making KA rev to 8k rpms.... how hard?

[Veriest1]

And the times when it will come into play are almost completely irrelevant when taking the car as a whole. And it's certainly not 2 car lengths, that was just a ridiculous statement.

The times when it does come into play seem to be completely relevant to winning. I don't understand this "taking the car as a whole" euphemism you speak of.

What Alan did or didn't do is irrelevant to RPMs being relevant and "the car as a whole." Please don't group them together like that when debating my points because I never said such an event occured or didn't occur.

As for changing gearing based on the engine, take the viper for example. It redlines at 5k. A viper has a top speed around 200mph. That obviously requires very wide gearing. Specifically, a 3.07 final drive. While the redline is lower than say an NX or 240, a viper can do 0-60 in first gear.

My dads powerstroke redlines somewhere around four grand and it has a 4.10 in it. Therefore, I would say it's dependant on what you want to do versus what the redline is. I've never heard of anyone changing gearing in response to incerasing their rpms. I have heard of it being changed because of track layout.

No. You are soooo wrong. Your car ends up going slower at higher gears because of increased air resistance, which doesn't factor in when you're going the same speed. If you are making 300hp and you're in second gear and I'm making 300hp and I'm in third gear, and we're both going the same speed, in a car the same weight and aerodynamically the same, we have the same acceleration.

So you're second gear is the same size as your third gear? Aren't they normally lower? Have you ever tried to accelerate from a stop in 3rd gear then tried the same thing in second gear?

Quote »Torque itself means absolutely nothing for acceleration.[/quote]power = torque * RPM

I guess power and RPM don't either....

[InsanityInc]

The times when it does come into play seem to be completely relevant to winning. I don't understand this "taking the car as a whole" euphemism you speak of.

Euphemism? The point is, winning a 0-60 time because of a semantic difference in gearing tells you nothing about the capabilities of the two cars. If you beat me to 0-60 like that in a quarter mile or a road race, my overall acceleration will top yours. That's why it's irrelevant.

Quote »My dads powerstroke redlines somewhere around four grand and it has a 4.10 in it. Therefore, I would say it's dependant on what you want to do versus what the redline is. I've never heard of anyone changing gearing in response to incerasing their rpms. I have heard of it being changed because of track layout.[/quote]That's because it's a desiel truck. The entire point of the car is to make massive amounts of torque, so of course it's going to have a torqey gearing.

Quote »So you're second gear is the same size as your third gear? Aren't they normally lower? Have you ever tried to accelerate from a stop in 3rd gear then tried the same thing in second gear?[/quote]You know nothing of physics. Let me explain how acceleration works, because it's obvious you have no clue what you're talking about. Torque is a type of work. Work and Forces (work = Fd, if you didn't know) by themselves cannot move you, only the application of that force/work over time can, at which point it's called power. Reduction gearing changes your torque, not your power. The reason why it's difficult to accelerate from a dead stop in a higher gear (and thus with lower torque) is that all other FORCES acting on the car (inertia, friction, air resistance, etc) are all subtractive from your torque, while a reduction gearing system is a ratio relationship. Since you probably don't understand why that's important, I'll give you an example:

Say your car overall has a force of friction of 20N, and due to it's weight, has an inertial force of 200N, since you're stopped, air resistance obviously isn't a factor. That force subtracts from your torque. I'm going to really simplify this and just subtract numbers so you can see how it works, but this isn't an accurate representation of calculating a car's acceleration, so don't take it as that. Say you resist that with your torque in first gear, and have 200N left, meaning you started with 420N. For the sake of simplicity, lets say first gear is 1:1 and you have no final drive. Second gear is .5:1, so your car produces 210N to the wheels before other considerations. Your power would still be the same if you had no resistance to that force, due to the greater rotation speed of the wheels. HP = ftlbs*rpm/5252 If you use reduction gearing to increase or decrease torque, you increase or decrease wheel RPM by the same factor, so your power output is exactly the same. However, since forces subtract, that makes it not the same from a stop due to the inertial force. Your theoretical power is the same, but your car doesn't have enough torque to move at all. Now when you're in motion, the inertial force goes down really far, so that you can actually accelerate with less torque.

That's a pretty shiesty example due to the conversion between ftlbs and newtons being much worse than that, considering you have to work it into Nm but I'm guessing you get the idea? Power is always the same, torque differs, and you need enough torque to overcome other forces so your remaining torque can actually translate to power. Plus in a higher gear, you're running at a much lower RPM where your engine can barely work.

Quote »power = torque * RPM

I guess power and RPM don't either....[/quote]it's torque*rpm/5252 for horsepower, by the way. And that's why I said torque means nothing in and of itself. Power is what accelerates you because that's the definition of power, the application of work over time. When you're measuring bhp and btq, obviously it's the same regardless of the gear you're in. When you measure wtq and whp, wtq changes ratiotically with wrpm so you get the same power, but can achieve a higher speed without blowing up your engine.
Modified by InsanityInc at 7:09 PM 8/4/2005

[Nismo_Freak]

No. You are soooo wrong. Your car ends up going slower at higher gears because of increased air resistance, which doesn't factor in when you're going the same speed. If you are making 300hp and you're in second gear and I'm making 300hp and I'm in third gear, and we're both going the same speed, in a car the same weight and aerodynamically the same, we have the same acceleration. Torque itself means absolutely nothing for acceleration. The fact that you're trying to dispute this proves you haven't got a ****ing clue what you're talking about.

I don't need to even comment, if you understand this chart you will understand why it is you are wrong.

This is a chart put out by Porsche BTW.

Oh and air resistance is exponential as speed increases. You will notice that airflow resistance, and other frictional losses are minimal in effect until greater speeds are achieved.

So you're wrong in the theoretical realm, and I'll tell you why you're wrong in reality as well. If you rev your KA out to 7000, and I shift at 6400, then gain the same amount of speed lower in the RPM range, I will be accelerating faster than you. End of story. This is because engines do not have a perfectly linear powerband. The only way you'll show a faster time on paper is if that shift happened very soon before the end of the E.T. before the superior acceleration makes up the time differential from the shift, and/or before you yourself shift to make up that time difference.

See above chart as to why this is incorrect. You could rev the Boxster out to 8000 RPM and it would create more acceleratory force in 1st gear than at peak power in 2nd.

First, when did I say shifting after torque drops off? You decide your shift points based on a number of factors, one of which is when your power drops off, and another is where you will end up in your powerband after you shift from that point.

I was making a general point. Not really the overall response to your commentary.

No, you're actually being owned by logic and physics right now.

*Yawn*

Try again.

[AZhitman]

take the viper for example. It redlines at 5k.

Ummm, no. It produces 500 bhp @ 5600 rpm, and redlines at 6K.

[Veriest1]

Euphemism? The point is, winning a 0-60 time because of a semantic difference in gearing tells you nothing about the capabilities of the two cars. If you beat me to 0-60 like that in a quarter mile or a road race, my overall acceleration will top yours. That's why it's irrelevant.

If Car A beats Car B through a segment of a track by having the ability to accelerate a couple of seconds longer then Car A is faster in that segment of the track. This has nothing to with the amount of power it puts out. Both cars could be putting out approximatly the same power with the difference being Car A has better balanced internals and a higher redline.

Quote »That's because it's a desiel truck. The entire point of the car is to make massive amounts of torque, so of course it's going to have a torqey gearing.

Like I said, the gearing doesn't rely soley on RPMs. It's a function of what one wants to do with the engine he or she has and the engines characteristics.

You know nothing of physics. Let me explain how acceleration works, because it's obvious you have no clue what you're talking about.

I'm a Criminal Justice/Theology major... of course I know nothing of physics. This stuff is just a hobby. Not my lifes work.

Quote »Torque is a type of work. Work and Forces (work = Fd, if you didn't know) by themselves cannot move you, only the application of that force/work over time can, at which point it's called power. Reduction gearing changes your torque, not your power. The reason why it's difficult to accelerate from a dead stop in a higher gear (and thus with lower torque) is that all other FORCES acting on the car (inertia, friction, air resistance, etc) are all subtractive from your torque, while a reduction gearing system is a ratio relationship. Since you probably don't understand why that's important, I'll give you an example:

Say your car overall has a force of friction of 20N, and due to it's weight, has an inertial force of 200N, since you're stopped, air resistance obviously isn't a factor. That force subtracts from your torque. I'm going to really simplify this and just subtract numbers so you can see how it works, but this isn't an accurate representation of calculating a car's acceleration, so don't take it as that. Say you resist that with your torque in first gear, and have 200N left, meaning you started with 420N. For the sake of simplicity, lets say first gear is 1:1 and you have no final drive. Second gear is .5:1, so your car produces 210N to the wheels before other considerations. Your power would still be the same if you had no resistance to that force, due to the greater rotation speed of the wheels. HP = ftlbs*rpm/5252 If you use reduction gearing to increase or decrease torque, you increase or decrease wheel RPM by the same factor, so your power output is exactly the same. However, since forces subtract, that makes it not the same from a stop due to the inertial force. Your theoretical power is the same, but your car doesn't have enough torque to move at all. Now when you're in motion, the inertial force goes down really far, so that you can actually accelerate with less torque.

That's a pretty shiesty example due to the conversion between ftlbs and newtons being much worse than that, considering you have to work it into Nm but I'm guessing you get the idea? Power is always the same, torque differs, and you need enough torque to overcome other forces so your remaining torque can actually translate to power. Plus in a higher gear, you're running at a much lower RPM where your engine can barely work.

it's torque*rpm/5252 for horsepower, by the way. And that's why I said torque means nothing in and of itself. Power is what accelerates you because that's the definition of power, the application of work over time. When you're measuring bhp and btq, obviously it's the same regardless of the gear you're in. When you measure wtq and whp, wtq changes ratiotically with wrpm so you get the same power, but can achieve a higher speed without blowing up your engine.

Modified by InsanityInc at 7:09 PM 8/4/2005

According to this table on howstuffwork.com at 3,000 rpms in fifth gear the output from the transmission is higher than in first gear.

GEAR RATIO RPM at Trans Output Shaft with Engine at 3,000rpm1st 2.315:1 1,295 rpm2nd 1.568:1 1,913 rpm3rd 1.195:1 2,510 rpm4th 1.000:1 3,000 rpm5th 0.915:1 3,278 rpm

This is basically because the engine is turning a gear that gets larger as the gear number increases. In first the transmission gear has to turn 2.315 times to turn the out put shaft once. In contrast fifth gear turns less than once for each revolution of the output shaft at .915:1.

Traveling at 30 mph the wheels and differential will always be spinning at the same speed no matter what gear you're in. This is why the car lurches and jerks whenever a downshift is prformed without having the engine matched to wheel speed for the gear being selected.

When I'm traveling 30 miles per hour and I accelerate in second gear at 3,000 rpms I can accelerate much faster than doing the same exercise in third, fourth, or even fifth gear. The same is true while parked. This is confusing becasue according to the chart I pasted the revolutions of the output shaft should increase as gears increase. However, when ever I shift into 3rd at 3,000 rpms going 30mph my engine speed falls after the clutch is engaged and since the engine is having to turn a larger gear without the momentum of the cars speed translating through the wheels to aid it I can't accelerate as fast.

Hopefully I got that mostly correct. I may have just said the same thing as you in a less scientific manner though.

Lets look at some dyno graphs.

This is European Car Magazines project M3 and more details can be found on their website.



Notice that power was picked up everywhere and the curve itself was extended out a tad over stock once the car could breath better while making better power everywhere in the RPM range.

EDIT: Factory redline is at 6600. The AA tuning took it out to 7200.

Here's Pandas14's dyno run with ITB's

The motor isn't dead up high anymore and gains were seen over the avg. stock dyno throughout the graph. His car now has more useable power. Obviously some cams would help it but that's what the other thread is for.

Here's a stock KA dyno for comparison. The redline wasn't increased in this last example but the power curve was extended. So as long as the power curve keeps getting larger extending the redline will be benificial.

Modified by Veriest1 at 9:22 AM 8/5/2005
Modified by Veriest1 at 9:32 AM 8/5/2005

[Veriest1]

Here's the same car as was used in the stock KA graph with pulley, intake, and exhaust. I'd rather drive panda's smooth powerband myself.
Modified by Veriest1 at 9:35 AM 8/5/2005

[InsanityInc]

I don't need to even comment, if you understand this chart you will understand why it is you are wrong.

This is a chart put out by Porsche BTW.

Oh and air resistance is exponential as speed increases. You will notice that airflow resistance, and other frictional losses are minimal in effect until greater speeds are achieved.

That chart is ambiguous to the point of uselessness. Do you have a link to the page it came off of? Is that net force or gross force? Is it accelerative force or force enacted on something else? You can claim it's one or the other to support your argument, but that wouldn't make sense because it flies in the face of basic physics. Whether I'm producing 300hp with 100 torque or 300hp with 500 torque, I'm going to accelerate at the same rate. Regardless of whether or not the torque difference is due to gearing or just engine output.

Read this: http://www.houseofthud.com/car...r.htm

Hell, or just take those little dash mounted horsepower meters for example (they are semi-legit, they would work perfectly if you had no suspension). They use a device capable of measuring g forces to calculate power, not torque. Then it finds torque by working backwards.

Or from a physics perspective, it's very easy to see why torque doesn't equal acceleration. Torque is work, it has a distance and a force component. Power has a distance, force and time component. Acceleration cannot be measured without time. Torque cannot be converted to acceleration.

[cyrus240sx]

wow i didnt know what i thought was a simple question would start so much controversy... i just thought simply if you were able to go 1000 more rpm in your powerband in every gear, that you would be faster overall.... i didnt mean this to get all physical...

[InsanityInc]

Also, read this, Nismofreak:

http://www.yawpower.com/tqvshp.html

Very complete, and explains why the accelerative Gs from torque mean roughly d!ck compared to your actual accelerative rate.

[InsanityInc]

wow i didnt know what i thought was a simple question would start so much controversy... i just thought simply if you were able to go 1000 more rpm in your powerband in every gear, that you would be faster overall.... i didnt mean this to get all physical...

The problem is, with how badly the power drops off even to the stock 7k redline, revving out another 1000 would be pointless and actually slow you down.

[Nismo_Freak]

Bwhahahhahahahhahahahhahah...

Main Entry: g force

Part of Speech: noun

Definition: a unit of inertial force on a body that is subjected to rapid acceleration or gravity, equal to 32 ft. per second per second at sea level; also written g-force

Usage: also used attributively

Just give it up already! LOL.

[Nismo_Freak]

Also, read this, Nismofreak:

http://www.yawpower.com/tqvshp.html

Very complete, and explains why the accelerative Gs from torque mean roughly d!ck compared to your actual accelerative rate.

Note that at any point on the chart, the percent difference in the rate of acceleration is EXACTLY the difference in horsepower. For instance, at 200 mph, the Honda F1 engine is accelerating at a rate of .572 G’s, while the Cummins diesel is accelerating at a rate of .228 G’s. If we divide .228 into .572 we get 2.5, and so the acceleration rate of the Honda is 2.5 times greater than that of the Cummins

You just owned yourself pretty badly haha.

[InsanityInc]

You still don't understand. Trying to explain it to you is a completely lost cause. The rate at which the speed of your tires is increasing has little to do with how hard you are turning them. Not to mention you failed to take into account that the acceleration RATE is 2.5 times greater because the horsepower is 2.5 times greater, not the torque, so not sure how I owned myself.

If torque were really all that mattered, accelerative rates would be independent of horsepower, which is untrue. Also, 0-60 and 0-120 times of high powered cars would have a FAR greater disparity than they do, since air resistance is already quite a large factor.

[Nismo_Freak]

You still don't understand. Trying to explain it to you is a completely lost cause. The rate at which the speed of your tires is increasing has little to do with how hard you are turning them. Not to mention you failed to take into account that the acceleration RATE is 2.5 times greater because the horsepower is 2.5 times greater, not the torque, so not sure how I owned myself.

If torque were really all that mattered, accelerative rates would be independent of horsepower, which is untrue. Also, 0-60 and 0-120 times of high powered cars would have a FAR greater disparity than they do, since air resistance is already quite a large factor.

I don't know where you are getting this torque arguement you think I made.

Why dont you try looking up some of the previous torque vs. hp threads and understand where I stand on the matter, or you can re-read my previous statements, none of which indicate that peak torque is the peak acceleration point of the vehicle.

The point is the car is accelerating hardest in 1st gear because of the torque multiplication benefit the gearing employs. This means in X amt. of time the car will output more work because you have increased the "reactants" of the equation.

This is why diesel trucks have so many gears, and why the G-forces are exponentially greater in the lower gears.

And again, air resistance is exponential, as is any other frictional resistance.

Cd = D / (A * .5 * r * V^2)

The thing you will notice is that the acceration curve will follow an inverse relationship. Meaning the difference in g-force between 1st and 2nd is much greater than 4th to 5th. This is because of two factors, the resistance (bearing friction, drag, tire friction, rotational efficiency, etc.), and the fact that the gear ratios are much closer.

[Q45tech]

Remember all a simple chassis dyno does is measure the speed increase of the drum in 1/1000 sec increments and mathematically converts than increase into a torque number [how much torque it takes to speed the drum up in that time] and simultaneously creates a HP number by apply the 5252/ rpm ratio formula once the torque has been calculated.

All those dyno graphs posted trying to prove some point? Still not sure what the point is?..........other than the most torque at the tire road interface accelerates the fastest at a given EQUAL [tire] rpm [vehicle speed point]and weight and aero drag. A major reason "The Calculus" was invented to solve this type of problem-----acceleration in very tiny time increments.

A chassis dyno is very very very far from a real world on the road acceleration [timewise] due to aerodynamics and the differences in a road [asphalt] vs a steel drum.

The never ending saga "Engine HP vs Torque" from non engineers.

My " fill the blank" is BIGGER than yours, that males always engage in.

[InsanityInc]

I don't know where you are getting this torque arguement you think I made.

Why dont you try looking up some of the previous torque vs. hp threads and understand where I stand on the matter, or you can re-read my previous statements, none of which indicate that peak torque is the peak acceleration point of the vehicle.

The point is the car is accelerating hardest in 1st gear because of the torque multiplication benefit the gearing employs. This means in X amt. of time the car will output more work because you have increased the "reactants" of the equation.

This is why diesel trucks have so many gears, and why the G-forces are exponentially greater in the lower gears.

And again, air resistance is exponential, as is any other frictional resistance.

Cd = D / (A * .5 * r * V^2)

The thing you will notice is that the acceration curve will follow an inverse relationship. Meaning the difference in g-force between 1st and 2nd is much greater than 4th to 5th. This is because of two factors, the resistance (bearing friction, drag, tire friction, rotational efficiency, etc.), and the fact that the gear ratios are much closer.

Yes, I'm aware it's exponential. That's my point. The disparity between 0-60 and 0-120 would be FAR greater than it is if torque from gearing were all that mattered for acceleration. Also, accelerating "harder" and accelerating faster are two entirely separate things, which is what you're not getting.

[Nismo_Freak]

Yes, I'm aware it's exponential. That's my point. The disparity between 0-60 and 0-120 would be FAR greater than it is if torque from gearing were all that mattered for acceleration. Also, accelerating "harder" and accelerating faster are two entirely separate things, which is what you're not getting.

I am not repeating myself again.

If you can't read that acceleration chart I posted and understand it then you won't understand what I'm trying to say.

[Nismo_Freak]

Remember all a simple chassis dyno does is measure the speed increase of the drum in 1/1000 sec increments and mathematically converts than increase into a torque number [how much torque it takes to speed the drum up in that time] and simultaneously creates a HP number by apply the 5252/ rpm ratio formula once the torque has been calculated.

All those dyno graphs posted trying to prove some point? Still not sure what the point is?..........other than the most torque at the tire road interface accelerates the fastest at a given EQUAL [tire] rpm [vehicle speed point]and weight and aero drag. A major reason "The Calculus" was invented to solve this type of problem-----acceleration in very tiny time increments.

A chassis dyno is very very very far from a real world on the road acceleration [timewise] due to aerodynamics and the differences in a road [asphalt] vs a steel drum.

Dennis you aren't refering to the acceleration plot that I posted as well are you?

[InsanityInc]

I am not repeating myself again.

If you can't read that acceleration chart I posted and understand it then you won't understand what I'm trying to say.

I understand it just fine. You're completely missing my point. While how hard you are spinning everything is RELATED to your accelerative rate, it's not the end all of it. If torque is all that matters, explain why peak acceleration occurs at your horsepower peak? Horsepower is the application of torque over time, which is why it matters. If you apply 100FtLbs of torque 100 times/s, you have the same acceleration as if you apply 200FtLbs of torque 50 times/s, and reduction gearing can only change it ratiotically like that, so your actual accelerative rate will always (give or take a bit from various losses) be the same regardless of the gear you're in.

Not to mention you still didn't give me any of the information about the chart that I requested, or the page where it originally came from.

[InsanityInc]

Think of it this way. You're sitting in a wagon, I come up and shove you really hard and kind of slow, and you go 10 feet. Then I come up and shove you really fast 10 times, each time you go 1 foot. Say it takes the same time to do both. I applied more force in the first situation, but the same power in both. You accelerated the same, since you got there in the same amount of time. But which one do you think is going to give you a greater "thrown back in your seat" feeling?

[Nismo_Freak]

This arguement has ended.

You obviously lack the ability to read and comprehend.

It's like arguing with a deaf 4 yr. old

[InsanityInc]

This arguement has ended.

You obviously lack the ability to read and comprehend.

It's like arguing with a deaf 4 yr. old

Riiight. I'm the one who doesn't understand. All you've done is posted an ambiguous chart and said "Gs, Gs lol".

But whatever, think you're right all you want.

[Veriest1]

All those dyno graphs posted trying to prove some point? Still not sure what the point is?..........other than the most torque at the tire road interface accelerates the fastest at a given EQUAL [tire] rpm [vehicle speed point]and weight and aero drag. A major reason "The Calculus" was invented to solve this type of problem-----acceleration in very tiny time increments.

A chassis dyno is very very very far from a real world on the road acceleration [timewise] due to aerodynamics and the differences in a road [asphalt] vs a steel drum.

Heh, I didn't really conclude that very well. Sorry. My point was that I don't see how extending the redline wouldn't be benificial as long as the engine continues to breath efficiently.

How does this apply to this thread? A turbo can force the air in to make it breath.

Quote »The never ending saga "Engine HP vs Torque" from non engineers.

My " fill the blank" is BIGGER than yours, that males always engage in.[/quote]You sound like my dad (an electrical engineer) who is always saying things to the effect of, "Why don't you get a real degree?"

Anyway, I was trying to figure out if/why my assumptions and observations are correct and if you engineers could explain it in a less confusing manner that would be great. Almost all of my posts in this thread are stating observations I've seen and wondering why they may be incorrect or why I may be misinterpreting them. I used quite a few question marks actually.

Anyway, I, for one, would like to know what the graph nismo posted means. I haven't been able to find it/ figure it out yet.

[s13sr20chris]

im not an engineer but i listen carefully when those who are speak. i feel like i have this rather abstract lesson learned.

some corrections.

Torque is work, it has a distance and a force component. Power has a distance, force and time component. Acceleration cannot be measured without time. Torque cannot be converted to acceleration.

torque is not work. torque is a "force". forces cant even do work without a "rate". then you end up with "power".

in a car torque is the force, rpm the rate, and horsepower the power. it could just as easily be voltage, amps, and watts.

a car does accelerate hardest(from the perspective of rpm) at the torque peak. the reason to ride it out a little is to try to match acceleration to the next gear. the reason is that the next gear will give you less torque at the wheels if you shift at the torque peak. this is all very clearly illustrated(to me anyway) in alans acceleration graph(the "porsche" graph).

we have strayed from the original intent of the thread. sure your ka can rev to 8k. some ka builders have done it. if she can breathe at that rpm, the torque wont drop off(too much) and hp will continue to rise(or at least level out).

[s13sr20chris]

...all the threads that dont belong in the eng forum and this gets discussed here...

funny

[Nismo_Freak]

Anyway, I, for one, would like to know what the graph nismo posted means. I haven't been able to find it/ figure it out yet.

The graph has several points to notice.

1. RPM (right y-axis)2. Force of Acceleration (left y-axis) 3. Speed (x-axis)4. Overlayed gearing diagram (thin linear lines)5. Graphed Results (force in relation to gear, rpm, and speed - thick black lines)6. The verticle dashed line (denotes ideal shift point for the gear)

Basically with this graph you can plot out your ideal shift moments. The graph plotted out the force of acceleration for each gear as it's ran through.

Notice in 1st gear it is ideal to run it up all the way until redline, this is because even though the engine power has been reduced the gearing is still putting out more wheel torque (what dennis described as the bottom line point) than it would if you upshifted into 2nd.

These diagrams are in the back of every Porsche owners manual. I can get a picture of my friends 944 Turbo manual and show you some of the other graphs they have in the back. Porsche definately puts some cool stuff in their manuals! I also think his 1966 911 manual has the same info in it, I will have to dig it out and see.

The one I posted is from a Boxster forum's FAQ.

[s13sr20chris]

yeah, or you could do it the old way(as i did once for an sr20/s13/t25/14psi/avg dyno) by just charting the wheel torque per rpm before and after shifting.

[Veriest1]

torque is not work. torque is a "force". forces cant even do work without a "rate". then you end up with "power".

An example of torque without a rate would be the example of a one ft. stick in a vise with a one ft. string attached to the end holding up a one pound wieght. The wieght is excerting one ft. lbs. of torque on the stick but without the "rate" you can't calcualate any "power" so no work is actually being done.

So, Nismo_Freak, the graph modified to have all of its axis labeled (and some of your post copied into the bottom for my referance) would be like this attachment.

Niether first nor second gears "Force of Acceleration" crosses that of another gear. In contrast third gear becomes less effective than fourth at just over one hundred mph and the dotted line can be seen cutting off the top of third gear in the gearing overlay. Thus indicating an ideal shift point of around 7100 rpms in third gear and so on and so fourth for gears fourth, fifth, and sixth. With the shift points for fourth and fifth both coming slightly earlier than each preceeding gear. And for this Boxster only first and second gear should be held until redline.

That is freakin' cool.

So, extending the redline is only going to be benificial (with this level of power) until the first gear force curve intersects with the second gear force curve because the biggest gap lies between these two. Once these two curves meet the force curve, of at least first gear, will need to be raised or flattened in some way for an increase in RPMs to be benificiary.

Am I correct? Also, would it be improper to substitute the word "torque" into the above paragraph on place of the word "force?"

Now lets say, in theory, I've lightened and balanced and yadda yaddad this Boxsters engine internals and first gear can be held until it just touches second gears curve. What is the best way to go about changing the curve for the better? It seems to me lowering the gear ratio would be an option but it would work against how long I could hold first gear since first gear would wind out at a lower speed and the best way to further improve the engine and start this whole process over would be to add more power. Correct?

EDIT: Lowering first gear will just make first work harder but not longer and this is the reason my dads truck has a low 4:10 rear end. It maximizes the "force" * "rate" = "power" equation by raisin the rate, right?
Modified by Veriest1 at 10:14 PM 8/7/2005

[Nismo_Freak]

Right, pretty much.

That is a naturally aspirated motor, it gets harder to dictate faster (time wise) shift points when you are considering a turbocharged motor, due to differences in turbine response between RPM points and loading values (higher gears = faster spool in relation to RPM).

[InsanityInc]

torque is not work. torque is a "force". forces cant even do work without a "rate". then you end up with "power".

Torque is work. W = Fd. D in the case of torque is the distance from the centerline to the application point. Work can happen without movement in this case, because distance isn't referring to a moved distance. It's only a "force" in layman's terms.

http://en.wikipedia.org/wiki/Torque

Quote »a car does accelerate hardest(from the perspective of rpm) at the torque peak. the reason to ride it out a little is to try to match acceleration to the next gear. the reason is that the next gear will give you less torque at the wheels if you shift at the torque peak. this is all very clearly illustrated(to me anyway) in alans acceleration graph(the "porsche" graph).[/quote]No, it isn't at the torque peak. Go hop in your 240 and spin it all the way out to 7000 down some stretch of road. The car feels faster and faster until about 6000, at which point you can very plainly feel the acceleration dropping off. The acceleration very clearly does not drop off after the torque peak, nor does it in any car.