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girlpower

By girlpower,
in Engines & Forced Induction

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simpletool Mentor

simpletool

Posted
Posted

A combination of set figure (offset) and percentage (factor).

For a manual I estimate: flywheel(kw) = 20kw + (rwkw x 1.15)

Derived from stock figures. It's estimating it takes ~20kw to spin a gearbox at high rpm, plus you lose a percentage (15%) of gains as friction.

Seems to work well for a lot of cars, and is a hell of a lot better than straight up 25% loss claims!

ie. Stock GTST 140rwkw > 181kw flywheel

My current 230rwkw > 285kw flywheel

A random 300rwkw > 365kw flywheel

2011 Ford GS 256rwkw > 314kw flywheel, even seems to work for new automatics ;)

mad082 Grand Master

mad082

Posted
Posted

as people have said, it can't be a set figure. not just because of how hard the gears are meshing together. you have to remember that the more power it has the faster it is going to accelerate. the faster something accelerates the higher the forces acting against it. so while it might not be exactly a percentage it won't be a fixed figure either. there might be an exact figure of how much power it takes to turn a gearbox, drive shaft, diff, wheels, etc at a set speed, but when it comes to accelerating all that then it won't be a set number that applies to all power levels.

Adriano Grand Master

Adriano

Posted
Posted

A combination of set figure (offset) and percentage (factor).

For a manual I estimate: flywheel(kw) = 20kw + (rwkw x 1.15)

Derived from stock figures. It's estimating it takes ~20kw to spin a gearbox at high rpm, plus you lose a percentage (15%) of gains as friction.

Seems to work well for a lot of cars, and is a hell of a lot better than straight up 25% loss claims!

ie. Stock GTST 140rwkw > 181kw flywheel

My current 230rwkw > 285kw flywheel

A random 300rwkw > 365kw flywheel

2011 Ford GS 256rwkw > 314kw flywheel, even seems to work for new automatics ;)

That looks pretty good,cant we all agree to this

Rolls Grand Master

Rolls

Posted
Posted (edited)
But still Id think a 2000hp would quite easily lose 3-400 through driveline which is still close to 20% but once again the style of driveline would make a huge difference ,for instance a regular sedan with tailshafts compared to a car that has the engine/gearbox bolted directly to the diff..

It might show it on the dyno but there is absolutely no way it is actually losing 400hp, do you have any idea how much heat that would create? it would be enough heat to literally melt the gears if that was the case.

I think a large amount of it is how inertial dynos calculate their figures and they don't actually 'lose' the majority power that we see on a dyno, they just read less due to the way it is being measured, there are losses due to the extra weight being spun up, but they are not completely lost this way.

Edited by Rolls
  • 2 weeks later...
Thelen Rising Star

Thelen

Posted
Posted

A combination of set figure (offset) and percentage (factor).

For a manual I estimate: flywheel(kw) = 20kw + (rwkw x 1.15)

Derived from stock figures. It's estimating it takes ~20kw to spin a gearbox at high rpm, plus you lose a percentage (15%) of gains as friction.

Seems to work well for a lot of cars, and is a hell of a lot better than straight up 25% loss claims!

ie. Stock GTST 140rwkw > 181kw flywheel

My current 230rwkw > 285kw flywheel

A random 300rwkw > 365kw flywheel

2011 Ford GS 256rwkw > 314kw flywheel, even seems to work for new automatics ;)

That is basically the figures I have been using as well. I've always used 15% for manual, 20% for auto, then some % for age. ie, because the skyline is now 15 years old, you loose more power (compression loss, tune loss, just small bits here and there). Rebuilt/modded changes it though and brings it more closely down to the 15%, as does purely having more power (ie so the 20kw base means nothing).

I can't remember what the figures were on a rotary that was engine dyno built and then drivetrain (ie rear wheel) dyno'd, but certainly it couldn't have been even 15% otherwise the gearbox would have fried.

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