Why a shorter rear gear will accelerate the car quicker
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Now, let's pick any given rpm of the motor, we'll use 3,000 rpm's. The crankshaft is turning 3,000 rpm's, so 3,000 times 4 will equal 12,000 power pulses that the motor is producing in 1 minute. | Now, let's pick any given rpm of the motor, we'll use 3,000 rpm's. The crankshaft is turning 3,000 rpm's, so 3,000 times 4 will equal 12,000 power pulses that the motor is producing in 1 minute. | ||
| − | Now, divide the 3,000 rpm's the motor is producing by the gear ratio 2.73 and you find that the tire is turning 1,099 rpm's. If we assume a 88" circumference tire (28" tire diameter times 3.14159), then the tire is | + | Now, divide the 3,000 rpm's the motor is producing by the gear ratio 2.73 and you find that the tire is turning 1,099 rpm's. If we assume a 88" circumference tire (28" tire diameter times 3.14159), then the tire is traveling 96,712 inches at the given 3,000 motor rpm's with a 2.73 gear. (88 times 1,099). If it's getting 12,000 power pulses per minute at that motor speed, then dividing 96,712 by 12,000 reveals the tire is getting a power pulse each 8.059 inches of its circumference or a total of 10.9 power pulses for each 1 revolution of the tire. |
Using the 3.73 gear (shorter), the tire is turning 804 rpm's under the same conditions (3,000 divided by 3.73) and rolling 70,752 inches (88 times 804). Getting the same 12,000 power pulses, divide 70,752 by 12,000 and find that the tire is getting a power pulse each 5.896 inches of its circumference or a total of 14.9 power pulses for each 1 revolution of the tire. | Using the 3.73 gear (shorter), the tire is turning 804 rpm's under the same conditions (3,000 divided by 3.73) and rolling 70,752 inches (88 times 804). Getting the same 12,000 power pulses, divide 70,752 by 12,000 and find that the tire is getting a power pulse each 5.896 inches of its circumference or a total of 14.9 power pulses for each 1 revolution of the tire. | ||
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