So I have a bicycle computer that keeps track of my miles and speed. I changed out the tires with slightly different tire size than the previous.

I need to change the circumference setting. to match tire size, in the computer so it measures accurately. What I do is to place a mark on the pavement. I then try to push the bike as straight .as possible until one full rotation of the wheel. I then measure the distance between the marks to get the circumference. This seems to be the best way to get a rough measurement.

Later I go for a ride with my GPS. Or ride over a measured mile to fine tune the setting. It has been a while since I have done this and trying to remember what I do with that setting say if the GPS measures shorter than the computer reading.

I am thinking in that case the setting is too large. In other words the computer thinks the wheel is bigger than it really is so I need to adjust the number down.

Is this correct? For some reason I always have trouble wrapping my head around this kind of stuff.

 

Reduce your computer tire measured distance accordingly to match the actual distance.
Max.

 

So I have a bicycle computer that keeps track of my miles and speed. I changed out the tires with slightly different tire size than the previous.

I need to change the circumference setting. to match tire size, in the computer so it measures accurately. What I do is to place a mark on the pavement. I then try to push the bike as straight .as possible until one full rotation of the wheel. I then measure the distance between the marks to get the circumference. This seems to be the best way to get a rough measurement.

Later I go for a ride with my GPS. Or ride over a measured mile to fine tune the setting. It has been a while since I have done this and trying to remember what I do with that setting say if the GPS measures shorter than the computer reading.

I am thinking in that case the setting is too large. In other words the computer thinks the wheel is bigger than it really is so I need to adjust the number down.

Is this correct? For some reason I always have trouble wrapping my head around this kind of stuff.

You’ve got it. The error distance is proportional to the setting error. If you say the wheel is 5% larger than it really is, you’ll appear to travel 5% farther.

I’ve found that it’s pretty tough to get below a 1% difference between GPS and a bike odometer. You need a lot of miles and a consistent difference in one direction before you can be sure it’s not just the inevitable difference you’ll get by using different methods.

 

You’ve got it. The error distance is proportional to the setting error. If you say the wheel is 5% larger than it really is, you’ll appear to travel 5% farther.

I’ve found that it’s pretty tough to get below a 1% difference between GPS and a bike odometer. You need a lot of miles and a consistent difference in one direction before you can be sure it’s not just the inevitable difference you’ll get by using different methods.

Yeah, nearly impossible to get an accurate measurement, simply with trying to ride in a straight line.

What POs me is I have religiously recorded my mileage over the years in a spreadsheet. For some reason I have been getting lazy over the past couple of years and not recording. Well my battery went dead on the computer and so my mileage got wiped out. I think I am up around 23K miles but I am likely off by 100 miles or less one way or the other.

 

Yeah, nearly impossible to get an accurate measurement, simply with trying to ride in a straight line.

What POs me is I have religiously recorded my mileage over the years in a spreadsheet. For some reason I have been getting lazy over the past couple of years and not recording. Well my battery went dead on the computer and so my mileage got wiped out. I think I am up around 23K miles but I am likely off by 100 miles or less one way or the other.

I use Runkeeper now for my logging app. It does both the bike and running for me. After that I got a Garmin watch that does those also and even my swimming, along with sleep and heart rate. I abandoned my spreadsheet years ago.

 

Which speedometer do you have?
My Supercycle speedo can be programmed for any size wheel.
Circumference = pi x diameter.

 

Which speedometer do you have?
My Supercycle speedo can be programmed for any size wheel.
Circumference = pi x diameter.

I have the CtEye and I cab set the circumference. That was not my question.

 

I would have thought the manual of your Cateye tire sizes would have agreed with this website.

 

The way I have done it, I measure the diameter of the wheel and calculate the circumference.
I enter that value into the speedo. Then I ride around over calibrated distances and see if there is agreement. Then I adjust the setting if needed.

 

The way I have done it, I measure the diameter of the wheel and calculate the circumference.
.

Do you have a cloth tailors tape, if so you could measure the circumference more accurately maybe?
Max.

 

Static measuring is a waste of time. Start with the nominal value and then measure the percent error out on the road. Adjust the setting based on the percent error. Repeat if needed. In my experience you won't need a second adjustment until the tires age substantially, the pressure changes, you get fatter, and so on. Maybe not even then.

 

First, I don't believe the average user would "test the accuracy" of their odometer.

I looked at one tire in the mix, a 26 inch rim with a 1.5 inch tire. The user manual, the website I quoted, and the calculations all were different. Then I thought, well, we missed Mars once because of a conversion error ... why not miss something again.

here is the table ....



This shows the CatEye would exceed the measured mile by a little over 105 revolutions or about 800 feet or about 0.15 miles.

But then, is the measured mile, assuming a physical measurement by the government, really a mile? Is there correlation between the measured mile and the GPS?

Of course, this assumes no one is jamming the GPS at the times of the measurements.

Three sources of errors:
- the measured mile by the government
- GPS
- CatEye odometer

 

I would have thought the manual of your Cateye tire sizes would have agreed with this website.

No every tire size is a little different. Width, tread are all contributing factors to the size difference.

 

Static measuring is a waste of time. Start with the nominal value and then measure the percent error out on the road. Adjust the setting based on the percent error. Repeat if needed. In my experience you won't need a second adjustment until the tires age substantially, the pressure changes, you get fatter, and so on. Maybe not even then.

I never adjusted for wear but probably a good idea to at least check the accuracy ever once in a while.

 

No every tire size is a little different. Width, tread are all contributing factors to the size difference.

I guess that is why they have the sizes on the tire. We could also throw the air pressure and your weight into the circumference measurements if your going for the knat's ass accuracy.

One eight of an inch difference is only 3.175 mm different in the calculations. That would only be about 7 feet per mile difference.

I chose only one tire to look at, the 1.5 inch. The error existed on the 20 inch rims and 26 inch rims with a 1.5 inch tire.

How far from the calculated value was your measured value?

You could simply multiply the circumference by pi and enter that value into the CatEye. Do your measured mile test and compare results.

The difference in the mm circumference should equal the error your seeing. Break out your spreadsheet and have some fun with numbers.

 

The best measurement technique I’ve found is to roll over a line of ink, paint, whiteout, whatever and look for the spot the tire lays down on the next rotation(s).

At least this method uses a rider on a bike. But it’s still hard to get a precise mark.

 

I guess that is why they have the sizes on the tire. We could also throw the air pressure and your weight into the circumference measurements if your going for the knat's ass accuracy.

One eight of an inch difference is only 3.175 mm different in the calculations. That would only be about 7 feet per mile difference.

I chose only one tire to look at, the 1.5 inch. The error existed on the 20 inch rims and 26 inch rims with a 1.5 inch tire.

How far from the calculated value was your measured value?

You could simply multiply the circumference by pi and enter that value into the CatEye. Do your measured mile test and compare results.

The difference in the mm circumference should equal the error your seeing. Break out your spreadsheet and have some fun with numbers.

You mean multiply diameter by pi.

 

I went out for a 20 mile ride. I wish I had written the numbers down because I got myself messed up by using the wrong funtion in calculating the correction. I am pretty sure I was at 10.02 miles on the GPS and 9.54 miles on the odometer but not certain. But I was pretty darn close with my initial rough measurement.

Anyway I must have messed up the correction because it was wrong the other way on the way back. Then I got all screwed up on the way back.

I used Change of Percent.
https://www.calculatorsoup.com/calculators/algebra/percent-change-calculator.php

What I don't understand about this calculator is that you get different numbers depending on which on is entered first. I would think you should get the same when taking their absolute value.

In my case
V1 = 9.54
V2 = 10.02

Answer: 5.03145% increase

V1 = 10.02
V2 = 9.54

Answer: 4.79042% decrease


Not much of a difference but it is there.


and I think I should have used Percentage Difference.

V1 = 9.54
V2 = 10.02

Answer: 4.90798% difference

V1 = 10.02
V2 = 9.54

Answer: 4.90798% difference

Here they are the same. And I don't have this direct function on my calculator (orr at least I don't think so )

 

You mean multiply diameter by pi.

That's what I get for thinking at that time in the AM. Yeah, diameter. All calculations were correct, my words and I, failed me in that response.

@spinnaker it would depend on which number was the divisor. That causes the difference. If you used 1-(V1/V2) and the larger number was V2, it would be 4.79% while if you use the smaller number as the divisor it becomes 5.03%.

 

That's what I get for thinking at that time in the AM. Yeah, diameter. All calculations were correct, my words and I, failed me in that response.

@spinnaker it would depend on which number was the divisor. That causes the difference. If you used 1-(V1/V2) and the larger number was V2, it would be 4.79% while if you use the smaller number as the divisor it becomes 5.03%.

I understand the math. I am trying to figure out the value of the formula / function. Seem to me a change of percent value should be the same absolute value, no mater what number is entered first.