Really not looking for a guess or the standard answer, I am looking for the logic behind it.

Car makes 130hp at the flywheel, the transmission, weight of the axles, drag from bearings, weight of the rotating mass at the tires, and hell even some wind resistance (maybe) takes power.

So you end up with say, 110hp at the weels.

I understand that, got it, not dumb.

Here is my question.

Say I spray a 75shot, they claim that is power at the flywheel and only expect say 50-60whp.

What is causing the extra "drag" that is robbing the 15-25hp. Based off the stock calculation from the flywheel to the wheels, that line of parts takes 20hp to run. Why, when you add 75 more hp, does that stuff take an additional 25hp to spin? Extra friction from the additional torque? People full of crap about the hp increase of the mod you just spent a ton of money on... just curious.

 

hmmm jaws was spraying like a 80 shot and got 100hp on the dyno or somethin along those lines

 

Not just talking about spray, or jetting for X shot but seeing X+ more. That was just an example.

I always see "This is a 50hp gain", but on the dyno you see lower and the claim is it is drive train loss, but that loss was accounted for in the stock numbers. Unless I am missing something about how drivetrain loss works. If it took 20hp to drive that system stock, why is it going to take 45hp to turn once you increase the amount of hp the engine is putting out.

 

because its a precentage loss not a set number.. the drivetrain takes x percent to run so x percent of 100 is one number asnd x percent of 150 is another number...

so if u have 130 x.15 (drivetrain loss) is 110.5.. add a 75 shot and u have 205 x .15 = 174.5 . now the first time it took 19.5 hp for the loss, the second time it was 30.75 ... so its not that it took more hp to move the car it just the percent of hp it took from the engine to convert it to the wheels (which stayed the same..)

hope this clears it up a lil.

 

I am asking why is it a percentage and not a set amount. Does the power steering pump rob more power as you increase overall? What about an A/C compressor. Those take power to be run and can be described as taking a percentage of stock output, but that doesn't mean it stays at that set percentage as power goes up. The percentage stuff is just math. The loss is through the trans is due to friction, mass, etc. Extra torque changes that how? Unless the % is factoring slip? I can see that with an automatic/torque converter, as it might not transfer the same amount with more torque applied, but what about a stick with a clutch that isn't slipping?

That is also assuming nothing else changed. (Not trying to further complicate this by adding modifications to the trans that would/could change how much hp it takes to spin it)

Edit: I honestly think the % stuff is coming from the general estimate of how much the trans will take from a stock car. Example, the 15% for a FWD stick. Why, because as the power goes up so does the requirement for a beefier trans behind it. More metal to handle the power output / the more stuff that is needed to not break once the extra torque is applied, you get more mass, friction, etc.. so it takes more power to turn. So the 15% estimate should be close. I really want to hear the logic behind why it is a set percentage, not the generic this is how it is. To me the idea of the set percentage is flawed.

If you gain 15hp at the flywheel over stock from tuning I think you should see around 15 at the ground unless your causing slip somewhere or changed the speed your spinning stuff to get that power (increasing the redline could mean more drag from bearings in the trans, etc). Why? because the power to turn the transmission, etc has already been lost.

 

The issue is that it's harder to accelerate objects with mass faster. So you lose more because it takes more power to spin the same amount of mass from x rpm to y rpm in a shorter amount of time. Think of it in extremes, to accelerate a one ton flywheel to 10rpms in one minute would be easy even by hand. To accelerate it to 10rpm instantly would require an infinite amount of torque. Back off from instantaneous a bit and it still takes a lot of power to spin up the rotating mass faster.

 

Why, thank you sir. That is what I was looking for, I didn't think about it from that angle. I was thinking of RPM when I put speed into my logic, not accelerating those parts. Like you would be doing on a dyno. However, torque at a set rpm (such as cruising would equal stock drivetrain loss giving you a set amount of loss, not a percentage), but who gives a crap unless your trying for better gas mileage. So my logic was only half flawed. It is loss due to friction (constant set amount) + loss due to the acceleration of mass (mass constant + change in speed -the variable I missed) = more lost with extra power, im not going to figure that math out and stick with the guesstimate 15%.

 

hmm good answer 187

 

I knew I was missing something, but I could never get someone to give the why behind the common knowledge answer.

 

Think of it this way... HP is just a calculation of work done over time (P = W/t). To accelerate quicker, you have to do the same amount of work in less time. This is represented as a higher HP number. Because the calculation involves multiplication, anything that changes the rate of acceleration (frictional losses, inertial losses, etc...) is going to have a percentage effect on the HP number. These are just the rules of Algebra.