At 08:32 AM 4/4/97 -0400, you wrote:
>Thanks for the info, Jeff. I've got a set of 235/70's on the truck
>now. I imagine the 245/60's would probably be close to them in
>height, don't ya think? I like the height of the truck now, so I
>don't want to change it much. Thanks again.
BTW: I didn't notice if anyone mentioned this, but here's the
"formula" for tire sizes... It can be used for a bunch of stuff,
such as keeping two different size tires the same height, for
example...
The first number (i.e. 245) is the width of the tire in millimeters.
The second number (i.e. 70) is the ratio of sidewall height to tire
width. This ratio is expressed as a percentage. (70%) So, the sidewall
height is 70% of the width of the tire) So, to find the height of the
sidewall, you multiply the two numbers together. (That'll give you the
height in millimeters)
To find the height of the tire, take the sidewall height times 2, then
add the diameter of the wheel.
To convert those useless millimeters into nice 'merican inches, divide
by 25.4. (With apologies to all our Canadian members.) ;)
So the heights of the tires in this example would be:
235/70:
(235mm * 70%) = (235mm * .70) = 164.5mm
164.6mm / 25.4 = 6.476377952756in
(6.476377952756in * 2) + 15 = 27.95in
245/60:
(245mm * 60%) = (245mm * .60) = 147.0mm
147.0mm / 25.4 = 5.787401574803in
( 5.787401574803in * 2) + 15 = 26.57in
You'd have a difference of 1.38 inches.
Wether that's a lot or not depends on your point of view, I guess. :)
Keep in mind that any change in the circumference of your tire will
throw off your speedometer and odometer.
I think that the formula for circumference is Pi * diameter. (Its
either that or Pi * r squared, but I think that one is area) Hey,
its been a while! :)
Anyway, assuming I'm right about the formula, the circumference of
a 235/70 tire would be 87.81617260034in, and the circumference of a
245/60 tire would be 83.4872063454in. So, for every revolution, the
new 245/60 tire would travel about 4.33 inches less.
If we assume a speed of 60mph:
I took 60mph / 60 mins per hour * 5280 feet per mile * 12 inches per feet
to get 63360 inches per minute. Dividing this by the circumference of
each tire, we determine the revolutions per minute that tire has to turn
at to make the truck go 60mph:
235/70: 721.5071908037 rpm
245/60: 758.9186747712 rpm
If we're using the 235/70 as our baseline, we can divide that by 60mph to
determine the rpm nescessary to sustain 1 mph.
721.5071908037 / 60 = 12.02511984673 rpm per 1mph
So, every mile per hour needs about 12.025 rpm. If we divide the
rpm of the 245/60 tire at 60mph by 12.025 we see that the speedometer
would actually read 63.1 mph.
Of course, the Dakota's speedometer is calibrated for a 215/75 tire.
An rpm of 728.1735685438 will cause a 60mph reading on the speedo.
Doing the numbers again from a baseline given by the 215/75 we get:
Indicated speed Actual speed
215/70: 60 60
235/70: 60 60.55
245/60: 60 57.57
So, it would appear that the 235/70 is a pretty good match with the stock
tire. That's what I've got on my truck now. I never bothered to go
through with this calculation though. Looks like those of us who
picked those tires did a pretty good job! :) I just did a quick
calculation for a 245/65 tire, and that would read 59.66mph at 60mph.
(Height of 27.539 in.) That's even better than the 235/70 for accuracy,
but I have no idea if they make a 245/65.
Well, that's enough garbage out of me for one post! :) I apologize
for any mistakes I may have made above. If anyone sees any, I'm sure you
won't hesitate to let everyone know. :) (please do! Its been a while
since I've done anything like this.)
-Jon-
Jon Steiger - Network Administrator for Academic Information Technology
.- steiger@ait.fredonia.edu -- http://www.cs.fredonia.edu/~stei0302/ -.
| DoD# 1038, EAA# 518210, NMA# 117376, USUA# A46209, KotWitDoDFAQ |
| '96 Dodge Dakota SLT V8, '96 Kolb FireFly 447, '91 Yamaha FZR600R |
`---------------------------------------------------------------------'
I do not speak for SUNY College at Fredonia; any opinions are my own.
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