drivetrain loss
#1
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2006 E55
drivetrain loss
why is it a percentage and not a fixed number? I would think it would be a set number of horsepower. Why would you lose more power in the drivetrain as you gain more from the engine? Thanks
#3
Notice I said torque. Horsepower is the result of torque.
#4
NERD ALERT
Anyway, the relationship is as follows: HP = TORQUE X RPM / 5252
(one consequence of this is that hp and torque should be the same at 5252 rpm ... if they're not, that dyno sheet is 'shopped)
![drive](https://mbworld.org/forums/images/smilies/driving.gif)
#5
I'm not sure I would call hp "the result of" torque, but you are 100% right in noting that they are both aspects of the same thing. Simply, hp is a measure of power, torque is a measure of force (not exactly the same thing)
Anyway, the relationship is as follows: HP = TORQUE X RPM / 5252
(one consequence of this is that hp and torque should be the same at 5252 rpm ... if they're not, that dyno sheet is 'shopped)
![drive](https://mbworld.org/forums/images/smilies/driving.gif)
Anyway, the relationship is as follows: HP = TORQUE X RPM / 5252
(one consequence of this is that hp and torque should be the same at 5252 rpm ... if they're not, that dyno sheet is 'shopped)
![drive](https://mbworld.org/forums/images/smilies/driving.gif)