There must be a slope that is equal to a gear reduction that could be used to evaluate launches with deeper ratios before making the actual gear change, so one can get a seat-of-the-pants feel for it. I tried launching on the down ramp at the parking structure, which is a 15%+ slope, if not 20%, and whoa yeah, she really launched nicely, just jumped out and went like all get out for about 20 feet and then of course I had to cool it because the ramp/flat/ramp.flat sequence is a bit rough at speed.
So the key then to using sloped pavement to evaluate a gear change (deeper or higher numerically final drive) is to figure out what pavement down slope equals a particular gear ratio change for testing purposes. I'll be launching with a manual transmission too. Ratio change will be achieved by smaller tires in the order of about 6% smaller total diameter.
So using various websites to find equations I cobbled together an analysis and post it here so it can be fixed because surely I got it wrong, but it looks interesting:
Bottom line is that I get a 1 percent slope (2 inch drop in 10 feet) being roughly equal to approximately 6% deeper gear. That is not much of a slope unless you are hand push starting a car, then you will appreciate it. Here is the "math":
Tractive Effort (TE) = torque * gear ratio * final drive ratio / tire radius
Torque estimated for 1400 rpm, first gear 3.72, final drive 3.73, tire diameter 27.3"
TE = 80 * 3.72 * 3.72 / 1.14' = 974 pounds on launch (never mind the vehicle weight as we only need to know the effort here presumably). Yes, that is right, only 80 pound feet around 1400 rpm, got it from a Ford brochure torque curve for my 2.3L Duratec DOHC 16-valve engine in a 2001 Ranger. Remember this is a stick shift so no torque multiplying slush box.
Increase in TE due to slope = Weight of vehicle * sine of the angle.
Vehicle weight is 3300 pounds with me in it (estimated from dry weight listing).
For a 1 degree angle the increase in TE is 58 pounds. Then 58 / 974 gives about 6% increase which is what my tires would give me. The slope (rise/run) is the tangent of the angle or about 0.0175.
0.0175 * 10 foot run = 0.175 feet or a little over 2 inches.
But somehow I must be missing a piece of the puzzle. A 1 percent slope is very common on roads even in flat country like Detroit is in. You would think I'd notice these great power surges every time I take off on a slight down-slope, but I don't. There must be a missing factor in the above calculations.
So the key then to using sloped pavement to evaluate a gear change (deeper or higher numerically final drive) is to figure out what pavement down slope equals a particular gear ratio change for testing purposes. I'll be launching with a manual transmission too. Ratio change will be achieved by smaller tires in the order of about 6% smaller total diameter.
So using various websites to find equations I cobbled together an analysis and post it here so it can be fixed because surely I got it wrong, but it looks interesting:
Bottom line is that I get a 1 percent slope (2 inch drop in 10 feet) being roughly equal to approximately 6% deeper gear. That is not much of a slope unless you are hand push starting a car, then you will appreciate it. Here is the "math":
Tractive Effort (TE) = torque * gear ratio * final drive ratio / tire radius
Torque estimated for 1400 rpm, first gear 3.72, final drive 3.73, tire diameter 27.3"
TE = 80 * 3.72 * 3.72 / 1.14' = 974 pounds on launch (never mind the vehicle weight as we only need to know the effort here presumably). Yes, that is right, only 80 pound feet around 1400 rpm, got it from a Ford brochure torque curve for my 2.3L Duratec DOHC 16-valve engine in a 2001 Ranger. Remember this is a stick shift so no torque multiplying slush box.
Increase in TE due to slope = Weight of vehicle * sine of the angle.
Vehicle weight is 3300 pounds with me in it (estimated from dry weight listing).
For a 1 degree angle the increase in TE is 58 pounds. Then 58 / 974 gives about 6% increase which is what my tires would give me. The slope (rise/run) is the tangent of the angle or about 0.0175.
0.0175 * 10 foot run = 0.175 feet or a little over 2 inches.
But somehow I must be missing a piece of the puzzle. A 1 percent slope is very common on roads even in flat country like Detroit is in. You would think I'd notice these great power surges every time I take off on a slight down-slope, but I don't. There must be a missing factor in the above calculations.
