- #1
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Hi everyone,
I'm interested to know how much kilograms force an average car can produce. I made the following calculations to answer my question but I'm not sure if I'm right or wrong!
Imagine we have a 1.8L engine that can produce a max torque of 155 nm @ 4250 RPM.
1st gear ratio = 3.455 Final drive ratio = 4.529 Tire size = 185/65R14 (Rolling Tire Radius = 0.289 meters)
It is a FWD car so we have a drivetrain loss of around 10%.
I calculated the acceleration force as follows:
Wheel torque = 155 x 3.455 x 4.529 = 2425 nm Assuming 10% drivetrain loss, we will end up with 2183 nm of torque at the wheels.
Now, to calculate acceleration force, I divided 2183 by tire radius (0.289):
2183 / 0.289 = 7553 Newtons of acceleration force.
Finally, to convert Newton to kilograms force:
7553 / 9.8066 = 770 kgf
Does it mean that this car can lift 770 kg of weight if its engine and transmission (both as a single unit) are used as an industrial elevator?
I'm interested to know how much kilograms force an average car can produce. I made the following calculations to answer my question but I'm not sure if I'm right or wrong!
Imagine we have a 1.8L engine that can produce a max torque of 155 nm @ 4250 RPM.
1st gear ratio = 3.455 Final drive ratio = 4.529 Tire size = 185/65R14 (Rolling Tire Radius = 0.289 meters)
It is a FWD car so we have a drivetrain loss of around 10%.
I calculated the acceleration force as follows:
Wheel torque = 155 x 3.455 x 4.529 = 2425 nm Assuming 10% drivetrain loss, we will end up with 2183 nm of torque at the wheels.
Now, to calculate acceleration force, I divided 2183 by tire radius (0.289):
2183 / 0.289 = 7553 Newtons of acceleration force.
Finally, to convert Newton to kilograms force:
7553 / 9.8066 = 770 kgf
Does it mean that this car can lift 770 kg of weight if its engine and transmission (both as a single unit) are used as an industrial elevator?