# How to calculate width of 28“ tire to 26” with constant damping factor?

Let's say you have a 28" tire with X width.

The question is how to calculate the width for 26" tire to get the same damping factor? I am looking only for solid solutions.

I will assume that calculating it backwards is simple reformulating the equation.

# Update

"Solid solution" = "I tried it, I know it, so I say it".

Damping = it is like "user experience", you ride and either it feels comfortable or not. After all I hit both kinds of obstacles, rails for example and holes in the street (Poland is famous because of them).

Tires -- what I have in mind, is exactly the same tires, same manufacturer, same type, etc. Currently I use 26" 2.35" (Fat Frank and Super Moto) tires, and I am wondering (approximately) what 28" tires would be the equivalent.

• What exactely would be a "solid" solution, or what wouldn't be a solid one? – heltonbiker Feb 27 '12 at 17:19
• Maybe I'm missing something in your question, but surely you just adjust the tire pressure to get the dampening factor you want? – Tom77 Feb 27 '12 at 17:38
• @Tom77, (1) I can adjust the tire in "original" tire as well; obviously you have to make everything else constant, otherwise you won't end with "what if..." (2) you cannot adjust pressure freely, because there is upper and lower limit, and besides, pressure affects the speed (see (1)). – greenoldman Feb 27 '12 at 17:59
• I'm guessing you'd get 2" more dampening with the 26", since you'd be 2" closer to the water on the ground. ;) (Well, actually only 1" closer, I guess.) – Daniel R Hicks Feb 27 '12 at 19:34
• damping x = (-1)^(n-1) x^(2n-1) / (2n-1)! – Daniel R Hicks Feb 28 '12 at 13:30

From my experience, dampening depends too much on non-dimensional factors. That makes very difficult to come with a formula that would work reliably.

Of course the width of the tire is a very important factor, and balloon tires always will dampen better than skinny ones.

But the hardness of the carcass (be it the layers within or even the rubber hardness itself, or the thread pattern) plays such an important role that I think it counts more than the difference in diameter from 26 to 28 inch wheel sizes.

I have ridden wide tires which were very bumpy and uncomfortable, and narrow ones that just made the harshness of the road disappear.

Also, and this might help to formulate a real answer to your question, are you referring to the transversal dampening, where the tire's cross section counts a lot, or to the longitudinal dampening, where the wheel diameter counts more - and which is responsible for "rolling over" obstacles being smoother on bigger wheels - ?

As a very first "thumb suck" I'd guess your "damping effect" is roughly proportional to the "footprint". And that is controlled by air pressure and tire stiffness more than tire diameter/width.

In theory, as tire diameter goes up, pressure can reduce slightly for the same tire width and still maintain the same vertical compression ("squat") of the tire. Or, alternatively, the "spot" will become more oblong (in the direction of travel) for a given pressure as tire diameter increases. But this effect is slight -- if you worked out the formula for a chord of a circle I'd guess it's only a few %.

Of course, another thing to think about is that the larger tire can span minor road imperfections better than the smaller one. Though again the effect is slight given the small difference between the two sizes.

It is not solid solution which I asked for, but since there are no better answers at the time of writing this should fit.

If you spot here an error or over-simplication let me know.

I will assume tire is torus-shaped. It is not true, but it has to stay, because real shape is to computational expensive to handle.

# Comfort

## Crossing holes

Like crossing the rails, here diameter matters.

``````26 + 2 * w26 = 28+2*w28
``````

So you get `w28 = w26-1` (translating into English -- you can have width 1" narrower than with 26" tires to get the same effect of rolling over holes).

## Crossing obstacles

Another assumption -- when we hit an obstacle only pressure is changed, and the force from the tire material is negligible. Not even going to formula for force of changing pressure of the air, it is sufficient to notice we are not seeking the formula, we are seeking the same effect for 26" and 28". So the only factor that matters here is the volume of the tires -- it has to be identical to get identical effect.

Volume of torus: `2 * Pi ^2 * R * r ^ 2`, here `r` is half of the width of the tire, and R is half of the size of the wheel.

``````2 * Pi ^ 2 * 26"/2 * (w26/2)^2 = 2 * Pi ^ 2 * 28"/2 * (w28/2)^2
26" * w26^2 = 28" * w28^2
sqrt(26/28) * w26 = w28
``````

The scaling factor is here ~ .96, this is more tighter constraint than the first one.

## Conclusion

For above 2" width tire (my case), I can have 0.1 narrower tire in 28" to get the same feeling when riding.

# The effort

If I want to ride comfortably I have to consider air resistance.

``````resistance ratio = w28*(28+2*w28) / w26*(26+2*w26)
``````

It tells how much resistance will change:

``````= sqrt(26/28) * w26 *(28+2*sqrt(26/28) * w26) / w26*(26+2*w26)
= sqrt(26/28)*(28+2*sqrt(26/28)*w26) / (26+2*w26)
= sqrt(26/28)*28+2*(26/28)*w26 / (26+2*w26)
= sqrt(26*28)+(52/28)*w26 / (26+2*w26)
=~ 26.98+1.86*w26 / 26+2*w26
``````

For 26", 2.35" tire the ratio would be 1.02.

Please note there is constant speed assumed. You usually go for 28" tires to get more speed, and air resistance is proportional to the square of the speed!

Now, for gusts of the wind from the side:

``````resistance ratio
= 2*Pi*(28/2+w28)^2-2*Pi*(28/2)^2
/ 2*Pi*(26/2+w26)^2-2*Pi*(26/2)^2
``````

I don't count for spokes, fork, they are important factors, but it will make computations even more complex (and besides, the effect will be different for blade spokes, and for regular).

``````= (14+w28)^2-14^2 / (13+w26)^2-13^2
= (14+w28+14)(14+w28-14) / (13+w26+13)(13+w26-13)
= (28+w28)*w28 / (26+w26)*w26
= 28*w28+w28^2 / 26*w26+w26^2
= 28*sqrt(26/28)*w26+(sqrt(26/28)*w26)^2 / 26*w26+w26^2
= 28*sqrt(26/28)*w26+(26/28)*w26^2 / 26*w26+w26^2
= 28*sqrt(26/28)+(26/28)*w26 / 26+w26
=~ 26.98+0.93*w26 / 26+w26
``````

Let's say `w26=2.35`. The ratio is 1.03 (a bit of surprise).

So, there is no free lunch after all ;-) However if you forgive discomfort when hitting obstacles, you can build 28" bike faster in regard of air resistance and yet the one which rolls over with much more ease than 26".

## The weight

Because of the difference of the radius, even the same weight is more harder to to move in case of 28". I will calculate this factor some other time.

## My choice

If I go for 28" bike (I didn't yet decide) I will change my 26" 2.35" tires to 28" 2" -- they should be faster (~0.9 less resistance), they will go more easily over the holes, and they will be just a bit stiffer when hitting an obstacle.

I thank Slovakov from BikeForum.pl for spotting the errors in equations. All still existing errors are only mine of course ;-).

• I have no idea where you get that first equation from -- it's totally bogus. (Easily demonstrated by making the 26" tire 1" wide.) – Daniel R Hicks Mar 3 '12 at 14:26
• @Daniel R Hicks, it depends on assumption. If you agree that tire can be approximated by torus, you have to calculate total height of the tire which is size of the tire + doubled tire "height" (=width; see torus assumption). And the outcome is actually solid, and easily provable -- with the exactly same rims and tires, there is 2 inches difference between the rims + tires. – greenoldman Mar 4 '12 at 7:19
• "you can have width 1" narrower than with 26" tires to get the same effect of rolling over holes" -- So if I have 26x1" tire then I can switch to a 28x0" tire?? – Daniel R Hicks Mar 4 '12 at 13:08