Anyone Fast Running Over 4" Stroke?
#21
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Now that we have established torque, we can move on to more fun things like how much torsional shear stress a driveshaft can take assuming dumping the clutch can be expressed as an instantaneous torsional impact and the shaft responds elastically to the impact. Since the driveshaft must absorb the energy of the flywheel, we must calculate the rotational potential energy contained in the flywheel, U = 1/4 pi (radius of the flywheel)^4 (flywheel thickness)(flywheel material density)(rotational speed of flywheel)^2. Our shear stress can now be expressed as twice the square root of (U)(Modulus of rigidity of driveshaft (72 GPa for aluminum or 207 GPa for steel)) / (volume of the driveshaft). Pretty cool eh? <img border="0" title="" alt="[Big Grin]" src="gr_grin.gif" />
#22
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I'd like to add my $0.02: if you have an CVT transmission (like audi a4), then torque is irrelevant, neglecting changes in drivetrane losses with different gearing.
hp=torque*rpm. We need max torque at the wheels, two engines, one making 100ft*lb at 10000rpm, and another making 200ft*lb & 5000rpm , will have the same hp and after first one is geared down twice more than the second, they will make IDENTICAL torque at the ground and produce identical acceleration results.
It IS torque at the wheels that applies the force at the wheels, and, =>, moves the car at any instance, but less torque produced at larger rpm, thanks to gearing, results in identical acceleraction.
NOW - real world transmissions have power band - you go, f.e., from 4000 to 6000rpm in each
gear. In each particular gear curve area - torque integrated over rpm is important. That is related to hp, though hp is a function of torque with value at any rpm, it's not an integral. HOWEVER - it is not important where exactly in the 4000-6000rpm torque reaches it's maximum, as long as in each particular gear the area is max. In fact, if torque has a well-defined peak, in our example it's beneficial to have the peak around 5000rpm.
Now - in the above example, given constant maximum torque, max hp is going to be produced if tq maximum is at 6000rpm. That does not necessarily create the maximum area under the curve. Depending on a curve, it might be beneficial to have curve maximum at some lower point. Therefore, max hp in itself, as a function of torque, is generally irrelevant. Integral of torque over each gear rpm range IS relevant, however it's NOT hp, which is just a convinient function. Hope that helps <img border="0" title="" alt="[Smile]" src="gr_stretch.gif" />
hp=torque*rpm. We need max torque at the wheels, two engines, one making 100ft*lb at 10000rpm, and another making 200ft*lb & 5000rpm , will have the same hp and after first one is geared down twice more than the second, they will make IDENTICAL torque at the ground and produce identical acceleration results.
It IS torque at the wheels that applies the force at the wheels, and, =>, moves the car at any instance, but less torque produced at larger rpm, thanks to gearing, results in identical acceleraction.
NOW - real world transmissions have power band - you go, f.e., from 4000 to 6000rpm in each
gear. In each particular gear curve area - torque integrated over rpm is important. That is related to hp, though hp is a function of torque with value at any rpm, it's not an integral. HOWEVER - it is not important where exactly in the 4000-6000rpm torque reaches it's maximum, as long as in each particular gear the area is max. In fact, if torque has a well-defined peak, in our example it's beneficial to have the peak around 5000rpm.
Now - in the above example, given constant maximum torque, max hp is going to be produced if tq maximum is at 6000rpm. That does not necessarily create the maximum area under the curve. Depending on a curve, it might be beneficial to have curve maximum at some lower point. Therefore, max hp in itself, as a function of torque, is generally irrelevant. Integral of torque over each gear rpm range IS relevant, however it's NOT hp, which is just a convinient function. Hope that helps <img border="0" title="" alt="[Smile]" src="gr_stretch.gif" />
#23
FormerVendor
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Hey Jason, I think that Mikey ran a 4.125 stroke on a 4.080 bore. Its undersquare, but that is what I remember reading in GMHTP LONG ago <img border="0" title="" alt="[Cool]" src="gr_images/icons/cool.gif" />
I know that you are trying to kill that little 346 of yours, so is this brainstorm of whats to come? <img border="0" title="" alt="[Big Grin]" src="gr_grin.gif" />
I know that you are trying to kill that little 346 of yours, so is this brainstorm of whats to come? <img border="0" title="" alt="[Big Grin]" src="gr_grin.gif" />
#24
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Thats awesome if Mickey is running a 4.125 stroke! I was just curious, I'll probably stay 4" stroke, just kicking stuff around <img border="0" title="" alt="[Smile]" src="gr_stretch.gif" /> That 346 has forged pistons, its pretty tough <img border="0" title="" alt="[Smile]" src="gr_stretch.gif" />
Jason
Jason
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Jason
Co-Owner, Texas Speed & Performance, Ltd.
2005 Twin Turbo C6
404cid Stroker, 67mm Twins
994rwhp/902lb ft @ 22 psi (mustang dyno) www.Texas-Speed.com
Jason
Co-Owner, Texas Speed & Performance, Ltd.
2005 Twin Turbo C6
404cid Stroker, 67mm Twins
994rwhp/902lb ft @ 22 psi (mustang dyno) www.Texas-Speed.com