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Torque is equal to friction force (F) * radius (r). Torque is also equal to moment of inertia (I) * Angular acceleration (a). Angular acceleration is equal to rotational velocity * 2Pi/time, which is 2 seconds. So, F = IRa/r, or 45.63 Newtons
No, if you are driving and press the accelerator down to increase the fuel and air mixture & increase RPM then you will accelerate forward for a time before you come stop accelerating and just keep going
Feed Rate = No of Cutting teeth x RPM x Feed per tooth.
a = v^2/r.r = 0.25 metresv = ((0.25 * 2 * pi * 20) / 60) = 0.5236 metres / secso:a = (0.5236^2) / 0.25.a = 1.097 (m/s)/s
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Horsepower: 140 @ 3800 RPM torque : 215 @ 2400 RPM
260 hp @ 5000 rpm 330 tq @ 2400 rpm
245 Horsepower @4400 rpm and 345 ft lb torque @2400 rpm
That depends what you want to compare. In any case, 3G is more acceleration than 1G.
My Chilton book shows ( 170 horsepower at 3800 RPM and 270 ft-lbs torque at 2400 RPM )
The operating range of rpm, for a C 12 Caterpillar diesel engine is 800 RPM to a maximum of 2400 RPM. The engine should be run at no more than 1800 RPM for a long period of time.
120 horsepower at 4400 RPM and 190 ft/lbs torque at 2400 RPM
250 HP @ 4400 rpm, 350 ft lbs torque @ 2400 rpm.
as for crankshaft rotates for a certain rpm ...... and with respective rpm we need it to come to rest and the brake applied to it with much power..... determines how much power we need to stop it at a certain rpm..... these brake power is known as brake horse power..
2.2 vin 4, 130 ft-lbs. @ 2800 rpm4.3 vin W (Vortec), 235 ft-lbs. @ 2400 rpm 4.3 vin Z, 235 ft-lbs. @ 2400 rpm
My 115 EFI mercury outboard would only generate 4200 RPM. The power gradually reduced and finally levelled off at 2400 RPM. I have had other problems in the past relating to power settings. Fuel injectors and impellor have been eliminated as potential causes.
According to my Chilton repair manual : 160 hp @ 4200 RPM / 220 ft.lbs torque @ 2400 RPM (Helpfull)