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Another math question
I was looking back at the formula Old Tele man gave about the relationship between speed/RPM/gearing and wondered how tire size would factor in. We can change our RPM at a given speed by changing trans. gearing or rear end gearing or tire size. Here is the formula I started with.....
MPH = [60/(G x A)] x [RPM/rpm] MPH = miles per hour RPM = revolutions per minute rpm = revolutions per mile of tire 60 = conversion constant G = gear ratio of highest forward gear A = axle drive ratio I found this formula and wondered if it would be closer to the "real world". It was given as this: Vc = (Ct x Ve) / (GRt x GRd) Vc = car speed Ve = engine speed Ct = tire circumference GRt = transmission gear ratio GRd = differential gear ratio I morphed it using the same lettering OTM used (with one exception) to come up with this: MPH = (Ct x RPM) / (G x A) I was thinking that Ct might work better than revolutions per mile because the tire circumference is different when the tire is "loaded". Also it is different at 35 psi than it is at 50 psi. Probably not enough to make any real difference, but I thought I would just throw it out there. The thing I am having trouble with is the 60. I think in the original formula it's a way to "equalize" the rev's per MINUTE and the miles per HOUR. If it is...then why is it not used in the second formula? I'm no math genius. I can take a formula and do the sum, but as to how a formula is derived I don't have a clue. Is the second one different than the first or am I just looking at two sides of the same coin? |
the 60 is needed because engine RPM is in RPM, not RPH.
alternatively, the 60 is needed because vehicle speed is in MPH, not MPM. math tends to be..... interchangable like that, depending on what you want to end up with. |
The equations do the same thing but in different units: The first formula gives speed in miles per hour. The second gives speed in whatever units you used for tire circumference, per minute, e.g. inch/min, ft/min, miles/min, etc., since the units are not specified in the description.
i usually have to use inches and millimeters to calculate tire circumference in inches based upon the bast*rdized units of tire size, e.g. 175/70-13, then can divide by 12 to convert to feet for calculation purposes. e.g. Ct= 3.14 x [13 + [2 x (175 x .7 / 25.4)]] inches |
the reason it usually uses revs per mile is because it doesn't change even when highly loaded. That is because the steel belt is the main unchanging length in a tire. Someone tested the difference between new and worn tires and nothing changed. I think it was cfg83.
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FWIW -- there's a small 'problem' with relying upon tire DIAMETER, in the real world, because of vehicle weight loading, it LIES to you:
1) With NO load, tire diameter is accurate... ...but... 2) With VEHICLE load, "true" rolling diameter is ALWAYS SMALLER than the unloaded diameter -- typically about 3% smaller, or 0.97:1 -- and this value VARIES from manufacturer to manufacturer, depending upon tire size, load capacity, sidewall height, etc.. Michelin tires are about 0.97:1. |
Because the thread spreads out flat on the tarmac.
The grooves in the thread become narrower as they no longer spread out. It also means there is not that much difference between a new tire and a worn one. The circumference of a new tire may be bigger, but when the thread folds flat it has just the same footprint. |
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