How to REALLY calculate wheel diameter

I was looking through the owners manual for my micro computer and it states that when you calculate the wheel diameter, you can roll out the tire to get the circumference.

Yeah, okay, that’s a no brainer, however it also mentions that you
do it as you sit in the saddle. Doesn’t this in effect reduce the diameter of the tire?

Hmmm, from some people, I am sure they could care less.
Those people ought to go read another thread now.
However, I was intrigued by this. I always calculated the diameter by rolling it out without rider. I am not a physics major, in fact, I failed it twice in college and had to retake beginning calculus three times. But, i do find this stuff most intriguing.

Anyone want to take this up? Are percentages of crank arm length to diameter all off by a wee bit because everyone does the calculation without the rider on the uni?

Consider psi, especially with a Gazz 3.0, in this and weight of rider - any significant difference when taking these into consideration?

Here’s my thought. You sit on your unicycle, then get a friend to measure the tire at the bottom where it is touching the ground–that is the distance between the rim and the ground.

Take that distance and multiply by two, then add that to the diameter of the rim. Then multiply that total by pi (3.142)
to get the circumference.

Or more clearly:
circumference =
((Compressed tire measure X 2) + Rim diameter) X 3.142

I think it’s also a good idea to roll out the tyre with, say, 5 revolutions along a straight line (eg tennis court marking) and then divide that distance by 5 to minimise measurement error.

The problem is that you have ride in a perfectly straight line for 5 rotations and you have to start and stop measuring when your cranks are at exactly the same position after 5 rotations. In order to ride in a perfectly straight line, I have to already be moving. So somebody would have to measure 5 rotations while I am moving. This is very hard to do very precisely. If you took a digital video, then analyzed it you might be able to do this. You can precisely start and stop the image when the cranks reach as certain angle in the rotation. But then you have to have distance markers on the ground where you were riding, so you could measure the distance in the video. We did something like this in my physics class 10 years ago. I was a very expensive setup just to measure something very simple.

We did a quick role-out on a 24x2.5 Arrow DH knobby inflated at less than 20PSI. The diameter measured approx. 24 1/2 inches, role-out measured at:

No Rider 76.5" for calculated diameter of 24.3"

180 lb Rider 72" for calculated diameter of 22.9"

75 lb rider 74 1/2" for calculated diameter of 23.7"

That’s about a 6% difference for the heavier rider.

Re: How to REALLY calculate wheel diameter

1.) Yes, it effectively reduces the diameter of the wheel.

2.) The calculations you mention are off by a wee bit only by those who do the calculation with the rider OFF the uni. Some people already knew that the wheel diameter is affected by tire compression.

3.) Rod, are you putting a 20 year old Cateye on a MUni with a Gazz? For off-road use? Why? The footprint of a tire, the area that is in contact with the ground, is determined loosely by the tire pressure and the weight of the uni and rider. The stiffness of the sidewall comes into play at some time. The limit of compression is equal to the diameter of the tire’s cross-section when fully inflated and unloaded. At some point the rim, and not the tire, is supporting the weight. The footprint does not increase without limit just because the weight goes to infinity or the pressure goes to zero, gauge. This all suggests a problem which is extremely non-linear and difficult to solve. But you can graph the solution by sitting on the uni and measuring the ground to axle distance versus pressure as you let air out of the tire. This will be the curve for one Rod+Rod’s uni. Find a guy who, when on your uni, weighs half of a Rod+Rod’s uni (a little fellow indeed). This is the half weight curve. Then go find a real porker who weighs 2 Rod+Rod’s uni when on your uni. Then fill in the other curves for folks of all diffent weights and post your results. If it’s interesting enough to Mikefule, we can get him doing tire pressure, compression, weight curves because he’s almost done with crank ratios.

Maybe try using chalk and getting on a treadmill with the uni. don’t turn it on, but mark the start and stop. Either force it to move by peddling, or have someone turn it with their foot. Use the hand rail for balance.


I am bustin up. That was hilarious.


To your question, actually,I plan on putting it on my 28’er that arrives here on Wednesday. 90 percent of riding should be road with probably 10 percent off road, at first. I used the example of the Gazz, because it seemed the easiest to visualize in my description.

I thought compression from rider weight and psi, would play less of a role with a skinnier tire.

Like I said, all you would need is a straight line like that painted on a tennis court. You would need to mark with a piece of chalk on the tyre where it started and correspondingly on the line in the tennis court. Roll out five times with someone supporting you to keep a straight line. Then mark it off where the chalk mark hits the tennis court line on the fifth revolution. Divide that distance by five and that would be about as accurate as you can get.

Re: How to REALLY calculate wheel diameter

In article <>,
amosbatto <> wrote:
)The problem is that you have ride in a perfectly straight line for 5
)rotations and you have to start and stop measuring when your cranks are
)at exactly the same position after 5 rotations. In order to ride in a
)perfectly straight line, I have to already be moving. So somebody would
)have to measure 5 rotations while I am moving. This is very hard to do
)very precisely.

Just ride over some water or chalk on the ground; then you can measure
the marks your tire makes.

If you want to allow for the wobbles and general tyre movements then go for a ride. Ride for a distance that you can measure (use a long tape measure or a short one several times) and re-calibrate your computer.

If you like you can count pedal revs but there is no need to.

I’ve always wondered about this “riding in a straight line” bit. Surely what you’re really trying to measure is your speed between point A and point B, not the rotational speed of your wheel? That will mean that you measure your point to point time and just ignore your wobble.

If you wiggle from side to side a lot and your average wheel speed is 10 mph, then cycling 10 miles will take you more than an hour. Maybe it will take only a few minutes more, but if you’re making the effort to measure your speed then surely you’d like it to be pretty accurate. I’m sure this would have a bigger influence on accuracy than tyre compression would.

That said, the problem with the distance/time method as opposed to the wheel rotation speed is that the faster you go the less you wiggle (for me at least). So if you set your odometer whilst going at, say, 8mph then if you ride at 12mph you’ll be wobbling less and therefore your odo will be reading slightly low (or is it high, Monday morning dopeyness :roll_eyes: ).

So what was my point again? :thinking: I think I’ll stick to unicycling computerless, that way I can make totally unsubstantiated claims “I came down that track at 20mph!”:wink:

Have fun!


You guys are missing something, though. While compressing
the tire may change its diameter, it does not change its
circumference, which is the important value when calculating

Think of it this way. You can take a piece of string a
meter long, tie it in a loop, and make it any shape you want
by pushing and pulling it. But you won’t change the fact
that the string is a meter long unless you go after it with
a pair of scissors.


good point, Jon

I think i’ll ask my physics teacher about this next time i’m in class (winter break right now. yay!) because I’m rather interested in getting a computer too.

I ride a united trials uni, but because of the snow, I’ve been doing more distance rides and trails than trials (The plow guys in my neighborhood just push all the snow against my favorite ledges and obstacles :angry: ).


I think that’s wrong though, because you’re constantly compressing the tyre differently so that the middle of the contact patch is describing a circle which is smaller than that of the outside of the circle.

The important thing is to note that the contact patch is more than a point on the circle, it’s basically a line which intersects the circle at two points.

To see this, take a ruler and put both ends on the tyre and trace the circle described by the middle of the ruler, it will have a smaller diameter and circumference.

Your argument would be right if the contact patch was a single point, but that would only occur with no compression.

Have attached a picture, black is tyre, green lines are contact patches, red circle is the circle that the middle of the contact patch moves through.


Put another way there is some distance of tread that must travel across the ground for one revolution of the pedals. This amount of tread doesn’t become appreciably less for an undreinflated tire. That is interesting. The distance from the axle to the ground is less, and stays less. I think the way this is explained is that with lower pressure there is more tire scrub. This seems to be the case with cars as an underinflated tire gets hot. Not for sure though.

I might almost think this thread is bait for me… but I’ll bite. ;0)

I can’t get too excited by it, though.

For crank:wheel ratio calculations, the difference is pretty negligible, given that the ratios only give a rough guide to how a uni will perform, and there are so many other variables.

take a ‘real’ 26 inch wheel [that’s with an inflated tyre of 26 inches diameter at rest] and let the tyre down so that it deforms sufficiently to reduce the rolling radius by 1/2 inch. That knocks an inch of the diameter, reducing the wheel from a ‘real’ 26 to an ‘actual’ 25. 25 is 96.15% of 26, so you are losing less than 4% of your diameter, and therefore of your circumference.

So put, say, 150mm cranks on.
25 inches = 635 mm
26 inches = 660.4 mm

The cranks are 23.6% of the diameter of the 25 inch wheel.
The cranks are 22.7% of the diameter of the 26 inch wheel.

The difference in use will be negligible, and heavily outweighed by the difference in rolling resistance of the softer tyre!

And for speedos? On cars and motorbikes, a speedo error of 10% is not uncommon. On cycles with correctly set speedos, the error is rather less. Those unicyclists who care about speed and distance are likely to ride with their tyres hard anyway. Those who ride with soft tyres are probably more interested in what obstacles they can surmount.

And as for measuring, it seems a pity to use tape measures and chalk and so on, about 2,500 years after the discovery of Pi and the various simple calculations which use it.

Sit on a tyre,and get someone to mark two dots at the edges of the contact patch and measure the distance (around the tyre). Now get up and measure the same distance between the two dots around the tyre. Won’t you find the unweighted distance is larger?

I think you’ll find you will. I’m not sure what tire scrub is, but basically by sitting on the tyre you’re squishing up the bit of tyre that makes a curve between two points into a line between the same two points.

btw, JonM, if you don’t believe in the effect, it’s pretty easy to verify, just measure a rollout, let your tyre down / get on the unicycle and measure again.


Re: Re:


You’re picture is only half-correct. The tire doesn’t just form a straight
line contained within the diameter of the circle. It deforms outside the contact patch, as shown in the diagram I’ve attached. The black is the
ideal circle, the red is the deformed tire shape. Notice that it bulges out
to the side of the contact patch.

My contention is that, for the purposes of the original poster (measuring tire circumference for the purposes of measuring velocity), compression has little effect on the circumference of the tire. However,
it does effect the true diameter of the wheel, which is manfested in
increased torque under compression (smaller diameter == more torque).



Hmm. I tried this, and indeed, you were right. With the tire down to 10psi, I could compress it about half an inch by putting force on
the seat with my hand.

And sure enough, compressed, the rollout was about 3/4 of an inch shorter.

This doesn’t quite make sense to me, though. There’s a fixed amount of
tread on the tire. All of that tread has to pass under the wheel during a revolution.

Odd. I’m suspecting elastic effects at play. Then again, it’s been a good
7 years since I took physics in college…