So you're telling me that if you hook up an engine dyno to an engine, take one dyno run, then pull the stock flywheel and put on the lightened flywheel, you'll make more on the the second run on the engine dyno?

Last I checked, flywheels were part of the drivetrain, as opposed to the engine.

My statement was referencing the combustion effects of adding a flywheel as opposed to adding things like nitrous, bigger port, or turbo. I would think that the difference with the lightened flywheel will not show up on an engine dyno, but will show up on the wheel dyno. I would imagine that the power peak on the wheel dyno would be the same, but there would be slightly more power along the curve.

I would imagine you're correct; lightening things like the flywheel will increase the engine's ability to accelerate, and acceleration is one of the factors when calculating HP. Anyway, I'm not an expert, but I do know this:

HP ~= mass * velocity * acceleration

Actually, I don't think acceleration has anything to do with it...

http://www.g-speed.com/pbh/torque-and-hp.html


-Ted

port and polish the throttle bodies
 

i've heard of some people porting and polishing the trottle bodies out to gain low to mid range torque increases is this true ??

sorry about the poor grammar im fixing a coolant leak and trying to write lol

He was also contridicting himself. IF you read his entire post, he says to shift AT the peak torque curve so the RPMs fall at the peak torque curve which is impossible.

He then decided to argue with me saying I was wrong by saying shifting at your peak rpm with power. On my engine, thats around 7500-7800. My peak torque curve would be around 5500-6000. Why shift there?

How is that possible?

It does. There's the familiar torque * rpm method, but there's also the mass * velocity * acceleration method. I've used this method myself with awesome results. From the g-tech site (http://www.gtechprosupport.com/support/HP1.htm):

edit: in case you're interested:

v=d/t
F=m*a
W=F*d
P=W/t

P=m*a*d/t
P=m*a*v

Hmmm...interesting...
Cause acceleration can be manipulated through gearing.
The formula does not take into account gearing.

You'd get two different results when doing a 1st gear run versus a 5th gear run?


-Ted

Exactly, the peak HP values would be the same between the two runs, but the time it took to get there would be different. For a car, that formula tells you wheel HP, not engine HP. Aerodynamic drag, etc. is also baked into the result, so it's more like "usable HP" than wheel HP, and will usually be lower than dynos because of it.

Is "v" defined as "terminal" velocity at the instantaneous time when you're trying to calculater horsepower?


-Ted

I believe you would do runs with the Gtech just like you do on a dyno. 4th gear runs at 1:1 ratio.

Ted, if you were referencing my post earlier, my point was that

bhp X hp loss due to drivetrain = whp

All a lighter flywheel or a carbon fiber driveshaft would do is lessen the hp loss. However, I have no proof, but I believe the loss savings will converge to zero as rpm and hp get closer to the hp peak. That's just my intuition speaking there.

....

Quote:
Originally Posted by Project84 https://www.rx7club.com/images/buttons/viewpost.gif
Then shift at a RPM where the gear you are going into will have the engine RPMs at a point where it is just going in to that peak torque area.

That is what I said. I think you are getting hung up on the part that I highlighted in orange.

I never said "shift at the peak torque curve so the RPMs fall at the peak torque curve". The only way that would be possible is if the two gears had the exact same ratio. You completely fabricated the statement, then decided to say that I said it and that contadicting myself and arquing with you. You need to re-read my post, understand it, and stop trying to stroke your giant internet-ego or what ever it is you are doing because I was not and I am not trying to argue with you.

Instantaneous everything. It looks at HP, acceleration, mass, and velocity at a particular instant. If you logged HP over time, calculating HP continuously with that formula, you'd get what looks like a stretched/squished dyno graph, but with unknown RPM.

edit: if you did a smooth zero- to redline- run

Your right, I must have misread your post and some how thought you said to shift at your peak torque.

YAYE! FREE MODS! I always tighten up some of the slack on the throttle cable because they come conservative from the factory for average usage, but since you want more response just get rid of a little bit of slack and you will see a difference. Also it's easier to get out of first, at least to me it was because there are alot of hills around my area.

Sorry, you're right.
I keep getting confused, cause I automatically assume velocity is a derivative fraction of distance over time. :P
You can get instantaneous velocity at one point.
Stupid "dt" kept pointing me in the wrong direction!


-Ted

Before this gets too heavily into torque, gearing, etc., I'll give the simple answer. Shift to maximize horsepower. The answer to all the torque, gearing, etc. calculations will equal horsepower. As for acceleration, acceleration = power / (mass * velocity). Lighter cars accelerate faster and you accelerate quicker at the top of 1st gear (~40mph) than you do at the top of 2nd (~60mph), 3rd, etc. You don't get infinite acceleration at near zero velocity b/c your power is also near zero.

As for throttle response, there are two things to quicken: foot to throttle body and getting the air from outside to the engine. That may be why newer cars are drive-by-wire: quicker throttle response b/c you don't have to deal with the stretchiness of a cable. Might also be quicker b/c of computer control on the throttle, I dunno. Advancing the timing makes the spark plugs fire sooner, so I suppose that's why that would improve throttle response. But, like most tuning, that's not the best thing for the longevity of your engine.