RWHP vs. Fly Wheel HP

[97 horse snake]

does anyone know the equation for turning rwhp to fly wheel hp? or vice versa?

 

[Booyah]

RWHP is about 14 - 16% less than BHP with a manuel transmission. So for a 15% estimated loss you would take RWHP and divide it by .85, IE - RWHP= 391 with 15% drivetrain loss. So estimated BHP would be 460(391/.85=460). And to do BHP to RWHP it's BHP multiplied by the drivetrain loss then subtract that amount from the BHP number. For example: 460*.15(15%)=69. 460-69=391. Hope this helps, automatic transmission loss' are usually around 20% some higher some a little lower all depends on the converter they have.

 

[E. Green Cobra]

Now not to catch anyones flames, as everyone likes to say that a car losses a certain percent due to parsitic losses. From what I've seen in relation to flyweel/rwhp is ~ 40hp loss in stick cars an 50 or slightly more in auto. Essentially a '99 cobra would be 320/280. A '99 Gt 260/220. 01 z06 385/345. Think about it if you say that a certain car looses 15%. Well then my '99 goes from 320 crank to 272rwhp right? for a loss of 48hp. Now lets just slap on a blower or other poweradder, now lets say my cobra makes 380rwhp, so multiply that by 1.15 and you get 437hp(flywheel) a difference of 57hp. Why does the more powerful car loose more hp if its through the same drivetrain. Since its a percent the larger(hp) number is going to loose more, as opposed to the lower hp number. Hard to explain but get what I mean?

 

[Booyah]

Originally posted by E. Green Cobra
Now not to catch anyones flames, as everyone likes to say that a car losses a certain percent due to parsitic losses. From what I've seen in relation to flyweel/rwhp is ~ 40hp loss in stick cars an 50 or slightly more in auto. Essentially a '99 cobra would be 320/280. A '99 Gt 260/220. 01 z06 385/345. Think about it if you say that a certain car looses 15%. Well then my '99 goes from 320 crank to 272rwhp right? for a loss of 48hp. Now lets just slap on a blower or other poweradder, now lets say my cobra makes 380rwhp, so multiply that by 1.15 and you get 437hp(flywheel) a difference of 57hp. Why does the more powerful car loose more hp if its through the same drivetrain. Since its a percent the larger(hp) number is going to loose more, as opposed to the lower hp number. Hard to explain but get what I mean?

It's simple numbers. If you take 15% of 100 you get 15 if you take 15% of 400 you get 60. You can't argue with math man. And it's not what any people say it's from the people that design dyno's either for brake horsepower or wheel horsepower that give the percentages. Also by trying to multiply by 1.15 isn't accurate either because you are trying to add 15% of an already lower percentage number to itself. Meaning 380 rwhp should be 447 and some change because it 85% of 447. RWHP/(100-percentage loss) meaning 380/(100%-15%=85%(.85))=447.05 BHP. By using the 1.15 factor you are trying to see a 15% increase of 380 which isn't the same as a 15% loss of 447. Hope that clears that up.

 

[97 horse snake]

oh

 

[Wildpony]

Let's see: 483rwhp/.85=568- I like it: :burnout:

 

[E. Green Cobra]

ok Booyah, I understand what you are saying, I also see the part about the mult of 1.15. Thanks for correcting my math skills (I'm an engineering major hehe) I'll let ya know which bridges I build so ya can avoid them. However my main point is still the same. 15% of 447 is 67.05 right? 15% of 320 is 48. Now if this was the same car (say before an after a blower install) Why would the hp loss numbers be different? If its the same car with the same conditions why does one car loose 67hp and the other only 48hp? Wouldn't they both lose the same number?

 

[chassebr]

One of the reasons why a percentage is used...

Simple, viscous friction. The faster you try to move a viscous fluid, the more resitance you get. It may not be perfectly linear, I'll give you that, but yes the drivetrain losses DO increase the more power you put through them. This is why lighter oils will improve efficiency but will break down much quicker.