Jason 98 TA

iTrader: (4)

Anyone know of a car with more than 4" of stroke running good times? I'm just wondering if more than 4" of stroke is just hurting performance. I think if I was building a motor it would be 4.050 bore 4" stroke iron block, anyone running more than 4" of stroke with good luck?

Colonel

That's a very general way to ask a little more complex question...

Take my car for example. Gimme a 4.1, 4.2, or even a 4.25" stroke and I'll make the car pretty close to as fast as it is now. I might change the gearing, converter, cam, or all of the above. It's the total package that matters. But, IMO, my 4" stroke is better for an all out effort due to the extra RPM at the same HP that it affords. This leads to the benefit of utilizing more converter and especially more gearing. In other words I'll be faster with a 525 RWHP engine that turns 6800 than a 525 RWHP engine that only turns 6100. I'm able to use 4.10 gears and a 4600 stall rather than 3.73s and a 4000 stall. This translates into quicker ETs.

But yes, you can still go with what would be, in most people's opinions, very quick even with a less than perfect stroke. <img border="0" alt="[Burnout]" title="" src="graemlins/burnout.gif" /> The longer stroke will in turn give you better drivability and a smoother idle for a given HP level.

<small>[ October 01, 2002, 11:20 AM: Message edited by: Colonel ]</small>

JimmyKash

</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by Colonel:
<strong>

IMO, my 4" stroke is better for an all out effort due to the extra RPM at the same HP that it affords. This leads to the benefit of utilizing more converter and especially more gearing. In other words I'll be faster with a 525 RWHP engine that turns 6800 than a 525 RWHP engine that only turns 6100. I'm able to use 4.10 gears and a 4600 stall rather than 3.73s and a 4000 stall. This translates into quicker ETs.

</strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">FINALLY....someone who agrees with me that cubes aren't always king

racer7088

You'd be faster at the lower rpm because the lower stall converters and taller gears are more efficient anyday. The fact is that the cylinders aren't really long enough for 4.250 stroke in the LS1 anyway with any rod and a piston that will fit it so it's really a moot point for the most part. The BBC is another story and again the BIG BBCs whoop *** on the smaller ones and run taller gears and lower stall converters as well. There is a limit to what will clear and balance with 4+ inch strokes in the LS1 before hitting the diminishing returns area and I agree with the Colonel's estimate of the 4 inch crank being a great practical displacement while still retaining reasonable strength in the pistons and yet still balancing pretty easily.

Colonel

IMO, more gearing and more converter = more leverage over the ground which generally = quicker ETS all else being equal.

You can have just as an efficient 4600 stall as a 4000 stall. My TPE4600 lost 10 RWHP on the dyno. That's less than a stock converter would lose.

Gearing inefficiencies are a small and almost mute point that really only matters on the dyno. The extra TQ multiplication FAR outweighs the slight loss of efficiency.

"The fact is that the cylinders aren't really long enough for 4.250 stroke in the LS1 anyway with any rod and a piston that will fit..."

I don't know what modifications had to be made but it's already been done and will be done again. My '02 Z28 will have a 4.25" stroke.

BTW, I have alot of respect for racer7088's knowledge and opinions. The above are only opinions of mine. <img border="0" alt="[Burnout]" title="" src="graemlins/burnout.gif" />

<small>[ October 01, 2002, 11:17 AM: Message edited by: Colonel ]</small>

racer7088

Colonel,

Generally when using strokes at that length you'd need a near 6 inch cylinder to keep the pin in the hole at the bottom. You know how I like cubic inches but you have to run SERIOUSLY short pistons and you basically run out of room everywhere to put stuff in the LS1 engine case at this size. Reliability then starts going downhill. The Ford Windsor for instance has another .300 more cylinder and 9.500 deck height to hold the pin than the 5.500 long and 9.235 deck LSx does and the race versions of this block have even more at 6.00 inch cylinders to give an example. If it works though I like it! <img border="0" alt="[cheers]" title="" src="graemlins/gr_cheers.gif" />

SStrokerAce

iTrader: (2)

</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by Colonel:
<strong> In other words I'll be faster with a 525 RWHP engine that turns 6800 than a 525 RWHP engine that only turns 6100. I'm able to use 4.10 gears and a 4600 stall rather than 3.73s and a 4000 stall. This translates into quicker ETs.

</strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">Colonel,

Not to be a ***** but that part of your statement is just something I gotta talk about.

First if you have 525rwhp @ 6100 then you have 452 rwtq at that same RPM, if it's at 6800 then you lose TQ and go down to 405rwTQ. That's a difference of 11.6% You would need more than 11.6% increase in gear to make up for that loss of TQ. Even a jump from 3.73's to 4.11's will not get you there. Then you might be adding a extra shift in your run and the ET's would go down because of that. (maybe)

Now if you can make the same HP at a lower RPM then you will have more TQ, and that's what will get you down the track faster. It is much harder to have a engine make more TQ at the same cubes than it is to make more HP at the same cubes, all you gotta do then is spin the MFer faster.

BTW this is all moot anyways, the engine that is going to win is the one with the best average power/torque. Not the best peak numbers. The sad thing on here is the cam dicussions that keep adding duration to high flowing heads, man if you wanna have a freeking hard hitter motor that is a sleeper with TONS of TQ back down on that cam. You might drop a mph but gain in ET and that's always what wins.

Bret

<small>[ October 04, 2002, 05:03 PM: Message edited by: SStrokerAce ]</small>

Colonel

Without getting into the whole TQ v/s HP thing to much...

It's not TQ that moves you down the track exactly (since TQ in itself does not move and thus does no work.) It's HP to the ground that accelerates the car. Since HP is a measure of TQ and movement (in this case the movement being RPM), TQ certainly is important in that it is half the equation of making HP (again, RPM being the other.)

So, when accelerating down the track is is average HP to the ground that I'm concerned about. Not at what RPM my peak HP or peak TQ or what my peak HP or TQ is. It's laying down the most HP to the asphalt that matters. Overall gearing, transmission gearing, TQ curve (or you could say HP curve, doesn't matter), and the TQ converter all play a hand in determining how much average HP is applied to the ground over the course of the 1320'.

One more consideration is the power of inertia that a spinning engine holds when sitting at speed waiting to launch. In other words (albeit an extreme example), an engine sitting at 6000 RPM would hold roughly 4 times the inertia that an engine sitting at 3000 RPM would. With a suitible converter and traction we can take advantage of this stored power. <img border="0" alt="[Burnout]" title="" src="graemlins/burnout.gif" />

Through proper gearing we can shift the same amount of times reagardless or if our engine turns 6000 or 9000 RPM.

<small>[ October 06, 2002, 02:49 PM: Message edited by: Colonel ]</small>

Jason 98 TA

iTrader: (4)

You guys are reading to far into this post!! We can toss out theories all day long, I'm wondering if there is anyone running good times with more than 4" of stroke.

Jason

Fenris Ulf

</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by Colonel:
<strong>IMO, more gearing and more converter = more leverage over the ground which generally = quicker ETS all else being equal. </strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">Very true, but you are forgetting that leverage over the ground is a function of gearing/verter AND the input torque. That said, the longer crank will show more input torque but at a lower rpm and preferably transfer it to the ground through numerically lower gears, and the 4.0" stroke will have less input torque but at a higher rpm so the leverage won't necessarily favor either setup. Using the stock LS6 intake definately restricts rpm power of the shorter stroke motors, so its really a question of how much stroke you can stuff into the car to take advantage of the max rpm band the intake allows without accelerating the piston to a precarious speed and risk failure. All assuming that the intake hinders high rpm power, but I just thought I would point that out.

P&DZs

iTrader: (1)

Jason 98 TA. I think your thread got derailed/hijacked. <img border="0" title="" alt="[Wink]" src="gr_images/icons/wink.gif" /> <img border="0" title="" alt="[Big Grin]" src="gr_grin.gif" />

Maybe just temporarily diverted. <img border="0" title="" alt="[Smile]" src="gr_stretch.gif" />

<small>[ October 08, 2002, 10:02 PM: Message edited by: P&DZs ]</small>

WeatherGuy

</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by Colonel:
<strong>Without getting into the whole TQ v/s HP thing to much...

It's not TQ that moves you down the track exactly (since TQ in itself does not move and thus does no work.) It's HP to the ground that accelerates the car. Since HP is a measure of TQ and movement (in this case the movement being RPM), TQ certainly is important in that it is half the equation of making HP (again, RPM being the other.)</strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">To get into this hijacked thread a little . . .

Colonel, something confuses me about what you just wrote. Torque has units of force (within a radial coordinate system here). Since F=ma, and taking the normalizations of the vectors so that we can dispense with vector quantities, we obtain a = F/m. In other words, the acceleration of the vehicle at any point is exactly equal to the net torque (successfully) applied scaled by the mass.

Now, power and work have equivalent units, so that HP = work done. It then holds that the average acceleration can be backed out by the area under the curve arguments, but that the acceleration at any point on the power curve is precisely given by differentiating the net applied power - i.e., the torque.

Have I made a mistake here, misunderstood you, or am I writing dribble at the end of a 13-hour work day?

racer7088

Weatherguy, power is a time based view of torque. Static torque does nothing. Torque being applied at a rate is power and that's what accelerates stuff in real time. Acceleration can only occur through time and so we're concerned with power since it also is a time based measurement. Torque is just a component and RPM is the time based component that must go with it to give it any real relevance. Together they are linked to tell you what POWER the engine makes.

<img border="0" alt="[Fluffy]" title="" src="graemlins/fluffy.gif" /> Tq X RPM / 5252 = POWER ! <img border="0" alt="[Fluffy]" title="" src="graemlins/fluffy.gif" />

Colonel

You guys talk way over my high school educated head! LOL!

Restating for us laymen what Eric so eloquently wrote...

TQ in itself does NO work. It doesn't move. It is a measure of force NOT work. Movement HAS to take place for work to have happened. Movement nor time is involved in TQ. If I push against a wall as hard as I can for a week I am applying TQ to the wall but NO work has taken place UNLESS the wall moves. This work could be measured in HP but not in TQ. (BTW, this is not a completely accurate analogy but only in the sense that TQ is in reference to force that attempts to rotate. Again, work only takes place once this rotation takes place.)

Think of HP (work) as a measurement of TQ (force) and motion (RPM in the case of an engine.)

"Very true, but you are forgetting that leverage over the ground is a function of gearing/verter AND the input torque."

More accurately, the input TQ at a given RPM (or HP) since movement is taking place. And no, I'm not forgetting that. <img border="0" title="" alt="[Smile]" src="gr_stretch.gif" />

<small>[ October 09, 2002, 02:02 PM: Message edited by: Colonel ]</small>

horist

My head hurts... time for a <img border="0" alt="[chug]" title="" src="graemlins/gr_chug.gif" /> <img border="0" title="" alt="[Big Grin]" src="gr_grin.gif" />

Pro Stock John

iTrader: (34)

Average HP iz da bomb if you make say 450rwhp from 4500-6500 that's better than a motor that peaks at 475rwhp but only makes 425rwhp at 4500... Peaky stuff is peaky.

I think 4.250 stroke stuff is too out there as far as block modifications, piston velocity, side loading the pistons etc... Heck 4.000 stroke is a lot by 4th gen terms.

I run 4.075 stroke and my car has never gotten a great pass out of it but let me redyno with new AF ratios and then we can pass final judgment.

John 2:13

KTamez

To give extreme examples of whats being talked about look at Pro Stock in NHRA and IHRA. 500ci vs 800+ ci leaves the two types of cars within only a couple tenths of each other. Why? One of the reasons is directly related to this:

An NHRA P/S'er turns ~9500rpms, but uses DEEP gears, and so accelerates well into the 6s, even though it "only" has ~1200hp.

An IHRA motor only spins ~7500rpms, makes some more power AND torque, but yet doesn't get the job done much faster, as it doesn't have the gearing advantage.

Somewhere in between you could place our own 2200lb 63 Vette. ~1150hp 632. ~7500rpm. However even with the same hp, as NHRA P/S it can't run the Number, as w/o the RPM it can't run the DEEP gear the P/S car does.

This is just my take on this sort of comparison. In other words I'm with Stephen on thos one. <img border="0" title="" alt="[Smile]" src="gr_stretch.gif" />

Toxic Texan

My head is starting to <img border="0" alt="[kaboom]" title="" src="graemlins/gr_bomb.gif" />

Nine Ball

iTrader: (38)

nerds! NERDS!
Nerd-Alert!
How Nerdical
Nerd-A-Mongo!

<img border="0" alt="[jester]" title="" src="graemlins/gr_jest.gif" />

<img border="0" title="" alt="[Embarrassed]" src="gr_emb.gif" /> I actually understand all that stuff <img border="0" title="" alt="[Embarrassed]" src="gr_emb.gif" />

-Tony

WeatherGuy

</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by racer7088:
<strong>Weatherguy, power is a time based view of torque. Static torque does nothing. Torque being applied at a rate is power and that's what accelerates stuff in real time. Acceleration can only occur through time and so we're concerned with power since it also is a time based measurement. Torque is just a component and RPM is the time based component that must go with it to give it any real relevance. Together they are linked to tell you what POWER the engine makes.

<img border="0" alt="[Fluffy]" title="" src="graemlins/fluffy.gif" /> Tq X RPM / 5252 = POWER ! <img border="0" alt="[Fluffy]" title="" src="graemlins/fluffy.gif" /> </strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">OK - I am more awake tonight than last night.

It turns out that we are saying the same thing, but using different words.

To clear up:

F = ma ==> a = F/m The acceleration at any point, period, is going to be exactly equal to the force in pounds applied scaled by the mass. This is Newton's second law, and is not debatable. Note, however, that torque is measured in ft-lbs - the foot unit creeps in because we are measuring a force (torque) in radial units, but that is the only difference.

Now, as written above, power is the time based view of this. By the work-energy theorem, and simplifying a little, average power applied = work done. But recall that power is a time based measurement (joules/sec in MKS units). Thus work = force*distance = average power applied. To make this simpler:

force*distance = average power applied ==>
(using F=ma)
m*a*distance = average power applied ==>
average accel = average power applied / (m*distance)

Thus the average acceleration can be obtained from the average POWER applied, while the instantaneous acceleration is always directly proportional to the torque applied.

</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by Colonel:
<strong>TQ in itself does NO work. It doesn't move. It is a measure of force NOT work. Movement HAS to take place for work to have happened. Movement nor time is involved in TQ. If I push against a wall as hard as I can for a week I am applying TQ to the wall but NO work has taken place UNLESS the wall moves. This work could be measured in HP but not in TQ. (BTW, this is not a completely accurate analogy but only in the sense that TQ is in reference to force that attempts to rotate. Again, work only takes place once this rotation takes place.)

Think of HP (work) as a measurement of TQ (force) and motion (RPM in the case of an engine.)</strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">True and not true. True - no work is done if the wall does not move no matter how much you torque it. Not true - the example is a little misleading. In the case of the wall, F=ma. You are applying a force to the wall F, but since the wall is not accelerating, an equal force (the static tension of the wall) is counteracting your force, so the wall does not accelerate. Hence no movement and no work. What you said is technically not wrong, but slightly misleading. However, you do note that above.

The problem is that we cannot seperate F=ma from power and torque. If net F = 0, no work is done no matter what torque is applied. If F ~=0, we do work, and then we can define torque as a derivative of the instantaneous amount of work done.

</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by Nine Ball:
<strong>nerds! NERDS!
Nerd-Alert!
How Nerdical
Nerd-A-Mongo!

I actually understand all that stuff

-Tony
</strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">Yup - we are. We are nerds who are doing well in our careers and have fast cars. I knew there was going to be a time to get back for all of the poundings I took in elementary school <img border="0" title="" alt="[Big Grin]" src="gr_grin.gif" /> <img border="0" title="" alt="[Razz]" src="gr_tounge.gif" />

Incidentally, it is cool to have a discussion like this. It sure beats the heck out of the "opposition", like the Ford guy who put the trailer hitch on my Durango. To quote him - "The first thing I did when I got that motor was to take the EFI off and slap a carb on it." Or the Autozone guy I got a Mobil 1 filter from:
Him: A Z-28? What year is your truck?
Me: It's a Camaro
Him: Oh yeah, that's right. Is that the 6 or 8 cylinder engine?
Me: It's a Z-28
Him: Would that be the 8-cylinder then?

Greg