Tire Diameter VS MPG'S

is it because the tire od effects the odometer?

 

I am not sure I see your logic. As a tire wears and has a smaller diameter, your car will not go as far for each rotation of the tire. But the odometer and speedometer will think that nothing has changed. So if you want to increase your mileage, either increase the tire pressure (to increase diameter) or go to a larger diameter tire.

 

Smaller diameter tire will make the car think you've gone further. Which will "improve" your mpg, both displayed and calculated too, as long as you're using the odometer to calculate.

That said, I don't think the tire OD reduction is as great as indicated in initial post. At most, assuming initial tread depth of 10/32", worn down to the wear bars at 2/32", that's a radius diff of 8/32", diameter 16/32", aka 1/2".

In other words, 25" to 24.5", which works out to 2% circumference reduction.

Still, most people have the sense to NOT run tires down to 2/32", so maybe 1% circumference reduction at most.

For an example, with 1% tire circumference reduction, 50 mpg would become 50.5 mpg.

 

Ignore the tire's new vs worn OD, that doesn't correspond well to the tire's real rolling circumference.

Instead, look at the specified RPM (revolutions per mile) on the spec sheet, and use this to figure the real rolling circumference. On the sample of tires I have checked, this gives a rolling circumference of a bit less than the zero-tread OD (or (OD minus (2 * tread thickness)) *2 * pi). I.e. the rolling circumference is set mostly by the length of the steel belts just beneath the thread. These belts do not noticeably change length as the tire wears and ages.

Remember that under normal loads and inflation pressures, the contact patch on the ground is flat, not round. Thus when attempting to figure changes in rolling rate as the tread wears, the normal circle equations just don't work right.

 

You'd need to drive a measured course, when the tires are new, and well worn.

 

There is also changes in the tire rolling resistance as the tire wears. It gets better.

Going further, installing taller tires than OEM changes the final gear ratio of the car. It will go further per revolution of the drive train. The potential improvement is countered by the increased frontal area, though.

 

Is this regardless of the tread depth when the tire is new? My PRIME came with Dunlop Enesave OEM tire 195/65-15. The tire has only 8/32 tread depth when new. Some of the tires I am considering to replace has 11/32. With deeper tread when new, are they bigger than OEM Dunlop?

 

The static height of a tire with no load on it is not the critical parameter. Revolutions per mile with the tire installed on the car and loaded is the critical parameter.

 

In light of what @jb in NE is saying it's maybe a moot point, but I would think if two tires are spec'd as the same size, the OD should be the same (theoretically), regardless of new tread depth. In other words, it's the bottom of each tread depression that varies, if the tread depth varies.

 

Look up the particular model's spec sheets at TireRack.com, they will answer this question. This tire is shown at: https://www.tirerack.com/tires/tires.jsp?tireMake=Dunlop&tireModel=Enasave+01+A%2FS&partnum=965SR5ES01AS

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Separately, let me use this tire to illustrate my earlier point. This tire is listed as 25.0" diameter, 8/32 tread depth, and 836 revs per mile.

If the normal circle equations worked, a new tire would have (5280feet/mile * 12 inches/foot) / (pi*25.0inches) = 806.7 revs per mile.

A completely worn-out tire, zero tread remaining (diameter = (25.0 - (2*8/32) = 24.5 inches), would have 823.2 revs per mile. But both of these are outside the listed spec of 836.

Computing backwards from that listed rpm spec, I find an effective rolling diameter of just 24.125 inches. That puts the effective rolling surface 14/32nds underneath the original outer tread, and even 6/32nds underneath the surface even after all the tread is shaved off.

This is why I'm saying that the normal circle equations just don't work for this sort of figuring. Has anyone found a passenger car tire spec that doesn't fit this pattern?

 

Well, I don't profess to be an expert on anything (…with the exception of the Art&Science of Pipe-Smoking (Tobacco)), but what I would really like to know, and perhaps all of you "experts" out there might be able to help me, is precisely how may angels can dance on the head of a pin?

 

When the tire is loaded, the (edit - static, not effective) rolling radius is from the hub center to the ground. All tires have compression under load, which is why comparing static unloaded diameters is meaningless.

 

I've tried computing from that too. It produces a result that is sensitive to both inflation pressure and load, and still doesn't work.

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Try figuring your claimed effective rolling radius with a 195-width tire, inflated to the Prius front tire spec of 35 psi, and loaded with 1000 pounds. I'm finding 815.8 revs per mile with full new tread, and 832.8 revs per mile after all the tread is worn off. This still doesn't match or even straddle the spec sheet's listed 836.

These calculations assume a completely flexible sidewall, zero stiffness. Any sidewall stiffness will only increase the mismatch.

What do your calculations indicate? Are you getting different number outputs?

On my Subarus with manual transmissions and known fixed gear ratios, tire revs-per-mile computed from these circle equations simply haven't been able to explain the observed engine RPMs. But the listed revs-per-mile on the tire spec sheets have matched up well with observations.

 

That is why the revolutions per mile are used. A static loaded radius is not the same as the effective rolling radius.

I corrected the error in my earlier reply.

 

Here is the technote from TireRack. The diameter seems to be measured at the top of tread as I expected, not at the bottom of the tread. If a very small change in the tire size affects mpg, those with deeper threads will be able to shed more tread to get smaller than the shallow thread one. For me, my OEM Dunlop Enasave tire in two years shed 2/32 to 3/32 inches of tread, but I am not seeing dramatic or consistent changes in mpg, up or down, when driven on HV.

 

I added some computations before your edit arrived here. I don't like using any form of radius at all for this puzzle. But if you are going from spec-sheet revs-per-mile, as I am, then we are effectively on the same page.

 

I would think that any minor changes in revolutions/mile over the life of a tire are well within any inaccuracies in the fuel economy calculations. The dash display is typically optimistic, and the fuel filler rarely stops at exactly the same tank level, driving conditions on that tank are different, etc. If you look at a typical graph of measured fuel mileage over time, there is a lot of scatter in the results.

 

--quote--
Since rolling resistance is caused mostly by internal friction of the tire, then there are a number of things that can be concluded:
....
2) The more the material in a tire - especially in the tread area - the more the rolling resistance.


This means that new tires are going to have more rolling resistance than otherwise identical, but worn out, tires. So when you buy a new set of tires, you should expect a loss in fuel economy.
.....
This also means that higher speed rated tires - with their additional cap plies - are also going to have higher rolling resistance.
--end quote--
Barry's Tire Tech