The Dodge Challenger Message Board

This is a bit heavy and not all my own words. I do not have a Physics or Mechanical Engineering Degree!

But here goes….

First of all you have to get to grips with the basics. Before talking about Torque lets understand the basic unit which is Work.

Imagine you have a one pound weight bolted to the floor and you try to lift it with one pound (or 10 or 50 or whatever) of force. Remember it is bolted to the floor so it will not actually lift.

In this case you have applied force and exerted energy but no work will have been done.

Now you un-bolt the weight and apply just enough force to lift the weight one foot.
You have now done one foot-pound of work.

If it took you one minute to lift the 1lb weight one foot then you were doing work at the rate of one foot-pound per minute. If it took you just one second to lift the weight then you were doing work at the rate of 60 foot-pounds per minute, and so on.

James Watt observed that the average horse at the time could lift a 550lb weight one foot in one second. So it was performing work at the rate of 550 foot pounds per second or 33,000 foot pounds per minute. So he made this rate of work ONE HORSEPOWER.

When you apply this to a running engine you find it is impossible to measure directly. What is measured on an engine dynometer is TORQUE so any calculations have to use this unit.

Torque is a twisting force. Imagine you had a steel rod in your hand held horizontal at arms length. The rod has a one pound weight hanging one foot along from your hand.

If you disregard the weight of the rod you will need to apply a foot-pound of torque (twisting force) from your arm to the rod to hold it horizontal.

OK?

So now say if we rotated that one pound force on its one foot rod in a circle a whole 360 degrees you have moved the end of the rod against a one pound force for a distance of 6.2832 feet (Pi x a two foot wide circle) so you have done 6.2832 foot pounds of work.

Now see if you can follow this.

Watt said 33,000 foot pounds of work per minute was ONE HORSEPOWER.
So if we divide the 6.2832 foot pounds of work into 33,000 foot pounds we get 5252.
SO!
One foot pound of torque at 5252 rpm is equal to 33,000 foot pounds per minute of work and so is equivalent to one horsepower.

So if we only move that one pound force at the rate of 2626rpm we are making ½ a horsepower.

So to calculate horsepower from torque you use this formula;

HP = (Torque x RPM) divided by 5252

You may be also able to see that at 5252rpm horsepower will always exactly match torque because 5252 divided by 5252 cancel each other out so at that exact rpm HP=Torque.
Look at any torque/horsepower dyno sheet for any engine and you can see the two lines always cross at 5252rpm.

You can also see that torque will always be higher than horsepower below 5252rpm.

So what is the real difference between torque and horsepower in the ‘real’ automotive world?

Well as a driver you can only ‘feel’ torque. It is that push in your back. You cannot feel horsepower.
Any car in any gear will accelerate at a rate that exactly matches its torque curve (disregarding air and rolling resistance etc). So a car will pull hardest at its torque peak in any gear.

Also a car making 300ft-lbs of torque will accelerate you just as hard at 2000rpm as a car that had the same torque at 4000rpm. However if you use the formula you can see that the horsepower is double at 4000rpm.

In contrast to a torque curve horsepower rises rapidly with RPM. More so when torque is also climbing. However horsepower will continue to climb until well past peak torque as long as the RPM keeps rising until a point where torque is falling faster than RPM is rising. It is after that point that HP drops off.

OK so lets get practical with this and see how this works in our field of interest.

To increase the power of our cars we normally fit a camshaft of a longer duration (the valves open longer). This has the effect of moving peak torque higher up the rpm range. We don’t have to go into how at this stage. Because I don't know! Just accept it does.

But why do we need to make torque at higher rpms? I just said that the car will pull just as hard no matter what rpm it makes peak torque at.

So lets look at something that makes a lot of torque at low RPM to see how it performs. How about a big steam engine?

Let us imagine a steam engine chugging away at 12rpm. Now a real big old steam engine like that could generate something like 2600 foot pounds of torque!

If you hooked that engine up to the drive wheels of a car with no gearing so it was driven directly it would go from 0-12rpm in a flash and the engine would hardly notice!

BUT 12rpm with normal car wheels is about 1mph. Not very fast is it? So why not just gear up the output speed of this steam engine to give a higher top speed?

The trouble is that as you increase the output speed you decrease the output torque proportionally. So if you double the output speed you half the torque at the output.

To make this car go 60mph you would have to multiply the wheel speed with gears x60. That will reduce the torque at the output x60 to about 43 foot pounds! That is less torque than a car would need to get pushed along at 60mph.

So what horsepower would this steam engine have?

12 x 2600 divided by 5252 = 6HP

So you can see that despite having an enormous amount of torque this steam engine is limited in the amount of Work it can do over Time. Or to put it another way it has no Power.

What happens if you make torque at a higher rpm?

Let us take a performance 426 hemi making, say, 600lbs/ft of torque at 5,500 rpm.

Again lets us drive the wheels directly with no gears at all so that the wheels rotate at 5,500rpm. That rpm at the wheels of a typical car is 426mph! But again the amount of torque we have would not actually push a car along at that speed. Air resistance alone would limit its speed.

What is its horsepower?

600 x 5,500 divided by 5252 = 628HP!

What if we fitted an axle with 4.10:1 gears to reduce the speed of the wheels and the mph?

Now we are only doing 103mph but the amount of torque at the wheels has gone up by the same proportion to 2460lbs/ft of torque! That is easily enough force to push even a very heavy car or truck along at 103mph.

So gearing the speed up makes less torque and you cannot go fast. Gearing the speed down makes more torque but you need to make the torque at high rpms to be able to do that and still go fast.

So you see if you want to go fast it is better to make torque at higher rpms than lower. Because you can take advantage of gearing to multiply torque.

If you then take this 426 hemi engine and fit a gearbox between it and the axle to reduce the gearing even more you can see that with a typical 2.45:1 first gear ratio as well as the 4.1 axle gear you will make 3930lbs/ft of torque at the rear wheels. That is enough to lift the average muscle car! But that is launching at 5,500rpm mind.

This is also why we use lower gearing in the axle on drag cars. I went from a 3.23:1 ratio to a 3.91:1 ratio a while ago. That is about a 17.5% reduction in rear wheel speed. But that meant a 17.5% increase in rear wheel torque so the car accelerates faster! Since then I have fitted a 4.3:1 gear to make the car leave even harder.

Some racers even fit a lower 1st gear in the trans.

What about top speed or a land speed record car?
Well here is where HP is even more important because you need to be at peak torque to overcome all the forces trying to slow you down. So if you want a high top speed you need that torque at as high an rpm as possible so you can use less gearing.

OK
So it is easy to make an engine perform on a drag strip or at the speed trials isn’t it? You just make a huge amount of torque at a high rpm and away you go!

Ah but nothing in life is that easy. There are many forces at work to stop you doing that.

Have you noticed when you look at engine torque curves that torque seems to drop off very quickly at high rpms? Want to know why?

Well one of the reasons is Volumetric Efficiency (VE). The higher the rpm the less time the cylinders have to draw in air/fuel. The amount of air/fuel you burn is what makes the torque. If the cylinder only fills 75% of its total capacity (VE is 75%) that is the same as reducing the displacement of the engine by 25%. Which means a proportional loss in torque.

Let me explain that another way. Airflow into the engine usually increases with rpm (or more accurately – piston speed) until it reaches the maximum flow capacity of the carb, or intake runners, or ports, or valves (whichever is smallest) and then it levels off and becomes constant despite increases in rpm. The engine is then flow limited and VE (and therefore torque) drops off.

Another rpm limiting factor is mechanical losses. This is made up of stuff like piston ring friction. Bearings and seal friction, valvetrain and spring pressure, accessory drag, pumping losses (dragging the air in and compressing it) etc. The losses induced by these things increase with engine speed. But they increase with the square of engine speed. That is to say that you double engine speed and you quadruple the mechanical losses.

Also included in these losses is the inertia of the moving parts. Any part that has to change direction robs power. Things like the pistons/rods, even the lifters, rockers and valves. The heavier the part the more power it robs. Again – these losses go up as the square of engine speed. 4 x the engine speed and you get 16 x the losses.
This is the reason that engines are not made by the factory to be high rpm screamers. Engines are simply a lot more efficient at low rpms. So the manufacturer has to balance the need for power against the need for efficiency.

If you managed to read this far you are either a very sad person or very intelligent. Unfortunatly I am more the former.

My brain hurts. Can I stop now?

There is a rough calculation you can use to work out the power potential of a pair of V8 cylinder heads from the intake flow.

It is;

HP = flow @ 10" X 3.44

So basically for every 1cfm of flow you can make 3.44HP.

Now the thing is that some people measure flow at 10-inches of vacuum and others at 28-inches.

So to convert 28" flow numbers into 10" numbers so you can use the formula multiply the 28" flow by 0.598.

Here is an example:
Hughes Engines Stage 2 340 LA cylinder heads flow 245cfm on the intake at 28" of vacuum.

(245 X 0.598) X 3.44 = 504

So these heads can make up to 504HP. However this is a rough estimate of power based on an average engine. It is possible to make up to 30% more power than that if your engine is very efficient with a great induction/exhaust and lots of compression.

For flow infomation on heads try this web site http://users.erols.com/srweiss/tablehdc.htm#Chrysler but bear in mind that most of these heads have been ported even if it does not say so and the flow is at 28".

OK so you know how much power your engine could make. Now you want to know how that translates into 1/4 mile performance?

Start here. MPH = ((HP/10) - (weight/100)) + 103

So let us use the example above. First of all you need to deduct losses from your crank HP to your wheels. As a rule of thumb with street driven automatics I deduct 75HP for the trans and water pump/alternater.

So 504HP - 75 = 429HP

429/10 = 42.9

4000lbs/100 = 40

(42.9 - 40) + 103 = 106mph

Now although many people use this simple formula I find it a bit low. So I tend to use the spreadsheet I have at;

http://www.challenger440.pwp.blueyonder.co.uk/maths.xls

I use this for most of my calculations. Just plug in the numbers and play around with it. From the rear wheel power you can find out what MPH and ET you can run.

Here are some charts that you might find useful too. The "Power-To-Weight Ratio" chart is based on rear wheel power and not engine power.

I personally have found some of these charts to be a bit off. But they give you a basic idea at a glance and so are useful as a rough guide.

Here is a VERY handy and accurate conversion tool.

http://www.challenger440.pwp.blueyonder ... onvert.exe

From the "Transmission" section here are these two spreadsheets that will tell you what speed you can do in each gear and at each rpm.

http://challenger.mpoli.fi/web/4SPEEDchart.xls is for 4-speed users.

http://challenger.mpoli.fi/web/AUTOMATICchart.xls is for automatic users.

If you have a dyno readout graph of your engine power (either real or from a computer desktop dyno) you can work out what the best shift points are for your combo.

The ideal shift point is someway past peak HP. You need to shift at the point past peak power far enough so that when you engage the next gear, and the RPMs drop, you will be at a point on the power curve where the engine is making exactly the same power as when you shifted.

To work this out you need to know how far the rpms drop when you shift gear.

A typical 727 automatic has the following gear ratios;

2.45:1 (first)
1.45:1 (second)
1:1 (third or drive)

To work out the rpm drop we need to know the ratio between each gear.
The ratio between 1st and 2nd is 2.45 divided by 1.45.

2.45/1.45 = 1.69

So if you decide you want to shift to second at 6500rpm just divide 6500 by 1.69 and you get the rpm the engine drops to.

6500/1.69 = 3846rpm

Going from 2nd to 3rd the ratio is 1.45/1 = 1.45

Let us see what the effect of shifting from 2-3 at 6000rpm will have on this hot 440 engine.

6000/1.45 = 4140rpm

As you can see from the graph the shift was too early. The engine is now at a much lower point of the power curve than it was when we started the shift.

Let us try shifting at 7500rpm.

7500/1.45 = 5172rpm

As you can now see the shift was too high. Now we are making more power than we were before we shifted.

In this next graph you can see that both the 1-2 shift and the 2-3 shift have happened at the correct (for this engine) points.

You can also see that the ideal shift point is different for each gear.

1-2 @ 7400rpm and 2-3 @ 6800rpm.

Remember. This is for THIS engine. Yours will be different.

I guess the easy option in the above case would be to average the shift points and shift at 7100 both shifts or save the engine and short shift the 1-2 at 6800 same as the 2-3.

Dave,
Pretty cool theoretical analysis. So, where DO you shift your 440 on the track?

You mean where do I (or Linda) shift?

We shift at 6500rpm. It should be higher but to keep costs down we kept the stock rods which are the (very) heavy six pack rods.
I do not feel comfortable shifting higher with that amount of mass changing direction that many times per second.

I might push it a little harder one day....

Thanks for taking the time to explain this Dave. It's really useful stuff. A couple of follow- on questions, if I may :-

What's the reasoning behind trying to have the rpm the same after the shift as before? Is it so you don't shock the engine/ drivetrain too hard at WOT, or more to do with quicker runs?

What's the optimum rpm to shift at when you've only got a single- stage shiftlight. The highest rpm of the two shifts, the lowest rpm of the two shifts, or somewhere between the two?

What's the reasoning behind trying to have the rpm the same after the shift as before? Is it so you don't shock the engine/ drivetrain too hard at WOT, or more to do with quicker runs?

Quicker runs. It will shock the drivetrain more if you normally shift too early. If you shift early your rpm drops to where you are making less power and you have to build up power again.

Can you see that your average HP over the quarter mile is going to be maximised by shifting at the right point? Average HP in each gear will be PEAK HP. Shifting at any other point will result in a lower average.

By shifting like this and gearing your car so that it crosses the finish line at your max rpm in top gear you will run your quickest ET possible.

What's the optimum rpm to shift at when you've only got a single- stage shiftlight. The highest rpm of the two shifts, the lowest rpm of the two shifts, or somewhere between the two?

I ask that question myself above. I think it might depend on how well you have the car geared for running the quarter mile. If your gears are too tall for the strip you will not be making the best of third gear anyway. If you are rpm limited you might be better sticking to the lower of the two.

Trial and error is the only way with an unknown combination.

Whatever you do don't shift higher than is safe for your engine. Most stock engines will rev to 6500rpm OK as long as everything (and you KNOW everything) in the bottom end is in good order. Otherwise stick below 6000.

By luck (as opposed to skill and judgement on my part), with 3.91 gears my car goes through the top end at about 5800rpm. This is on the old 50 series tyres. The 60 series Quicktimes are due to be fitted in the next few days, which should give me back a few more rpms, BUT hopefully will also improve 60ft times resulting in more mph/rpm's through the lights.

Ideally, I'll then be going over the line between 5800- 6200?

Heeding your advice Dave, I think I'll try shifting at 6200rpm and see how the engine holds up. If nothing goes boom I'll up it by 100rpm each run, and see what the ET's are doing.

Oh great. So if it throws a rod I am going to feel great about this aren't I? I should just keep my mouth shut!

What's a rod?

Don't worry Dave, I'm inherently conservative with this sort of thing. Remember what a wuss I was when you offered me the use of your slicks?

Here is a handy way to get an idea of the fuel requirements of your engine.

To determine required fuel flow, multiply your total expected horsepower
by a bsfc (brake specific fuel consumption) of 0.65. This will give your
fuel flow requirement in 1bs/hr. To convert to gallons per hour, divide
this figure by 5.87.

You should pitch "total expected horsepower" a bit above what you realisticly expect to make from your engine.

For example;

If you are expecting your engine to make about 400hp you might be wise to shoot for 450hp.

So 450 x 0.65 = 292.5lbs/hr

292.5 / 5.87 = 50 gallons per hour.

So you need to select a fuel pump and install fuel lines/filter that will deliver at least that amount.

Here are a few more tools to help you.

http://www.juric.org/mopar/tools/index.php

Wow, so far I have read the first post and I am still trying to put it all together.

I have one question though, you say that you want to make tq higher-as then you will have more HP. Why is it that on STREET cars, builders often want to make low end power? The way I see it, your STREET car would spend more time at lower RPM's rather than higher rpm's. In that case, all that high end power would hardly be used while driving around town, and the car would feel like a slug.

Now lets say you get into a friendly street race . At the track, you would have time and room to hit those high speeds, so I am guessing you would be at a higher RPM more % of the time than if you were racing lets say an 1/8th mile, which could be looked at as an average stoplight street race. So if your at a lower RPM more percent of the time in an 1/8th mile compared to the 1/4, wouldn't it make more sense to have LOW tq curve? That way you will get off the line much faster (assuming you have some stuff done to get the car to hook) and you will have your peak power in your low RPM's where you are spending more time in a short race.

Im may be 100% wrong, im really still trying to grasp all of this. I just thought i'd throw that out there lol.

Thanks for the great posts!


Edit: well I forgot to take in stall speed converters. The higher you are making peak tq, the higher you want the stall speed converter to be at, correct? In that case, you will keep the car at a higher rpm, closer to your high peak tq and it will be fine. On the street however, in stop and go traffic a very high stall could get a bit annoying. So it seems that you would still want lower peak tq so you can avoid having to use a monster tq converter with a high stall speed? I think my head is going to explode! lol

Yeah you really need to read through it a few times before you get the idea.
Yes for a street car that never sees high speed you want low end torque. Like in an electric car. But if you want to go fast and race people you need high rpms.

Look at a Formula 1 car. I think they red line at around 18,000rpm these days? Producing what? 850hp or more from a 3000cc engine with no blowers etc?

Some great tech here;

http://victorylibrary.com/mopar/mopar-tech.htm

Great post, i've been really starting to grasp all of it lol.

One more question/comment about the converter, gears and tq in a street cars though.

So you mention that volumetric efficiency, how safe the motor is at high rpm's ect... are all limiting factors in how high you can have your TQ curve. Now ideally, you'd want that tq curve as high as possible, correct? So say you have that HUGE cam, a nice single plane manifold and good carb and you are making power from 4000rpm's all the way up to the moon with a solid cam. That would be ideal for the 1/4. BUT, You'd also need the appropriate converter and gears to make it all work, otherwise you'd be launching at 2200rpm's and not making any power. So now you need a very high converter to make the best of this power. That converter would probably be too much for the average (maybe daily driven) street car that sees racing use as well, but would be fine for a strictly race car.

Basically what I am trying to ask is wouldn't there be a limit to how high you'd want to make tq on a car that sees actual street use as well as race use simply because you just just wouldn't want to use a super high rpm converter on a street car? If you didn't use that high coverter, the car wouldn't move off the line until you started making power, which would come on high due to the tq curve being raised. I guess im just trying to point out a reason of why you need balance for cars that aren't strictly race cars.

Also, im thinking that you cant ignore low end power either, even if you do want to raise peak tq to as high as you can. I think this would be even more true on a heavy street car. Isn't this one of the reasons why people build stroker motors (aside from VE)? I have always thought that you WOULD want peak tq as high as possible, but you also wouldn't want to sacrifice the tq BELOW peak just to achieve that.

For example, lets say you have two motors, both with similar peak tq numbers. One makes power peak tq at 4000rpm's and the other at 5000rpms. The one that makes it at 5000rpm's barely has any tq on the curve before that, but the one that makes peak tq at 4000rpms has a relatively flat tq curve all the way to peak. In a light weight race car, that can use a super high converter-the motor that makes peak at 4000rpm's I would think is a better choice. But for a heavy car like my 4000pound 96 Impala, that motor would never have enough tq to get me off the line very well, I would guess that the first motor with peak tq at 4000rpm's but a flat tq curve before that would power a boat like the impala much better, all while allowing me to use a converter that would be appropriate for a street car.

Please correct me/school me if im wrong. Im just trying to learn all of this and figure out where the stuff I already know about heavy cars needing low end tq fits in.

My head hurts just reading that!

I will have to sit down and think about this one when I have the time.

Im getting a little confused myself reading it back

I guess what I am trying to ask, simply put, is:

Because we all cant run super high converters on our mostly street driven cars, and also because most street cars are heavy, wouldn't the need to maintain low end tq, and tq BELOW peak tq, be very important on the quest to raise the TQ curve? Meaning if you can raise peak tq a few hundred rpm, but at the same time you lose low end tq before the peak of the tq curve, it may in fact be worse for a car that needs to stick with a mild, streetable converter and lots of weight.

In short, if what I am saying is right, the goal on a street car would be to use a parts combo that makes the most tq in the low rpm's while still raising peak tq. You would not want to raise peak tq as high as possible without paying attention to what you may be doing to the tq curve below peak. It would seem that you would end up with a car that is a slug off the line because it can't get its weight moving being a heavy street car. And because its a street car, you cant just make up for all this by slapping on a 4000rpm converter so you can launch closer to your power curve.

im not sure if im bringing up valid points here, just wanted to throw it out there for debate.

What you say there is right. But you can make up for low rpm torque in a heavy car by using lower rear gears. Remember that the gears multiply the amount of torque actually at the wheels.

But as you suggest it dose not make for an idea daily driver.

If it was simple to build a daily driver that runs 10s on the strip we would all be doing it.

There are lots of people with very fast cars that will tell you their car is very streetable. Most of these people think streetable means it will run down the main street to the shops and back.
I doubt that would take their car and drive it hundreds of miles into the hills for a camping trip. Sit in traffic jams on hot summer days etc.

Thanks! I guess im starting to get it then. i've learned a lot from your posts.

That was one thing that I always wondered about. I've always thought to myself, well whats stopping me from using some high revving race motor for a daily driver as long as the MOTOR could handle the abuse. Then I realized that to make the best of a combo like that, I'd need a converter so high it would be rediculous for the street, and if I were to have a motor like that in a car like my 4000 pound impala, i'd need some really steep gears. If i didn't do all of that, i'd end up with a car that might even be slower than a motor that was setup for more low end power if I had to stick with a mild converter and gears. Using your motor to the fullest by alllowing the converter, gears and motor all work together as well as they could seems more important than simply making lots of peak power, if you cant make the bet of that peak power by using the right converter and gears.


I think I may have stated to obvious as well by saying low end power is good. I mean of course if you have two motors that make the same peak tq at the same RPM, who wouldn't want the one that ALSO makes a very flat tq curve below peak tq. It seems that the above would be more of an important issue with heavier cars, heavier cars that are also driven enough that you wouldn't want to run more than 4.10's. I mean lets say you have a bigblock in a 3400 pound car, no matter what you do you will probably always have more than enough low end tq to get the car moving. Stick a smallblock in a 4000 pound car though and its a different story.

You got it.

Just for info though. My Challenger weighs just under 4000lbs with me in it. 500 and something horsepower 440 incher. 3000 stall converter. 4.30 gears + overdrive. Runs high 11s and will run up to 145mph on 26" tires. Also handles like a kart. Well OK maybe not that good but pretty good.

And yeah. I do over 500 miles to the track and back. It has also been on holidays to the hills etc etc.

Not that i brag or anything.

Thanks to a mate I have a formula for you to calculate the size of K&N filter you need for your engine.

First if all you need to know how much airflow your engine needs.

CFM required = CID x Maximum RPM / 3456

This assumes 100% volumetric efficiency at high rpms and also tells you what the minimum size carb should be on your engine.


To find circumference of a circle multiply diameter by 3.1416
To find diameter of a circle multiply circumference by .31831


CFM Formulas for Filters

Round: Diameter x Height x 6 x 3.14
Flat Panel: Length x Width x 6
Universal Round Taper: (Base + Top) / 2 = A
Take A x Length x 6 x 3.14
Universal Round Straight: Diameter x Length x 6 x 3.14

Square inches of filter required for an engine = CID x Maximum RPM / 20839

I have measured my 6 pack air filter to be 58-inches around and 2-inches tall. This equates to just 696cfm of airflow.
But my engine needs around 850cfm!

So I have been looking at a 3-inch tall version K&N sell and that works out at 1044cfm so hopefully one of those will fit under the R/T hood.

I have used this formula on a few K&N filters where they advertise the flow rate.

I have found their advertised flow rates to be + or - 10% from the calculated rates as stated above.

Basically their filters flow on average 6cfm per square inch of filter material. However design differences make some filters flow better than others.