Need Help with a Formula to convert Change in RPM to torque applied

[soundengineer]

I need some help coming up with a formula..
This involves a Vehicles Engine

I know how fast the rate of change is (example being 3000 rpm over 3 seconds-an increase from 3000rpm to 6000 rpm in that 3 second period)
I know how much weight this RPM has moved (3700 lbs)
and I know the Gear Ratio that it used to move this weight..(5.495:1)

I just need help coming up with a Formula that will calculate the average torque of the engine that moves this weight(which is a vehicle)

if I need more Info.. I can supply that as well..
vehicle speed, tire size


sorry if this is a newb type question.. but I am not great at math, and I'm trying to build a software program to interface with a vehicle and show your torque and horsepower output in real time or to save the data to print a graphic display later.

 

[russ_watters]

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

 

[soundengineer]

sorry... that really doesn't mean much to me...
I'm looking for somebody who can walk me thru it...
gladly share any data needed to use formulas that anybody can supply to help me understand this so I can make a working software program.

 

[Lsos]

I need some help coming up with a formula..
This involves a Vehicles Engine

I know how fast the rate of change is (example being 3000 rpm over 3 seconds-an increase from 3000rpm to 6000 rpm in that 3 second period)
I know how much weight this RPM has moved (3700 lbs)
and I know the Gear Ratio that it used to move this weight..(5.495:1)

I just need help coming up with a Formula that will calculate the average torque of the engine that moves this weight(which is a vehicle)

if I need more Info.. I can supply that as well..
vehicle speed, tire size


sorry if this is a newb type question.. but I am not great at math, and I'm trying to build a software program to interface with a vehicle and show your torque and horsepower output in real time or to save the data to print a graphic display later.

How fast did the car accelerate when going from 3000-6000 rpm? You already know the time (3 sec), now just find the change in velocity (Acceleration = Velocity/ Time)

Once you know the acceleration, you can figure out how much force the tires were putting out (Force = Mass x Acceleration).

Once you know the force of the tires, you will also need to know the size of the tires. You can then figure out their torque (Torque = Force x Lever arm).

Once you know the torque at the tires, you can use the gear ratio to work out the torque at the engine. Keep in mind that usually there's two gear ratios that engine torque goes through: transmission and differential.

Also, if you know the size of the tires you might be able to figure out the speed of the car mathematically...

 

[sophiecentaur]

What you have told us, so far, is not enough to solve your problem.
In the end, it's torque (or, rather Power = engine speed times torque) that tells you the power available to give the car kinetic energy in a given amount of time (or carry it up hill).
If you know the distance moved along the road for each rev of the gearbox output (or engine rev, in a particular gear) then you can relate torque to the force on the road, which will allow you to relate acceleration to mass, force, engine power or torque. So far, we don't know the connection between the rotating world and the world of moving along the road. It will be different for each car.

There are two formulae:
One describes the 'linear' world:
force = mass times acceleration

The other describes the 'rotational' world':
torque = Moment of inertia times angular acceleration

It would probably be easier to work in terms of motion along the road so you 'just' need to translate your torque to motive force on the road and engine speed to peripheral wheel speed. So metres per engine rev will takes us there if you can find it out for us. This can be done if you know all the ratios involved (differential included) and the wheel diameter.

 

[soundengineer]

I do know the MPH Associated with it...
or I can easily figure out the MPH the vehicle is traveling using RPM and Gear Ratios and tire size...
Hence why I gave the gear ratio in the first post..but I forgot to mention that tire size, which is 27.61" Diameter.again... I seem to be Math deficient right now...its a new step that I am not used to figuring out and I need help

if somebody could walk me thru with the known values I have, and point out the variables I need to supply...
once I get one thing solved and can see the variables in the equation with results, I can easily do it myself after...

I am a Visual Learner.

 

[soundengineer]

at 3000 rpm it should be 44.844 MPH
at 6000 rpm it should be 89.688 MPH

it too 3 seconds to accelerate a total of 44.844MPH

Tire Size is a Diameter of 27.61"

it is irrelevant what my actual individual rearend gear and transmisson gear is...the value I have provided is the total gear ratio. (5.495:1)

total vehicle weight is 3700lbs

 

[Lsos]

at 3000 rpm it should be 44.844 MPH
at 6000 rpm it should be 89.688 MPH

it too 3 seconds to accelerate a total of 44.844MPH

Tire Size is a Diameter of 27.61"

it is irrelevant what my actual individual rearend gear and transmisson gear is...the value I have provided is the total gear ratio. (5.495:1)

total vehicle weight is 3700lbs

In that case the three formulas I provided could solve this.

The only thing...I always stay away from imperial units. First thing I do is convert mph to meters/second and lbs to kilograms...otherwise you might have to use slugs and dynes and who knows what other savage units.

Acceleration = Velocity/ Time
Force = Mass x Acceleration
Torque = Force x Lever arm

You have Velocity and Time (20.05 m/s, 3 sec). You have Mass (1678 kg) and you found Acceleration. You have Lever Arm = radius of tire (.351 m) and you found Force. You basically have all the variables. Just plug and chug...

 

[soundengineer]

so...
(20.05)/(3) = 6.683333
(1678)x(6.683333) = 11,2214.63277
(11.2214.63277)x(0.351) = 3936.336.104that isn't even remotely correct to measure force put out by an engine...
this is a vehicle that should be somewhere in the 400-600 ft/lbs torque Range

 

[Lsos]

so...
(20.05)/(3) = 6.683333
(1678)x(6.683333) = 11,2214.63277
(11.2214.63277)x(0.351) = 3936.336.104that isn't even remotely correct to measure force put out by an engine...
this is a vehicle that should be somewhere in the 400-600 ft/lbs torque Range

Please go up to my first post where I explained what the formulas mean. You'll realize that the value you got IS the torque...but at the wheels. And of course it's metric units (Newton meters).

Since the torque at the wheels is simply the torque at the engine multiplied by the gear ratio, you simply divide your final value above (3936 N-m) by the gear ratio to get the torque at the engine. Then you convert that into foot-pounds (I'll let you google the conversion) to get the final answer that you're interested in.

 

[soundengineer]

Please go up to my first post where I explained what the formulas mean. You'll realize that the value you got IS the torque...but at the wheels. And of course it's metric units (Newton meters).

Since the torque at the wheels is simply the torque at the engine multiplied by the gear ratio, you simply divide your final value above (3936 N-m) by the gear ratio to get the torque at the engine. Then you convert that into foot-pounds (I'll let you google the conversion) to get the final answer that you're interested in.

thank you... I knew I was missing something...
that is why I needed somebody to point that out...
I would have seen that if somebody had actually placed the #'s in an equation to get me there..so my math gets me 3936 / 5.95 = 661.5
convert from N-M to Ft-Lbs = 487.5 ft/lbs

that makes more sense...

once again... Thanks to everybody who helped out
I'll now see if I can apply this to my running data to see if it makes sense