since comparisons in this thread were for anti-rollbars made from steel and of the same length, then using I to compare two anti-rollbars used in a dynamical system is perfectly alright.

 

all the formulas im seeing have the denominator as 4, wheres the 32 coming from?

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by dfoxengr &raquo;</TD></TR><TR><TD CLASS="quote">http://en.wikipedia.org/wiki/L...ertia
use the mass ones (with an m in it)

here. wiki is awesome.
-derek</TD></TR></TABLE> The one with mass, I = mr^2, is used for calculating the rotational inertia of a spinning cylinder, i.e. the *flywheel* effect of the torsion bar as it rotates back and forth a few degrees. While I'm sure that it must have some measurable value, I'm equally sure that it would be completely insignificant relative to the rest of the unsprung weight of the suspension.

The formula for the Area moment of inertia, I = pi/2(r<FONT SIZE="1">o</FONT>^4 - r<FONT SIZE="1">i</FONT>^4), is the one used to calculate a torsion bar's resistance to a twisting force. Assuming that he isn't changing the length of the blades or the length of the bar itself, it is perfectly valid to use just the (r<FONT SIZE="1">o</FONT>^4 - r<FONT SIZE="1">i</FONT>^4) part of the formula to compare the relative stiffness of bars with different OD's and ID's.

Edit: oops, I originally had I = pi/4(blah blah blah) but that is for the area moment of inertia of a disc whose normal axis is perpendicular to the axis of rotation, like a throttle butterfly. pi/2 is correct for the polar area moment of inertia along the central axis of a shaft.



<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by dfoxengr &raquo;</TD></TR><TR><TD CLASS="quote">all the formulas im seeing have the denominator as 4, wheres the 32 coming from?</TD></TR></TABLE> d = 2r, so (d^4)/32 = (r^4)/2.



<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by dfoxengr &raquo;</TD></TR><TR><TD CLASS="quote">nobody was talking about mr^2

i was saying use the second two shapes under the mass section, both Ix </TD></TR></TABLE>Again, anything with m in it is a MASS moment of inertia, a kinematic rotational inertia like a flywheel, NOT what we want to use to calculate a torsion bar's resistance to twisting. To calculate that we need to use the AREA moment of inertia of a round shaft about its central axis, which is I = pi/2(r<FONT SIZE="1">o</FONT>^4 - r<FONT SIZE="1">i</FONT>^4). This comes straight out of my old Mechanics of Materials textbook and I'm 99.9% sure that I'm reading it correctly. Can I get an AMEN and a backup/formula check from one of the other engineers here?




Modified by tjbizzo at 9:39 AM 5/3/2006

 

nobody was talking about mr^2

i was saying use the second two shapes under the mass section, both Ix

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by dfoxengr &raquo;</TD></TR><TR><TD CLASS="quote">nobody was talking about mr^2

i was saying use the second two shapes under the mass section, both Ix </TD></TR></TABLE>
I edited a mistake in my previous post and also answered this one. Please see above ^^^^

 

yeah im looking in my statics book right now and the denominator is 4 when about
x and y axes for bending moments

about the z axis, or for torsional forces it is over 2 though. youre right.

again you can look at areas even on wiki if you like.

 

Debating mathematical formulae is fun and exciting

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Tyson &raquo;</TD></TR><TR><TD CLASS="quote">http://www.whiteline.com.au/do...r.pdf
</TD></TR></TABLE>

http://www.hotchkistuning.com/...D.pdf

what both links say:

If a hollow bar and solid bar have the same OD, the solid bar will be "stiffer"

If you took a hollow bar and increase the ID of the bar, you will make the bar "stiffer" while retaining the same OD.

---------------------------------

Whiteline says hollow bars will have more stress loads placed on them which can cause them to fail or wear out

Hotchikis says hollow bars and solid bars will undergo the same amount of stress levels

I don't know if you guys covered that stuff already. I skipped the math stuff

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by thawley &raquo;</TD></TR><TR><TD CLASS="quote">Debating mathematical formulae is fun and exciting </TD></TR></TABLE>

yeah engineers are really gay. i get to know that every day even more as i attend class. and slowly i become one

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by azian21485 &raquo;</TD></TR><TR><TD CLASS="quote">
Whiteline says hollow bars will have more stress loads placed on them which can cause them to fail or wear out</TD></TR></TABLE>

I consider their statement to be rather bogus and basically false. My NSX comes stock from the factory with both front and rear hollow anti-roll bars (however my ITR came with solid bars, which are cheaper). As long as the applied stress on any component is held within elastic limits, the component will not fail (well at least not for a very long while). Whiteline is exaggerating the problem to make the point they want to make. People have been using hollow sway bars on race cars for a long time, and they work fine. They are lighter weight by a wide margin than solid bars of the same stiffness, and that is why people use them. They cost more to make, in particular if you have to bend them, but they are better because they are lighter.

Here for example are some track hollow sway bars for an NSX: http://www.daliracing.com/v666...=1407
Here is a link showing the say bar characteristics on an NSX Type R from Honda:http://world.honda.com/NSX/technology/t4.html

I think Honda knows what they are doing in terms of engineering a sway bar and keeping the stress under control in a hollow sway bar.

Here is a link to some racing hollow sway bars from Speedway Engineering:http://www.1speedway.com/swaybar32.htm