ls1tork

i realized that when i posted last time. all cars have a torque curve, some will have advantage down low and others up high. if you look at a torque curve on a dyno chart, and if you have the patience to do it, you could take the weight of that car, and divide that by force at the wheels in each gear sweeping from 2k to 7k rpms (in increments of 1000) to calculate force to weight ratios, and you will get different or fluctuating ratios all the way up. do two different cars (compare), and you could tell which one will tend to have an advantage over the other (theory on paper, assuming all other variables are equitable or close).

doing this calculation will mathematically prove why a bike accelerates faster than a car (which is physically obvious), and the same is true for comparing car to car as well (import philosophy vs american muscle). i'm not saying that rwhp isn't important, it definitely rules regardless of how it's geared (i'm learning this), but where there is rwhp, there is torque, and torque gets transferred to the wheels via gearing, which multiplies and becomes turning force in thousands of pounds at the tire treads.

1 FMF

iTrader: (1)

dude, really bad example. an airplane propeller is orders of magnitude different than a helicopter main rotor blade! You cannot compare the two. for a helicopter, the main rotor does not generate thrust, it generates lift; the main rotor blade is an airfoil (an airplane wing going in a circle).

Shooter_Jay

torque converter slip, tire spin, mass, aerodynamics...sure you can calculate it, will the results be anywhere near the reality? Depends on how much effort you properly apply. I would guess the effort will be much higher than the accuracy...

02 BLK WS6

iTrader: (11)

Honda doesn't have the torque

Cheatin' Chad

iTrader: (3)

What is required for flight?
Thrust and lift must be greater than weight and drag.


Also, when did the advanced tech section have so many people that should be in remedial automotive classes?

ls1tork

the only advantage they have, if you can call it that, is torque multiplication and very light weight. turboed, and some are NA, puttin out 350+ whp at 9,000 rpms (no torque) on 4.76 ratio gears (w- high tranny ratios too) on 2,400 lbs weight is very hard to beat, assuming they have traction.

civics are all about torque multiplication from very little engine torque. it's not how much tq they produce, it what rpms it's being produced at- sort of like F1 racing cars. They have extremely little torque too- like 200 ft lbs at 19,000 rpms, but yet they are 900 rwhp on a 1,200 lb chassis and respond like lightning itself.

FootLbs

iTrader: (7)

So does this mean we never have to street race anymore? Can we just have debates at Starbucks and mathematically calculate who would win?

ls1tork

the wave of the future... racing by math. we'll do this when the world's energy supplies have run out, then it will be the battle of the minds from leftover dyno sheets LOL.

Schwanke Engines

iTrader: (1)

This is not true. All you have to do to back up my statement is run the car in 2nd gear on the chassis dyno, then run it in 3rd, then 4th. You will clearly see a difference in power at the wheels. Actually, the most power you will make at the wheels is in the highest drive gear.

The reason for this is acceleration. The faster you are accelerating, the more power you are losing to accellerate the components. I just wanted to clarify this for everyone.

With reguards to the question. It depends on who can hook up the best!!!!

It's also a lot more complicated than just gears, and you touched on that with thte weights and aero drag. I know there are a lot of great books out there that can do all the calculations, and there are even computer simulations.

ls1tork

Thanks- this is alot to consider. i guess i am researching all this because 1. its fascinating, 2. this helps me to find what it is i really want out of my car vs reading the internet and just throwing up mods based solely on rwhp gains, 3. it helps me to understand why one car maybe faster than another in certain situations.

DanO

Possibly.. im not sure what you learned, but power is what matters

Being that you have a bachelors in engineering, you should be able to calculate what the force would be at the rear tires. The part that is tripping you up is that you are assuming a fixed gear ratio.. and yes, with a given gear ratio you will have highest acceleration at torque peak..

HOWEVER you are not taking into consideration what the wheel speed or vehicle speed is.

Do a calculation assuming you have a CVT and a constant vehicle speed.. say 40mph. Now you have 2 points to choose from, Peak torque 350lb-ft @ 4000rpm or Peak HP 350hp @ 6000 rpm. Calculate what gear ratios are needed for 40mph and what the force at the contact patch is for each scenario.

(hint, it should prove that peak HP will have more force accelerating the vehicle)

Old SStroker

Good posts, Fast. I've been down this road before (pun intended).

You might consider that some folks do not believe in Newtonian physics. That lack of belief or understanding does not change what happens or why, fortunately.

The good thing about banging you head against a concrete wall is that it feels so good when you stop. Keep on banging, guy. I'm with ya'.


Jon

DanO

Fast.. you agree with what i said.. However, you are working with a Fixed Gear RATIO and different vehcile speeds!! The vehicle will accelerate fastest in a given gear at its torque peak. You are right its simple math... BUT

if you want to optimize acceleration at any vehicle speed, you would gear it so that the engine is at peak power.

do your same calculation of force in 3rd gear by changing the 3rd gear ratio so the vehicle is going the same speed at peak power as it was at peak torque.

For me, i'll take power over torque any day in drag racing..

If i were in charge of racing a 10 speed transmission.. i'd be shifting to keep the rpms as close to peak power as possible (say +/-200rpms)... ideally sitting right at peak power the whole way down the strip

now you do the same in the same car with the same transmission, and shift to keep it near the torque peak (+/-200rpms)

Lets see who wins.. I'll be placing my bets now.. any takers?

DanO

Thanks for the credential explanation.. just a fair warning.. there are a few on here that have been around the block a few more times.

I believe we agree on many points.

-in a fixed gear, peak acceleration occurs at peak torque
-you typically want to go beyond the torque peak for fastest e.t. down the track


However, Your views on CVT are not 100% right. there are many designs out there and most dont rely on centrifugal force for gear changes.. unless your talking snowmobiles.. For example look at the CVT mechanism that Fomula 1 will be using in 2009..

Also, for arguments sake, purely hypothetical (the whole point i was trying to make above)

there is a theoretically perfect CVT in your drag race car

Scenario 1: Change CVT calibration to hold engine at Peak torque while making a pass

Scenario 2: Change CVT calibration to hold engine at peak power while making a pass

Who would have a better ET and MPH? My money is on #2.. I'll let you figure out why. (hint: its not about torque going into the transmission as much as it is about torque coming out...)

DanO

Well then i guess im just plain out wrong since everyone agrees...

I will let you ponder this. Think of an electric motor, peak torque occurs at a relatively low rpm.. you could say zero.. but for arguments sake lets say 1rpm. But its power peak is at 2000rpm but only has 1/4 the torque. If i hook this up to a single ratio transmission and can select a single gear but have the requirement that the transmission output shaft spin at 50rpms. would you gear it to allow the motor to spin at peak torque (aka 1rpm) or so that the motor spins at peak power (aka 2000 rpm but 1/4 the torque) if i wanted to maximize the transmission output shaft torque.

Torque is worthless without speed/rpm... i can apply 100lb-ft to my lug nuts, but doesnt mean i can accelerate a car down the track like an engine with 100lb-ft @ 8,000 rpm

I guess i'll have to write out the math you. since im the "ONLY ONE" having an issue. I thought this was engineering 101. You can muliply and divide torque but you cant multiply and divide power. Peak power is the point where the torque can be multiplied the most.

For example.. My engine has 350lb-ft of torque at 3000rpm and 350hp at 7000rpm (aka 262.6lb-ft) . Now we agree that the output shaft of the transmission is directly related to vehicle speed (this is where the VSS is, so i hope we agree). so i will calculate the maximum torque achievable at both peak power rpm and peak torque rpm for 2 transmission output shaft speeds.


Output shaft speed: 3000 rpm (lets just call this 40mph)
Peak torque calc
Torque Input: 350lb-ft @ 3000rpm
Gear Ratio Required: 1:1
Torque Output: 350lb-ft @3000 rpm

Peak Power calc
Torque Input: 262.6 lb-ft @7000 rpm
Gear Ratio Required: 2.3333:1
Torque Output: 612.73 lb-ft @ 3000rpm


Output shaft speed: 5000rpm
Peak torque calc
Torque Input: 350lb-ft @ 3000rpm
Gear Ratio Required: 0.6:1
Torque Output: 210 lb-ft @ 5000rpm


Peak Power Calc
Torque Input: 262.6 lb-ft @ 7000 rpm
Gear Ratio Required: 1.4:1
Torque Output: 367.64 lb-ft @ 5000 rpm


So my statement, is that for any given Vechicle speed, to maximize acceleration, you would want the engine to be at peak power and have a higher (numerical) gear ratio rather than have the engine be at peak torque with a lower (numerical) gear ratio. The problem with your calculations is that you were not factoring in vehcile speed, just merely looking for Force. I agree a higher force at the contact patch will accelerate the vechile faster which is why i'd run my engine at peak power with as much gear ratio "aka mechanical advantage" as possible.

Which is why, if i was a drag racer an had the optimal cvt, i wouldnt care if i had a 400hp @ 20,000 rpm engine or 400hp@ 6000rpm engine.. they would get me down the track equally fast because the 20,000 rpm engine can use more mechanical advantage and the 6000 rpm engine has more torque. In the end.. the force at the contact patch is the same!!

Lets do a quick calc..

400hp @ 20,000 rpm = 105.04 lb-ft (aka lug nut torque.. pretty weak eh?)
400hp @ 6000 rpm = 350.13 lb-ft (now thats more like it)

Lets select transmission ratios so they are both at peak power at 40mph (3000 trans output shaft rpm)

6000rpm engine
Trans input torque: 350.13
Gear ratio required: 2:1
Trans Output torque: 700.26 lb-ft @ 3000 rpm

20,000 rpm engine
Trans input torque: 105.04
Gear ratio Required: 6.666667
Trans output torque: 700.26 lb-ft @ 3000 rpm

Hmm.. same vehicle speed, same power, different engine rpm, different engine torque = SAME ACCELERATION

Power is what matters.. torque is useless without rpm...

Oh, and i would not count on old SStroker being on your side so quickly

Old SStroker

Isn't power just torque x rpm, or how quickly (or often) the torque is applied?

I side with my old Physics 101 Prof, I. Newton. Yeah, I'm that old.


Jon

DanO

Yes, i completely agree..

i'd rather have a low torque applied at 1000rpm than a high torque at 0 rpm

I want my torque applied very often!

Fast Caddie,

I look at it this way.. i want as much fuel being burned and utilized as efficiently as possible. Fuel = energy = power = Torque x rpm = acceleration. More fuel is being burned at peak power than at peak torque... so how will you go down the track faster using less fuel (aka energy)?

However, i will throw another wrench in the works with your CVT calibration. Most engines have a peak BSFC near the PEAK TORQUE rpm.. this is why you would want to keep the engine at the minimum BSFC (close to peak torque) for the most fuel efficient acceleration... but for maximum acceleration possible.. you'd want to be at peak power!


On a further note......
I'll explain my position in a story..

You are trying to build a wall (aka drive down the drag strip) and you want to get it built as quickly as possible (ET). You have a few people to help (rpm) and they can each carry a certain size rock (torque/force) at a rate of one rock each per minute

Gasp.. you have one group that cant get along with the other to determine how to build this wall the fastest. One team of 3000 people big surly men think they can get it done faster than the team of 7000 women. The team of men cary big heavy rocks (350lbs) and each one can move one rock every minute. The women are weaker so they carry slightly smaller rocks (262 lbs) and they also can each get one rock placed every minute.

Which team is building at a faster pace?

A: The team of men are building their wall at a rate of 1,050,000 lbs per minute, and the team of women are building their wall at a rate of 1,834,000 lbs per minute. This is calculated by multiplying the weight of the rocks by the number of rocks being placed

Moral of story.. the size of your rocks dont matter unless you have alot of helping hands to build the wall. A single 4000lb rock is impressive but if nobody moves it.. no wall will be built

Cascazilla

The most important point is the in the title of the thread. Mechanical advantage.

Adding differential gear ratio to something might make it faster, but not because the force at the wheels is higher.

If you make the same power 10% higher, then you can put 10% more gear in it, right? I'll be impressed if someone can explain how that would add torque at the wheels! I'll make it easy; it won't. If you make the same power 10% higher then you have 10% less torque at the engine, and 10% more gear=0 increase.

DanO

sorry for how i come off, I was merely trying to get you to figure something out without having to do it. (maybe im just lazy). no arrogance here, just trying to get a point across..

Old SStroker

Wouldn't the 10% extra gear multiply torque throughout the usable rpm range? That might help acceleration a little. Drag limited top speed still is a function of hp. So, with more gear you still get up to 187 mph on a long straight, but you get there perhaps a tenth quicker.

Hey we can fix that by putting in a gear (or rpm) rule to keep competition close and "save money"! Ever notice how well the "saving money" thing works?

I believe Cup teams can choose from 2 or perhaps 3 rear end ratios for the Datona 500. The difference between a 3.54 and a 3.70 is about 400 rpm. Which would you, as a crew chief, choose for the 500? Why?

Jon