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[rico's-srt]

Quote: Originally Posted by wonderboy

You can even shift a bit higher than peak power depending on your gearing and what your torque curve looks like.


The ideal way to look at it would be to get a dyno sheet of torque at the wheels by MPH in every gear and then overlay them. Where the graphs intersect, thats basically where you would want to shift.

But I think it would have to be done on a mustang dyno or something other than a dynojet. Dynojets have a difficult time simulation actual road loads with wind resistance etc..

This can also be a fun experiment.. Take two same model cars that are equal in performance. Run them against each other, - one shifting at peak HP, the other at peak torque. and see what happens.

you just made less sense than me, in an earlier post you said the equation for figuring horsepower was something (x) and that all hp and torque curves would cross at 5252, how would overlaying both curves help me figure my shiftpoint? furthermore, every car's shiftpoint is at 5252 according to what your saying.

[matthew00029]

Quote: Originally Posted by rico's-srt

you just made less sense than me, in an earlier post you said the equation for figuring horsepower was something (x) and that all hp and torque curves would cross at 5252, how would overlaying both curves help me figure my shiftpoint? furthermore, every car's shiftpoint is at 5252 according to what your saying.

NO he said they are crossing at 5252 the HP and TQ curves but you need to shift a little after your HP drops off or wherever you get the fastest times but somewhere around there. Your hp won't always drop off early like stock turbo's, it all depends on what turbo, engine and gearing you are using. More variables than just power drop off is what he is saying in a nut shell.

[wonderboy]

Quote: Originally Posted by rico's-srt

you just made less sense than me, in an earlier post you said the equation for figuring horsepower was something (x) and that all hp and torque curves would cross at 5252, how would overlaying both curves help me figure my shiftpoint? furthermore, every car's shiftpoint is at 5252 according to what your saying.

Ok. Let me try to explain this again.

5252 rpm is simply a point in the rpm range where torque= hp (mathematically)
At 2626 rpm torque is always 2X more (numerically) than hp.. Its just a linear relationship.

The hp formula and the shift points don't have anything to do with one another. It was my answer to a different question.

I said that I typically would shift at around peak HP. It's only a general rule of thumb. Sometimes you can shift a bit past peak HP depending on what gear I am in and what my torque curve looks like.

I then said.. the IDEAL way to do it is:

(lets TOTALLY forget about HP for a second).

make 3 runs on the dyno... lets say 3rd gear run, 4th gear run and a 5th gear run. Overlay those charts (just the torque curves) on the same graph.
Since they are based on MPH on the X axis, you will see a point where the end of 3rd gear will torque graph should cross the 4th gear run. It is at that point that torque to the wheels are = and that's when you should shift.

I saw three problems with your initial post.

Quote:

regardless of how the dyno figures the number, its all bullshit, when everyone goes to the dyno, look at the TORQUE line, see where its highest point is, and shift there. thats the fastest your gonna be able to make your car go

1. You didn't account for the torque multiplication of gearing. That's one of the main reasons why you shouldn't shift at peak torque. (unless you have some SERIOUS drop in torque right after peak torque)

2. You didn't account for what torque is available after you have shifted gears (ignoring torque multiplication from gearing). So maybe you have 400 lb/tq at 5000rpm and only 350 lb/tq at 5500. So you shift into a different gear at 5000rpm and encounter only 300 lb/tq at 3000 rpm.

3. You mentioned that your car won't accelerate past your peak torque number, which is not true either. I assume you mean drag limited. A car, given limitless gearing will accelerate until the amount of drag on the car is greater than the force propelling the vehicle (torque). That doesn't mean that a car can't accelerate past peak torque.

Don't take my word for it.. Go to the track and make one run with shifts at peak torque, and another with shifts at peak hp. See what happens to your times.

I'm not sure how else I can explain this to you.

[slidesquad]

Quote: Originally Posted by nutz

for a "drag" vehicle pure volume is needed, a land speed car is NOT a drag car usualy land speed is looking for alot of torque "good velocity" to offset wind resistance and are not looking for acceleration they got miles to get to 200mph.....apples and oranges my friend

thats my .02

its not all about cfm and size there are 3 procceses to the intake stroke (Overlap, Suction and Charging. ) one key thing is high velocity ....

What happens when a low velocity port fills the cylinder really well, but too early ?? The result is a slower intake charge that stops flowing into the cylinder. Then some of the charge gets pushed back out of the cylinder and into the port as the piston returns up the bore during the intake charging phase. The gain in flow doesn't offset the loss in port velocity.
. Here's an analogy:

It's just like a roadrace; you can lead for 3 laps, then run out of steam and end up in 7th place.

In racing, only the last lap counts, because the one who leads at the checkered flag wins.

In the intake cycle "race" the last lap is the charging phase,
and the checkered flag is the intake valves closing.


its all about a proper match to your setup