Dyno numbers and what they really say ??

[1 Alibi 2]

As winter drags on in the east, ( 9 deg. this morning ), I've had time to do some reading. My goal is to upgrade my blower, leaning toward a 2.9 Wipple, Ford's package for 2011. ( M-6066-MSVT29D ) A tuner in my neck of the woods, Evolution Performance, also markets several " packages " that boast huge hp numbers.........( 858 rwhp )....................The reading has me asking the following question:

.

If you've had your car dynoed, what SAE spec did they use for corrections, SAE J 1349 or SAE J 607 ( also refered to as STD on some dynos ) One mfg. claims that most OEM's use J 1349 for hp & touque #'s in their advertising, it being more conservitive, and reading as much as 4 % " below " J 607 corrected numbers......this 4 % difference may explain why stock 2011 GT 500'S are pulling 508 hp !!

Still in the " plan " stage of the upgrage, although funding is available..................hoping to get some feedback from this knowledgeable group, so I can make sense of the numbers.................thanks.

[JWG223]

Try not to get hung up on the dyno numbers. The dyno is a tuning tool. Pick the package that best meets your needs and the tuner that you trust the most.

[BDrool]

One thing that confuses me. I've been told that the drive train uses about 15% of the available power. Well, as the available power increases, why would the power train need more of that power if it never changed? Realistically, how much in actual TQ and HP is needed to spin the parts? Keeping in mind that HP is only a calculation bases on the true number of TQ, which it the real strength of a motor.

[JWG223]

One thing that confuses me. I've been told that the drive train uses about 15% of the available power. Well, as the available power increases, why would the power train need more of that power if it never changed? Realistically, how much in actual TQ and HP is needed to spin the parts? Keeping in mind that HP is only a calculation bases on the true number of TQ, which it the real strength of a motor.

 

 

Well, the measure of power is how QUICKLY it gets those parts spinning. The faster you want to accelerate the parts to the terminal velocity, the more power you will need to do it, and the more parasitic loss will occur.

 

Just check out the definition of horsepower.

http://dictionary.reference.com/browse/horsepower

[BDrool]

Well, the measure of power is how QUICKLY it gets those parts spinning. The faster you want to accelerate the parts to the terminal velocity, the more power you will need to do it, and the more parasitic loss will occur.

 

Just check out the definition of horsepower.

http://dictionary.reference.com/browse/horsepower

 

 

Read my post again...

 

I know what HP is, a math formula. Your response did not answer my question. Try again or let a real motor head answer...

 

Being that this is a GT500 Forum, I guess I'm refering to the drive train of the GT500.

[JWG223]

Read my post again...

 

I know what HP is, a math formula. Your response did not answer my question. Try again or let a real motor head answer...

 

Being that this is a GT500 Forum, I guess I'm refering to the drive train of the GT500.

 

 

As relates to the GT500, well, you would need engine-dyno numbers from a GT500's powerplant to determine that. Then the rest is just simple percentages. However, as you now know, the more power the engine produces, the more the driveline will rob before it gets to the ground.

[BDrool]

As relates to the GT500, well, you would need engine-dyno numbers from a GT500's powerplant to determine that. Then the rest is just simple percentages. However, as you now know, the more power the engine produces, the more the driveline will rob before it gets to the ground.

 

 

So, as you would rationalize, if you start with a 550 HP / 510 TQ motor, the drive line wil consume XX amount of energy to spin at a given RPM. But, boost that power to 640 HP / 630 TQ with a simple power upgrade with out changing the driveline and the said driveline will use more energy because the engine is more powerful. Wow, I guess a 4000 HP funny car uses 600HP just to turn the drive line, I don't think so. The mass of the driveline has not changed and therefore, does not require more energy to spin at a given rpm.

 

If I put a torque measuring device on a shaft and it turns with 10 ft. lbs. of force, would it require more force if I used a cheater bar for more leverage? No, it still turns with 10 ft. lbs. of force, it's just easier now.

 

If percentages are used, a 600 hp motor, at 15%, would lose 90 hp to turn the drive train at a set rpm within a time parameters. Given a 500 hp and the exact same drive train and rpm would only take 75 hp to spin. There is no logice here.

 

The amount of energy required to move a load a set distance in a give time remains the same no matter what the power is. If you change the load, distance or time, then the amount of energy will change.

 

In reality, it isn't the driveline that robs the power, it's the laws of physics because the amount of power required will increase exponentially for each increment of performance increase.

[JWG223]

So, as you would rationalize, if you start with a 550 HP / 510 TQ motor, the drive line wil consume XX amount of energy to spin at a given RPM. But, boost that power to 640 HP / 630 TQ with a simple power upgrade with out changing the driveline and the said driveline will use more energy because the engine is more powerful. Wow, I guess a 4000 HP funny car uses 600HP just to turn the drive line, I don't think so. The mass of the driveline has not changed and therefore, does not require more energy to spin at a given rpm.

 

If I put a torque measuring device on a shaft and it turns with 10 ft. lbs. of force, would it require more force if I used a cheater bar for more leverage? No, it still turns with 10 ft. lbs. of force, it's just easier now.

 

If percentages are used, a 600 hp motor, at 15%, would lose 90 hp to turn the drive train at a set rpm within a time parameters. Given a 500 hp and the exact same drive train and rpm would only take 75 hp to spin. There is no logice here.

 

The amount of energy required to move a load a set distance in a give time remains the same no matter what the power is. If you change the load, distance or time, then the amount of energy will change.

 

In reality, it isn't the driveline that robs the power, it's the laws of physics because the amount of power required will increase exponentially for each increment of performance increase.

 

 

Apply what is in bold

 

Horsepower is a measure of work. It involves time and distance and weight moved. When you increase horsepower you effectively move something heavier, faster, further.

 

So...if my car has 500 horsepower, to see that horsepower on the ground, I have to spin the tires against the dyno drums at X rate of acceleration, and I have (lets just say...) 6000rpm to get the engine to. So really, since the weight we are moving stays the same, and the mass of all parts stays the same, a dyno measures how quickly those parts can be brought up to speed. This is why all cars are dynoed in 1:1 gear. If you dyno the car in 3rd gear, it will "gain" horsepower on the dyno. Did it really? No, it just got the engine rpm up to 6K by reducing how far the rollers had to be rolled due to gearing. Giving an APPARENT increase in power.

 

Example: My G20 will spin those rollers up to 6K, too, but it will take a while getting there. Just because it moves the same roller to the same rpm does it mean it has as much power as your GT500? No.

 

However, your GT500 gets those rollers moving pretty quick and the rpms climb fast.

 

 

Now, if you add another 100 horsepower to that engine, it must accelerate the tires (and thus the engine rpm) quicker for you to notice/measure the power gain.

 

Uh-oh.

 

Something has now changed. You need to move the same transmission gears, tires,drive shaft QUICKER. That means they are going to take more energy to turn. That means you are going to "lose" more measured output. (time, distance, weight. We modified time, by mandating that the parts be moved faster, if we are to see the power increase).

 

Does this make sense? I couldn't tell whether you had answered your own question or not, based on the last few lines of your post, so I posted this anyway.

[AlexVogel]

Apply what is in bold

 

Horsepower is a measure of work. It involves time and distance and weight moved. When you increase horsepower you effectively move something heavier, faster, further.

 

So...if my car has 500 horsepower, to see that horsepower on the ground, I have to spin the tires against the dyno drums at X rate of acceleration, and I have (lets just say...) 6000rpm to get the engine to. So really, since the weight we are moving stays the same, and the mass of all parts stays the same, a dyno measures how quickly those parts can be brought up to speed. This is why all cars are dynoed in 1:1 gear. If you dyno the car in 3rd gear, it will "gain" horsepower on the dyno. Did it really? No, it just got the engine rpm up to 6K by reducing how far the rollers had to be rolled due to gearing. Giving an APPARENT increase in power.

 

Example: My G20 will spin those rollers up to 6K, too, but it will take a while getting there. Just because it moves the same roller to the same rpm does it mean it has as much power as your GT500? No.

 

However, your GT500 gets those rollers moving pretty quick and the rpms climb fast.

 

 

Now, if you add another 100 horsepower to that engine, it must accelerate the tires (and thus the engine rpm) quicker for you to notice/measure the power gain.

 

Uh-oh.

 

Something has now changed. You need to move the same transmission gears, tires,drive shaft QUICKER. That means they are going to take more energy to turn. That means you are going to "lose" more measured output. (time, distance, weight. We modified time, by mandating that the parts be moved faster, if we are to see the power increase).

 

Does this make sense? I couldn't tell whether you had answered your own question or not, based on the last few lines of your post, so I posted this anyway.

 

 

 

I think your logic is a little flawed, or I just can't follow it how its expressed.

 

It doesn't take more HP by %, it just takes more HP. If your engine is making 550 hp at the crank, your drive train is using 15%, that is a smaller number of HP than if your engine is putting out 770hp at the crank and drivetrain is using the same approximately 15% of the drivetrain loss. The number is only bigger because its 15% of a bigger number, not because the rotating/unsprung weight has increased in any way.

 

Now here is a monkey in the works...if you happen to increase hp while using a more efficient hp upgrade such as a TVS blower over the stock eaton, your actually introducing a reduction in powertrain loss % (less parasitic loss of hp).

[JWG223]

I think your logic is a little flawed, or I just can't follow it how its expressed.

 

It doesn't take more HP by %, it just takes more HP. If your engine is making 550 hp at the crank, your drive train is using 15%, that is a smaller number of HP than if your engine is putting out 770hp at the crank and drivetrain is using the same approximately 15% of the drivetrain loss. The number is only bigger because its 15% of a bigger number, not because the rotating/unsprung weight has increased in any way.

 

Now here is a monkey in the works...if you happen to increase hp while using a more efficient hp upgrade such as a TVS blower over the stock eaton, your actually introducing a reduction in powertrain loss % (less parasitic loss of hp).

 

 

We are saying the same thing, just in a different way.

 

You are reducing parasitic loss with the TVS blower, and I guess that is "drivetrain loss", technically.

 

As an interesting tid-bit, gears usually cost a few % when going from 3.27 to 4.10's or so.

[darthracer777]

We are saying the same thing, just in a different way.

 

You are reducing parasitic loss with the TVS blower, and I guess that is "drivetrain loss", technically.

 

As an interesting tid-bit, gears usually cost a few % when going from 3.27 to 4.10's or so.

 

 

It's all about the FRICTION.......

 

As the power is transmitted through the drivetrain, it's being transferred from one gear set to the next, in the transmission as well as in the rear end. As one gear tooth engages its counterpart, meshes, and then disengages, the surface of the tooth has to slip slightly with respect to its opposing tooth. Any time there's metal sliding, there's frictional losses. The friction force F is of course our old formula from college statics: F=uN where u is the coefficient of friction and N is the normal force. The more power that's going through a gear set, the higher the value N. (Different types of gears and different gear profiles have different formulas for the gear tooth forces, spur gears versus worm gears for example).

 

For any given rpm, increased torque increases the force acting between gears, and between shafts and bearings directly increasing the drag for any given coefficient of friction. And increased rpm for any given torque increases the loss due to kinetic friction between shafts and bearings, increases the frequency at which individual gear teeth contact one another, so on and so forth. In either case the power went up, and the losses followed suit.

 

Essentially, a mechanical device, such as a tranny, has a set amount of efficiency. For example, your average vehicle is 15% drivetrain loss, then the drivetrain is 85% efficient. Now regardless of how much HP is pushed through that drivetrain, it will only be 85% efficient.

[JWG223]

It's all about the FRICTION.......

 

As the power is transmitted through the drivetrain, it's being transferred from one gear set to the next, in the transmission as well as in the rear end. As one gear tooth engages its counterpart, meshes, and then disengages, the surface of the tooth has to slip slightly with respect to its opposing tooth. Any time there's metal sliding, there's frictional losses. The friction force F is of course our old formula from college statics: F=uN where u is the coefficient of friction and N is the normal force. The more power that's going through a gear set, the higher the value N. (Different types of gears and different gear profiles have different formulas for the gear tooth forces, spur gears versus worm gears for example).

 

For any given rpm, increased torque increases the force acting between gears, and between shafts and bearings directly increasing the drag for any given coefficient of friction. And increased rpm for any given torque increases the loss due to kinetic friction between shafts and bearings, increases the frequency at which individual gear teeth contact one another, so on and so forth. In either case the power went up, and the losses followed suit.

 

Essentially, a mechanical device, such as a tranny, has a set amount of efficiency. For example, your average vehicle is 15% drivetrain loss, then the drivetrain is 85% efficient. Now regardless of how much HP is pushed through that drivetrain, it will only be 85% efficient.

 

 

I agree, but isn't rotational mass also a part of the equation? Say, in the instance of an aluminum DS vs. a heavy steel one? You could say there is less friction on the bearings it turns on and this is the cause of the power increase, but I think rotational mass has a good bit to do with it as well, no? What about a set of heavy vs. light-weight rims? The weight on the wheel-bearings is the same as the car rests on them and the tires on the ground, so the ONLY thing they offer is less rotational mass.

 

has anyone dynoed a car with different weight but same-size rims/tires?

 

*Curious*

[KEYSTONE]

I guess the way to do this is to dyno a stock engine on the stand and then chassis dyno the same engine and calculate the loss due to the drive line. Next, modify the engine and repeat the same 2-part test as written above. Only then will you see if driveline loss is a constant value or a percentage of input power.

[speedyman]

Keep in mind that the drivetrain resistance(parasitic loss) for any given set up under any given conditions is NOT linear, it varies with RPM. There are other variables that further cloud the issue such as lubricant viscosities, tire compounds, and the list goes on from there. The 15% factor is a generally accepted rule of thumb for these type of street configurations. Could there be a (+) or( -) 1 or 2 from that point, sure.

[darthracer777]

I agree, but isn't rotational mass also a part of the equation? Say, in the instance of an aluminum DS vs. a heavy steel one? You could say there is less friction on the bearings it turns on and this is the cause of the power increase, but I think rotational mass has a good bit to do with it as well, no? What about a set of heavy vs. light-weight rims? The weight on the wheel-bearings is the same as the car rests on them and the tires on the ground, so the ONLY thing they offer is less rotational mass.

 

has anyone dynoed a car with different weight but same-size rims/tires?

 

*Curious*

 

 

Good question, JWG. As it turns out, rotational mass is a not as significant a parameter as you think. Rotational mass doesn't consume energy. It stores it and returns it to the system.

 

I found a good link that explains everything you need to know about rotational mass, acceleration, etc.

 

rotational mass effects

[BoneDoc]

Good question, JWG. As it turns out, rotational mass is a not as significant a parameter as you think. Rotational mass doesn't consume energy. It stores it and returns it to the system.

 

I found a good link that explains everything you need to know about rotational mass, acceleration, etc.

 

rotational mass effects

 

 

great topic. Thanks your explanations seem to make the most sense.

 

someone nailed the definition above. work is force times distance. horsepower is work per unit time. that is why HP is measued in ft-lbf/second.

 

i am a mechanical engineer from many years back so forgive me if i start using equations... i cant help it. I agree that in a moving mechanical system....

 

Force of friction= Normal force x dynamic coefficient of friction

 

...and the dynamic coefficient of friction doesnt change all that much once your gear oil is up to temp. i also agree that this occurs at every engaged tooth in the gears of the tranny and rear.

 

but, I do think that, if a car has more HP, it will generate a larger normal force (N) and therefore the product of the coefficient of friction and the normal force will be larger... in other words, the frictional force will get larger as the car has more horsepower and therefore the frictional force (parasitic forces) will increase as well. The percentage does not change significantly, however, because the HP is increasing at the same rate as the frictional forces so the ratio remains the same...

 

none of this has anythng to do with the dyno because it only measures HP "where the rubber hits the road"... brake HP. we extrapolate what we think crank HP is based on estimated drive train loss.

 

this is hurtin' my brain! am i wrong here?

[JWG223]

Good question, JWG. As it turns out, rotational mass is a not as significant a parameter as you think. Rotational mass doesn't consume energy. It stores it and returns it to the system.

 

I found a good link that explains everything you need to know about rotational mass, acceleration, etc.

 

rotational mass effects

 

 

I understand the concept of storage/return, and can see how this works (accelerating, then coasting down). However, a dyno is only concerned with acceleration, and not the deceleration, so wouldn't this hurt dyno numbers?

 

I did find this supporting my theory: http://ws6.com/mod-1.htm

 

Keep in mind a drive shaft has a very small distance from its center point, but this is still a sizable difference I think given that small diameter.

[1 Alibi 2]

As winter drags on in the east, ( 9 deg. this morning ), I've had time to do some reading. My goal is to upgrade my blower, leaning toward a 2.9 Wipple, Ford's package for 2011. ( M-6066-MSVT29D ) A tuner in my neck of the woods, Evolution Performance, also markets several " packages " that boast huge hp numbers.........( 858 rwhp )....................The reading has me asking the following question:

.

If you've had your car dynoed, what SAE spec did they use for corrections, SAE J 1349 or SAE J 607 ( also refered to as STD on some dynos ) One mfg. claims that most OEM's use J 1349 for hp & touque #'s in their advertising, it being more conservitive, and reading as much as 4 % " below " J 607 corrected numbers......this 4 % difference may explain why stock 2011 GT 500'S are pulling 508 hp !!

Still in the " plan " stage of the upgrage, although funding is available..................hoping to get some feedback from this knowledgeable group, so I can make sense of the numbers.................thanks.

 

.

After being on a dyno, I can answer my own question, maybe help others....SAE J1349 is a correction standard that applies ( 92.5 % ) to the results, it also uses 77 degrees in it's calibration. SAE J607 ( STD ) uses 96.9 in it's correction & 60 degrees, ( which is adjustable )..Ford uses J1349 in it's hp & tq #'s for advertising, which may be ( 4 % ) below ( STD ) ( J607 ).....Using Ford's #'s for the 2011 GT 500 ( 550 + 4 % = 572 ) ( flywheel ) you get ( - 16 % power train loss ) what I got on a J607 dyno plot.

.

.

Which is nowhere near the numbers posted by others on this site. ( car was stock, even put the Ford air inlet & filter back on )

But the math checks out...........................J1349 hp # would be 459.2..

The lowest magazine article hp #'s I saw was 497.5 for a 2011 MY ( guessing J607 )

Saw posts of 508 & 530 hp for " stock " 2011 GT 500's on this site...................................I don't get it !!!

[bring500]

.

After being on a dyno, I can answer my own question, maybe help others....SAE J1349 is a correction standard that applies ( 92.5 % ) to the results, it also uses 77 degrees in it's calibration. SAE J607 ( STD ) uses 96.9 in it's correction & 60 degrees, ( which is adjustable )..Ford uses J1349 in it's hp & tq #'s for advertising, which may be ( 4 % ) below ( STD ) ( J607 ).....Using Ford's #'s for the 2011 GT 500 ( 550 + 4 % = 572 ) ( flywheel ) you get ( - 16 % power train loss ) what I got on a J607 dyno plot.

.

.

Which is nowhere near the numbers posted by others on this site. ( car was stock, even put the Ford air inlet & filter back on )

But the math checks out...........................J1349 hp # would be 459.2..

The lowest magazine article hp #'s I saw was 497.5 for a 2011 MY ( guessing J607 )

Saw posts of 508 & 530 hp for " stock " 2011 GT 500's on this site...................................I don't get it !!!

 

Almost all pulished tests are run on Dynojet units. The Mustang dyno units give considerably lower readings, by about 13%. They do this by considering "real world conditions" due to things like wind resistance and many other things that reduce hp and torque. They claim to indicate closer to true hp while driving, by adding characteristics from a data base for each specific model, such as Mustang vs Taurus. The operator has to pick the correct model to compensate correctly. My 2010 GT 500 registered 449 on one of these, vs published results closer to 500 on Dynojet models. I could not find another dyno in my area to get results comparable to others. Hope this helps explain things.

[darthracer777]

Actually, the best 'dyno' is the dragstrip itself.......race in good air & good track conditions...use the best 'sticky' tires...shift correctly......then your ET & speed should tell you everything you need to know.

[JWG223]

Actually, the best 'dyno' is the dragstrip itself.......race in good air & good track conditions...use the best 'sticky' tires...shift correctly......then your ET & speed should tell you everything you need to know.

 

 

Even this doesn't seem to work the best.

 

For example, the GT500 traps 118mph-119mph usually.

 

The new GT-R (2012) traps 123-126mph from mag-tests I have seen.

 

The GT-R outweighs the mustang by a couple hundred pounds, and puts down less power.

 

However, the DCT and the closer gear ratios allow it to put down more "Average" horsepower over the run, while the AWD all but eliminates wasted power due to traction loss on shifts.

 

The only way horsepower can ACCURATELY be measured is on an engine dino.

 

You should see all the guys on the 'vette forum talking about how their car "Dynoed high" or "is there a problem, I only made..." ALL THE TIME!

Yet Katech has dynoed PLENTY of LS7's on their engine dyno, and reported that the consistancy between them is almost scary, and dead on at the rated 505.

[darthracer777]

Even this doesn't seem to work the best.

 

For example, the GT500 traps 118mph-119mph usually.

 

The new GT-R (2012) traps 123-126mph from mag-tests I have seen.

 

The GT-R outweighs the mustang by a couple hundred pounds, and puts down less power.

 

However, the DCT and the closer gear ratios allow it to put down more "Average" horsepower over the run, while the AWD all but eliminates wasted power due to traction loss on shifts.

 

The only way horsepower can ACCURATELY be measured is on an engine dino.

 

You should see all the guys on the 'vette forum talking about how their car "Dynoed high" or "is there a problem, I only made..." ALL THE TIME!

Yet Katech has dynoed PLENTY of LS7's on their engine dyno, and reported that the consistancy between them is almost scary, and dead on at the rated 505.

 

 

Using a dyno is informative and you can make many adjustments based on the numbers, but it's much more satisfying (and fun) having your ETs slips in hand. And, you can make chassis, carb, & timing adjustments at the track (the old fashioned way).

[Venom55]

I asked the same question when i saw 858 rwhp form Evolution. I had the same setup with my car and made 747 rwhp on a 100 octane tune with Jon Lund remote tuning it. He said the 858 was with C116 racing fuel a very aggressive tune and perfect run conditions. Fred Cook from Evolution is great to deal with, he set me up with the right excpectations of power with their package setup around 730 rwhp. All I know is 747 rwhp is hard to put down even with drag radials on the street, 858rwhp is crazy... Jon said it best a dyno is used as a measure of tuning the car. Let me know if have any questions before you buy any parts.. I have already tested different combos on my GT500,