Gears and the Dyno?
Thread Starter
Melting Slicks
Joined: May 2006
Posts: 2,891
Likes: 1
From: NJ..."the way I saw it, everyone takes a beating sometimes."
Gears and the Dyno?
I have read that when you dyno a car with gears that it will result in lower power numbers. Is there a standard in which the numbers drop or is it in correlation with the gear size?
For example:
3.42s = 340/340 whp/tq
Then the same car with 4.10s = ?
For example:
3.42s = 340/340 whp/tq
Then the same car with 4.10s = ?
Melting Slicks
Joined: Jan 2007
Posts: 2,633
Likes: 2
From: Evans Colorado
Yes.
The car doesn't actually lose power, but the way the calculations are done a car with lower gears can give you a lower figure.
I'm too lazy right now to search for it, but there's a great thread over on SVTPerformance in the New Edge Cobra section that addresses exactly why this is the case.
The jist of it has to do with the moment of inertia and the speed at which the rotor is accelerated. The computer has to "catch up" so to speek and shows a bit less power.
The same thing happens when you go to a lower (numerical) gear, but in reverse.
This is exactly why dyno's should be used as tuning tools rather than for bragging rights.
The car doesn't actually lose power, but the way the calculations are done a car with lower gears can give you a lower figure.
I'm too lazy right now to search for it, but there's a great thread over on SVTPerformance in the New Edge Cobra section that addresses exactly why this is the case.
The jist of it has to do with the moment of inertia and the speed at which the rotor is accelerated. The computer has to "catch up" so to speek and shows a bit less power.
The same thing happens when you go to a lower (numerical) gear, but in reverse.
This is exactly why dyno's should be used as tuning tools rather than for bragging rights.
Melting Slicks
Joined: Jan 2007
Posts: 2,633
Likes: 2
From: Evans Colorado
Okay, I stopped being lazy. The following is a question posed directly to dynojet with their response on the subject.
QUESTION:
A car having 3:27 rear end gears is placed on a DynoJet chassis dyno and has a dyno performed. Then this same car is taken off the DynoJet chassis dyno and has the rear end gears changed to 4:10. The car is then placed back on the DynoJet chassis dyno and another dyno is performed. Will the second dyno show a loss of horse power caused by the 3:27 to 4:10 gear change?
ANSWER:
Yes
EXPLANATION:
"The 4:10 gear will show less horsepower than the 3:27. The reason is due
to rate of acceleration changes. The rate of acceleration is quicker with
the 4:10 because of torque multiplication being increased to the rear
wheel. The horsepower will show less because the targeted RPM is met
before the horsepower has a chance to overcome the rotational mass (dyno,
drive line, etc.) or moment of inertia in speed. Because the speed is
decreased and the RPM is met faster, the horsepower never has a chance to
catch up with itself, so to speak. The overall ratio of 1:1 will always
produce the most horsepower on the chassis dyno. Having said this, a
similar problem can occur with horsepower loss when the rear gear is too
high. The horsepower is being absorbed in just trying to keep the
rotational mass spinning. Please keep in mind that your engine's
horsepower never changes but what gets to the dyno or drive surface
does. If you have any further questions please don't hesitate to
ask. Thank you."
QUESTION:
A car having 3:27 rear end gears is placed on a DynoJet chassis dyno and has a dyno performed. Then this same car is taken off the DynoJet chassis dyno and has the rear end gears changed to 4:10. The car is then placed back on the DynoJet chassis dyno and another dyno is performed. Will the second dyno show a loss of horse power caused by the 3:27 to 4:10 gear change?
ANSWER:
Yes
EXPLANATION:
"The 4:10 gear will show less horsepower than the 3:27. The reason is due
to rate of acceleration changes. The rate of acceleration is quicker with
the 4:10 because of torque multiplication being increased to the rear
wheel. The horsepower will show less because the targeted RPM is met
before the horsepower has a chance to overcome the rotational mass (dyno,
drive line, etc.) or moment of inertia in speed. Because the speed is
decreased and the RPM is met faster, the horsepower never has a chance to
catch up with itself, so to speak. The overall ratio of 1:1 will always
produce the most horsepower on the chassis dyno. Having said this, a
similar problem can occur with horsepower loss when the rear gear is too
high. The horsepower is being absorbed in just trying to keep the
rotational mass spinning. Please keep in mind that your engine's
horsepower never changes but what gets to the dyno or drive surface
does. If you have any further questions please don't hesitate to
ask. Thank you."
Thread Starter
Melting Slicks
Joined: May 2006
Posts: 2,891
Likes: 1
From: NJ..."the way I saw it, everyone takes a beating sometimes."
Yes.
The car doesn't actually lose power, but the way the calculations are done a car with lower gears can give you a lower figure.
I'm too lazy right now to search for it, but there's a great thread over on SVTPerformance in the New Edge Cobra section that addresses exactly why this is the case.
The jist of it has to do with the moment of inertia and the speed at which the rotor is accelerated. The computer has to "catch up" so to speek and shows a bit less power.
The same thing happens when you go to a lower (numerical) gear, but in reverse.
This is exactly why dyno's should be used as tuning tools rather than for bragging rights.
The car doesn't actually lose power, but the way the calculations are done a car with lower gears can give you a lower figure.
I'm too lazy right now to search for it, but there's a great thread over on SVTPerformance in the New Edge Cobra section that addresses exactly why this is the case.
The jist of it has to do with the moment of inertia and the speed at which the rotor is accelerated. The computer has to "catch up" so to speek and shows a bit less power.
The same thing happens when you go to a lower (numerical) gear, but in reverse.
This is exactly why dyno's should be used as tuning tools rather than for bragging rights.
Thanks. I get the idea of how/why it happens. I'd just like to know if there is a set number or a way to calculate how much lower it reads.
Thread Starter
Melting Slicks
Joined: May 2006
Posts: 2,891
Likes: 1
From: NJ..."the way I saw it, everyone takes a beating sometimes."
Let me see if I can clarify. Most of us agree that the LS2 will make about 340rwhp on the dyno. So figuring that it's rated at 400bhp, we can calculate a 15% loss to the wheels. So my question is if we take a stock geared C6, dyno it at 340rwhp, then install 4.10 gears. What will it dyno? This will give us some sort of percentage to go off of, right?
Thread Starter
Melting Slicks
Joined: May 2006
Posts: 2,891
Likes: 1
From: NJ..."the way I saw it, everyone takes a beating sometimes."
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Team Owner
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From: NJ
St. Jude Donor '05-'08
Thread Starter
Melting Slicks
Joined: May 2006
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From: NJ..."the way I saw it, everyone takes a beating sometimes."
Team Owner
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From: Northern, VA
St. Jude Donor '15
"In honor of jpee"
So then, it is not linear, a 10 rwhp loss.
The delta betw. a 3.42 rear (first post) and a 4.10 rear is about 19%. And 19% of 342 rwhp is not 10 rwhp.
The delta betw. a 3.42 rear (first post) and a 4.10 rear is about 19%. And 19% of 342 rwhp is not 10 rwhp.
Thread Starter
Melting Slicks
Joined: May 2006
Posts: 2,891
Likes: 1
From: NJ..."the way I saw it, everyone takes a beating sometimes."
So basically we need a car dyno'd with stock 3.42s and then dyno'd again with 4.10s. Same car, same dyno, same day even, if possible.
Has anyone done something similar that can help? Any advice from a dyno operator?
Melting Slicks
Joined: Nov 2003
Posts: 3,375
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From: Northern CA
You would need to know the inertias of the dyno, your engine and to a lesser extent the inertias of your drive train and wheels.
Then you could calculate it fairly accurately.
The Dynojet explanation in an earlier post left me hoping there is still someone at Dynojet that understands haw an inertia dyno works. The person that wrote that sure didn't have clue.
If you can get ahold of one of the competent technical types at Dynojet, they might be able to give you some real rough ballpark numbers for the differances.
Then you could calculate it fairly accurately.
The Dynojet explanation in an earlier post left me hoping there is still someone at Dynojet that understands haw an inertia dyno works. The person that wrote that sure didn't have clue.
If you can get ahold of one of the competent technical types at Dynojet, they might be able to give you some real rough ballpark numbers for the differances.
Le Mans Master
Joined: Jul 2005
Posts: 5,328
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I found it hard to believe a person in his positiion could write such BS.
I'd like to think maybe he was trying to explain it so the layman could understand it and failed miserably.
Le Mans Master
Joined: Jul 2005
Posts: 5,328
Likes: 405
There are some dynos that measure HP first then calculate torque. Inertia type dynos do it that way. The speed of the weighted drum is monitored and HP is calculated by measuring the difference in speed over a very small time. Since the weight and moment of inertia of the drum is known, HP can be determined by the increase in the kinetic energy of the drum divided by the time interval. Using the RPM from a pick-up, torque is calculated by T=HP*5252/RPM.
FYI: HP is also heat or 1 HP=746 Watts...for a clear demonstration of HP, try grabbing a 60 watt light bulb that's been on a few moments and tell me how imaginary it is.
Drifting
Joined: Jun 2005
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But that would be 19% of 60 HP, the loss. Or about 11.4 HP more loss. Actually, it would be less than that, because the parts that are different are the flywheel, clutch, propeller shaft and rotation parts of the transmission. The effective inertia of the differential, half-shafts, wheels and tires remains the same. You'd have to measure all the parts and calculate it out to be exact, but I'd say that 10HP figure is pretty close.