tiganasfx
Known Member
Well take the above example:
Assuming car A has 160Nm or torque @ 4000RPM
Assuming car B has 160Nm of torque @ 6500RPM
160Nm = 118lb-ft
Car A = 118*4000/5252 = 89.9hp
Car B = 118*6500/5252 = 146hp
As expected car B has higher hp @ higher rpm with the same torque.
At rpm=5252, the Hp = Torque (lb-ft) and the two graphs should intersect.
Assuming car A has 160Nm or torque @ 4000RPM
Assuming car B has 160Nm of torque @ 6500RPM
160Nm = 118lb-ft
Car A = 118*4000/5252 = 89.9hp
Car B = 118*6500/5252 = 146hp
As expected car B has higher hp @ higher rpm with the same torque.
At rpm=5252, the Hp = Torque (lb-ft) and the two graphs should intersect.