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Please know first that my intent here is not to flame you. I think I get the jist of your point, but I beleive some corrections to your rationale may be possible.
First, a minor point:
"...Gislaved listed their 185/R14 as having a 196 mm section width, and a 662 mm overall diameter..."
If this is the case, the aspect ratio of the 185/R14 tire is 78. The math:
Overall diameter: 662mm/(25.4mm per inch) = 26.06"
Sidewall height: 26.06" - 14" = 12.06 (for both); 6.03" for "one" sidewall.
Tire width: 196mm/25.4 = 7.72"
Aspect ratio: 6.03"/7.72" * 100 = 78.14
"I converted the overall diameter to inches, by multiplying by 0.04..."
There aren't a whole lot of meausrements which have an "easy" or "exact" conversion between metric and English, but the linear measurement of distance is one of them. Per above, 25.4 mm per inch. This means 0.03937 inches per mm.
"...which gave me 26.48 inches, or 26.5 rounded off."
Per above, it's more like 26.06". For the sake of this discussion, let's call it 26".
"Now here is what I am not sure of. I figured that if 1/10th of an inch caused a tire to rotate 2 ½ times more per mile, then a 1" difference must be ten times that or 25 more revolutions per mile."
No, this isn't the case. Let's use some algebra to demonstrate.
Two assumptions:
First: All measurements are in inches.
Second: The diameter of the first tire is D(1).
The second tire is 0.1" smaller in diameter. Then the diameter of the second tire, D(2) can be calculated: D(2) = [D(1) - 0.1]
Remember the general equation for circumference:
c = (pi) * d
The circumference of the first tire: C(1) = (pi) * D(1)
The circumference of the second tire: C(2) = (pi) * D(2)
Substituting: C(2) = (pi) * [D(1) - 0.1]
Rearranging: C(2) = [(pi) * D(1)] - [(pi) * 0.1]
The third tire is 1" smaller in diameter. Then the diameter of the third tire, D(3) can be calculated: D(3) = [D(1) - 1]
Again, the circumference of the first tire: C(1) = (pi) * D(1)
The circumference of the third tire: C(3) = (pi) * D(3)
Substituting: C(3) = (pi) * [D(1) - 1]
Rearranging: C(3) = [(pi) * D(1)] - [(pi) * 1]
So for the second tire, the difference in circumference is acheived by subtracting one-tenth of "pi" from the circumference of the first tire, and for the third tire, the difference in circumference is acheived by subtracting "pi" from the circumference of the first tire. So the overall difference between the circumference of the second and third tires is not a linear multiple of 10.
"Here is where the gas mileage calculations came in."
"Consumer Reports and the EPA tested the 240 wagon at about 30 miles/gallon highway..."
Sheesh!! Is this the new 240 with the part-time electric motor?!? Just kidding.
"Okay, at 30 miles/gal/hwy., a wagon can go 474 miles on a tank of 15.8 gals."
Agreed.
"I also assume those wagons tested had OEM tires of 26.5" diameters on them. If those OEM tires spin 25.5 times less/mile, then they are turning 787.5 times per mile. That is 23,625 times for 30 miles/gal/hwy, and 354,375 for a 15 gal/tankful. By contrast, a 195/75 with an overall diameter of 25.5" turns 815 times/mile, or 24,450 times per 30 miles/gal/hwy, and 366,750 revs per/15 gal tank."
You kind of lost me in this paragraph. Here are my thoughts:
Back to the original 185/R14 tire... it's diameter is 26". This means:
Circumference = (pi) * 26" = 81.68". So this tire will travel 81.68" for each revolution (of the tire... and of the rear axle).
There are 63360" in a mile (5280' * 12 inches per foot = 63360"). This means the 185/R14 tire will turn:
63360 (inches per mile) / 81.68 (inches per revolution) = 775.71 (revs per mile). The rear axle also turns this same number of times (per mile).
In the case of a tire which is 1" smaller in diameter:
Circumference = (pi) * 25" = 78.54". So this tire will travel 78.54" for each revolution (of the tire... and, again, of the rear axle).
This tire (and the rear axle) will turn:
63360 (inches per mile) / 78.54 (inches per revolution) = 806.72 (revs per mile).
Let's assume that the "30 miles per gallon" figure is accurate. Then, for a given engine speed, transmission ratio, and differential ratio, the 185/R14 tire will yield:
30 (miles per gallon) * 775.71 (revs per mile) = 23271.3 (revs per gallon).
Now, we don't know the gas mileage for the second (one-inch-shorter) tire yet. However, let's assume the same engine speed and driveline gearing per above. Then we can plug the second tire's 806.72 (revs per mile) into the equation above and work backwards:
23271.3 (revs per gallon) / 806.72 (revs per mile) = 28.85 (miles per gallon).
SO:
Let's assume the car is driven 15,000 miles per year.
For the 185/R14 tire:
15000 (miles per year) * 1.70 (dollars per gallon) / 30 (miles per gallon) = 850.00 (dollars per year).
For the 1" smaller tire:
15000 (miles per year) * 1.70 (dollars per gallon) / 28.85 (miles per gallon) = 883.88 (dollars per year).
So there is a difference of 38.33 (dollars per year), which roughly supports your original calculation.
But realistically, would anyone put 1" smaller tires on their car? In the case of the 0.1" smaller tires (which, to me, seems quite possible), the result is much less dramatic. I'll spare you all the math details, but the key points are that this 0.1" smaller-diameter tire will give a gas mileage of 29.89 (miles per gallon), resulting in a fuel cost of 853.13 (dollars per year). Clearly not much different from the 185/R14.
I realize I'm not really modeling everything exactly correctly... most people in the world pay more than $1.70/gallon; different tires have different coefficients of friction; different tires wear at faster/slower rates; some people don't check their tire pressure or rotate/balance their tires as often as they should. But I believe the point is made: Vary the overall diameter of your tire (from stock) by +/- 0.1" or +/- 0.2" and there won't be an appreciable difference in operating cost.
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