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Cougar Pilot
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Just the horsepower of the engine itself (only) on an engine dyno, as opposed to the whole car on a chassis dyno.

Yes I think 205hp is overrated, hear it many a time. And when you dyno the car, then add 15%, it usually isn't 205.
 

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Im not being a dick

but if you take 15% of 205 its 174.25

and then if you take 15% of 174.25 and add it to 174.25 its only 200 so adding 15%

to the dyno shouldn't equal the crank...not being a dick just trying to help
 

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BlackCat94 said:
Just the horsepower of the engine itself (only) on an engine dyno, as opposed to the whole car on a chassis dyno.

Yes I think 205hp is overrated, hear it many a time. And when you dyno the car, then add 15%, it usually isn't 205.
Cordeez said:
Im not being a dick

but if you take 15% of 205 its 174.25

and then if you take 15% of 174.25 and add it to 174.25 its only 200 so adding 15%

to the dyno shouldn't equal the crank...not being a dick just trying to help
reason that doesn't add up the way you guys did it is because math of efficiencies doesn't work like that

174.25/.85 does equal 205....so you don't add the lost 15% by performing the multiplication 174.25*1.15.....you have to divide the actual HP by the efficiency percentage in decimal form

so for example, my car made 169.1 rwhp......that would make my losses 100%-(169.1/205*100%)=17.5%.......that makes sense since the solid axle/T-45 combo in a mustang takes ~12% from the crank HP, so our cars with the 4r70w/IRS-8.8" would be at a greater loss........so my post-SCT tune crank HP would be 182.7/.825= 221.5 crank HP
 

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MaleWhore
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are the 96-7's rated higher than the older ones...i know that i read somewhere that my 97 makes about 205 at the crank
 

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Here's the post Jerry tune graph of my '95 stocker.. the only "mods" were 22c plugs, 180*t-stat and 93 octane fuel.



-mike
 

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Hi, All--

I'm sorry, but I have to chime in here--

I have seen this posted time and time again, and every time, it is still wrong.

A chassis dyno measures rear wheel HP and Torque, that is true, but what is *NOT* true is that this "figure" is a percentage of actual crank HP and Torque.

Why, you ask?

Very simple--the amount of HP loss will be a constant, NOT a percentage.

For example, if we have an engine that makes about 250HP at the crank, and 200HP at the rear wheels, and we add parts onto it to the point where it now makes 500 at the crank, the car won't suddenly increase the amount of power absorbed by the transmission, just because the engine HP was doubled.

The "percentage" theory would indicate that at 250HP, the engine has a 25% power loss, by putting 200 of those horses to the pavement; HOWEVER--since we doubled the crankshaft HP, is it "magically" now going to absorb 100HP? No, it will absorb the SAME AMOUNT as it did before--50HP.

The power lost to drivetrain does not increase just because the power at the crankshaft increases.

A far more accurate figure is to measure the crankshaft HP and then the rear wheel HP, and figure out how much the drivetrain actually absorbed, and THEN use that number as the constant to determine actual crankshaft HP.

A good example is a 5.0 Mustang--it made 225HP from the factory at the crank, and most of the stock ones would dyno at around 180-195HP at the rear wheels (depending on the transmission used). There is your baseline--automatics (AOD's) absorbed about 45HP, and manuals about 30.

Now, if you increase your Mustang engine's power and now it makes 300 to the rear wheels, you can plug in the ACTUAL figures for what the transmission absorbs based on the stock engine, and come up with between 330 and 345HP at the crankshaft.
 

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Refrigerator Raider Hater
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Stonecoldtx said:
Very simple--the amount of HP loss will be a constant, NOT a percentage.
Your wrong. It is neither a percentage nor a constant, but a percentage is very close to correct.
 

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GreenBird said:


Your wrong. It is neither a percentage nor a constant, but a percentage is very close to correct.
yea.......the way i figure it is this......it don't believe it can be a constant because, say in my car, the difference between the factory rated crank HP and un-tuned RWHP was 35.9 HP.......if losses were constant, that would that mean it would take more than 35.9 HP to be able to make the tranny/driveshaft/diff. gears even start to move.....i for one don't think that is so.....am i close in my reasoning, GreenBird?
 

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guitar maestro said:


yea.......the way i figure it is this......it don't believe it can be a constant because, say in my car, the difference between the factory rated crank HP and un-tuned RWHP was 35.9 HP.......if losses were constant, that would that mean it would take more than 35.9 HP to be able to make the tranny/driveshaft/diff. gears even start to move.....i for one don't think that is so.....am i close in my reasoning, GreenBird?
Well it's not as simple as that when you consider the different gears. Plus it's really a certain amount of torque it would take to turn it, not power. In other words, it's too complicated to be answered that simply.

Like was said earlier, it's between a static amount and a percentage. The momentum of the drivetrain causes a percentage loss. (Remember Newton's law of momentum? Exert a force (or torque) on something and it will exert an equal but opposite force.) Friction, I would think, is mostly perentage based as well, but not completely. I'll have to go more in-depth into it when I have more physics classes.

Oh, and the loss from the torque convertor (when it's not locked) is inversely proportional to the engine speed, so that's the opposite of percentage based.
 

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kenshi said:
Well it's not as simple as that when you consider the different gears. Plus it's really a certain amount of torque it would take to turn it, not power. In other words, it's too complicated to be answered that simply.

Like was said earlier, it's between a static amount and a percentage. The momentum of the drivetrain causes a percentage loss. (Remember Newton's law of momentum? Exert a force (or torque) on something and it will exert an equal but opposite force.) Friction, I would think, is mostly perentage based as well, but not completely. I'll have to go more in-depth into it when I have more physics classes.
yea, i was simply using that as a simple example as to show how it could not be constant
 

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Boom.
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Stonecoldtx said:
Hi, All--

I'm sorry, but I have to chime in here--

I have seen this posted time and time again, and every time, it is still wrong.

A chassis dyno measures rear wheel HP and Torque, that is true, but what is *NOT* true is that this "figure" is a percentage of actual crank HP and Torque.

Why, you ask?

Very simple--the amount of HP loss will be a constant, NOT a percentage.

For example, if we have an engine that makes about 250HP at the crank, and 200HP at the rear wheels, and we add parts onto it to the point where it now makes 500 at the crank, the car won't suddenly increase the amount of power absorbed by the transmission, just because the engine HP was doubled.

The "percentage" theory would indicate that at 250HP, the engine has a 25% power loss, by putting 200 of those horses to the pavement; HOWEVER--since we doubled the crankshaft HP, is it "magically" now going to absorb 100HP? No, it will absorb the SAME AMOUNT as it did before--50HP.

The power lost to drivetrain does not increase just because the power at the crankshaft increases.

A far more accurate figure is to measure the crankshaft HP and then the rear wheel HP, and figure out how much the drivetrain actually absorbed, and THEN use that number as the constant to determine actual crankshaft HP.

A good example is a 5.0 Mustang--it made 225HP from the factory at the crank, and most of the stock ones would dyno at around 180-195HP at the rear wheels (depending on the transmission used). There is your baseline--automatics (AOD's) absorbed about 45HP, and manuals about 30.

Now, if you increase your Mustang engine's power and now it makes 300 to the rear wheels, you can plug in the ACTUAL figures for what the transmission absorbs based on the stock engine, and come up with between 330 and 345HP at the crankshaft.
Torque is equal to the moment of inertia of the system multiplied by the angular acceleration of the system. Just by looking at this formula, T=IA, I see that if I want to increase the angular acceleration of the drivetrain and thus for my car to be faster, more torque will be required to bring about the increase in ang. acceleration. As torque and horsepower are related in the manner that they are, at a given RPM an increase in torque necessary to spin the drivetrain at a certain velocity/acceleration will have an effect on the RWHP as well.

Maybe this isn't applicable to this situation. I can't see why it wouldn't be though.
 

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Refrigerator Raider Hater
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Yeah you three have it right.

Brian, your idea would work in theory, but it would be very expensive for relatively little gains.
 
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