How to calculate torque

[Guest daze]

A car engine produces a maximum power of 95 kw at 6000rpm how can i calculate the torque at the maximum power?

[Johnny5]

A car engine produces a maximum power of 95 kw at 6000rpm how can i calculate the torque at the maximum power?

 

Power has units of joules per second, which is energy per unit time.

 

Here is the definition of torque:

 

Torque = T = r X F

 

If we disregard the direction of the torque, we can focus on it's magnitude which is given by:

 

|T| = |r||F| sin(r,F)

 

Torque has units of energy. If you differentiate both sides with respect to time, each side will have units of power.

 

d|T|/dt = d/dt (|r||F| sin(r,F))

 

So, here is what happens in a car engine...

 

Gasoline is ignited, actually a mixture of gasoline vapor and air, regulated by the carburator, inside a chamber, by a spark which comes from the spark plug. So there is a tiny explosion which pushes the piston in a valve down, and then there is a firing sequence, if you have a four valve, there are four spark plugs.

 

So the plugs fire in an order, i think its 1324.

 

But regardless, as the pistons go up and down, the camshaft is turned.

 

A DOHC is the easist engine to learn first. Dual overhead cam.

 

So energy in the gas, is converted into rotational energy.

 

There are places on the web where you could investigate how a real DOHC engine works, they will have diagrams and you can see what I'm talking about.

 

But the principle is very basic, the explosions are used to make the wheel axis spin.

 

Camshaft animation

 

So the goal is to take the given information, and calculate torque.

 

You are told that the maximum power delivered is 95,000 watts. One watt is a joule per second.

 

Now, in that torque formula, you see the symbol r, you have to know what that is. I've always called it the "moment arm." I'm sure there are other terms for it. If you take a class, either in Statics or Dynamics, you will come across that formula.

 

here is one way to think about the torque formula:

 

 

You are walking towards a door, which is ajar. You want to enter the room beyond, so you have to push the door open. The door hinges are to your left, and the doorknob to your right, and the door opens inwards, that is you have to push. So where is the intelligent place to apply force?

 

You can either use trial and error, or you can use the torque formula and deduce the answer.

 

Here is the correct answer, regardless of your method of learning it:

 

Do not push near the hinges, push as far away from them as you can. You want to maximize the moment arm R, so that you can apply the least force possible, to produce a given torque.

 

Now, the hinges are located along the axis of rotation of the door.

 

A typical door has a width of about 4 feet, so we will call that the maximum radius, at which you can apply the external force F.

 

Now, when you push, you can either push perpendicular to the two dimensional plane which the door is in, or you can push at some other angle.

 

So what would be the best choice for the direction of your applied force?

 

Again, either you can use trial and error, or you can use the torque formula.

 

Now, the letter r, and the letter F, as well as torque T, that you see in the torque formula are actually vectors, and which i will now boldface, so...

 

t = r X F

 

I just switched to the Greek letter tau, instead of T, because it is common practice to use that symbol to denote torque.

 

Now, here is the definition of moment arm r, in the formula above, in the context of the door problem...

 

The vector r, has its tail located along the axis of rotation, its head is located at the exact place of contact, at which the force is applied, and r vector is perpendicular to the axis of rotation (which we know is located where the door hinges are). (From experience, you should already realize that if you push the edge of the door towards the axis of rotation, the door will not rotate at all, no matter how hard you push)

 

So let me take the time to explain the cross product briefly.

 

The cross product of two vectors, is another vector, which is perpendicular to the other two. And the direction is given by the right hand rule.

 

I should be able to find a place on the web with an illustration.

 

Here is a brief article on "right-hand rule" at Wolfram, I strongly urge that you read it, and then come back. As far as I know, the direction is by convention. Hence this is something you must memorize.

 

Now, I will use the diagram at Wolfram, to talk about the cross product of r with F.

 

As you can see in the diagram, the vector u X v is perpendicular to both u, as well as v. Let that axis be the axis of rotation, and let U be the moment arm vector, so that v is the applied force vector.

 

Thus, the direction of the torque points along the axis of rotation.

 

Now, here is the formula for the magnitude of torque again:

 

| t | = |r||F| sin(r,F)

 

sin(r,F) stands for the sine of the angle between r, and F.

 

if the angle between r,F is zero, then the torque would be zero, because sin(0)=0. Thus, there would be no "turning force" the door would not spin, and this corresponds to pushing the edge of the door directly towards the axis of spin. So clearly the door won't spin, if the angle between the moment arm r, and the applied force F is zero.

 

Now, sine takes on its maximum value when the angle is 90 degrees.

 

Thus, torque is maximized precisely when the angle between r and F is 90 degrees. In fact, sin(90)=1.

 

So now back to the car engine problem.

 

The force is being applied to a steel axis, namely the wheel axis.

 

Let me try to find the radius of an actual wheel axis, in a car assembly.

[swansont]

Power has units of joules per second' date=' which is energy per unit time.

 

Here is the definition of torque:

 

Torque = T = r X F

 

If we disregard the direction of the torque, we can focus on it's magnitude which is given by:

 

|T| = |r||F| sin(r,F)

 

Torque has units of energy. If you differentiate both sides with respect to time, each side will have units of power.

 

d|T|/dt = d/dt (|r||F| sin(r,F))

 

[/quote']

 

Dimensional analysis doesn't always give you the physically correct answer. (e.g. at² gives you meters, but not the correct displacement.) In this case the torque is constant, as are r and F and the angle between them, so your deriviative would be zero.

 

Work is T*theta, where theta is the angle through which the rotation occurs, analogous to W = Fd

 

So if you differentiate wrt time, you get dW/dt = P, and d(theta)/dt = w, or angular speed.

 

So P = Tw (w has to be in radians/sec)

[Johnny5]

Dimensional analysis doesn't always give you the physically correct answer. (e.g. at² gives you meters' date=' but not the correct displacement.) In this case the torque is constant, as are r and F and the angle between them, so your deriviative would be zero.

 

Work is T*theta, where theta is the angle through which the rotation occurs, analogous to W = Fd

 

So if you differentiate wrt time, you get dW/dt = P, and d(theta)/dt = w, or angular speed.

 

So P = Tw (w has to be in radians/sec)[/quote']

 

Alright so I won't differentiate. I just want to set power equal to torque divided by time, anyways thats what i had planned to do initially, before thinking about it. I had considered that since they said that the maximum power transfer was 95,000 watts, that this had something to do with setting the derivative equal to zero, and solving for something. Clearly r, and F are constant. So that dr/dt=0, and dF/dt=0, but I wasnt sure about d(theta)/dt.

 

I am recalling things as I go along. It might be the hard way to do it, yet it will work.

 

Now, here is what you say...

 

(see i do pay attention to others)

 

Work = W = T Q

 

So what the hell is your T? hmm (dont say anything)

 

Theta is going to have units of radians, hence T has units of Joules per radian.

 

Now, you have defined theta to be the angle through which rotation occurs.

 

So that I could use polar coordinates in the right reference frame if i want.

 

Power = P = dW/dt = dT/dt Q + TdQ/dt

 

Then you introduce omega for d(theta)/dt, so that:

 

Power = P = dW/dt = dT/dt Q + Tw

 

So, whatever T denotes, provided that it is constant in time we have:

 

Power = P = dW/dt = Tw

 

Which is another of your formulas.

 

So, using no thought, for power we have:

 

95,000 = Tw

 

The LHS has units of Watts.

 

Now, omega is angular speed, d(theta)/dt.

 

Now, we are told that the wheel axis is spinning at 6000 revolutions per minute.

 

So in one six thousandth of a minute, the wheel axis undergoes exactly one revolution.

 

The torque is of course zero, the rate of spin is not accelerating, it is a constant 6000 rpm.

 

So in amount of time 60 seconds/6000, the total change in angle is 2 pi radians.

 

60 seconds/6000 = 6/600 seconds = 1/100 seconds =.01 seconds

 

 

Thus, in one hundredth of a second, the wheel axis goes around once, in the appropriate frame of course, and so, in that frame, the change of angle is 2pi radians.

 

Therefore:

 

w = dQ/dt = 2p/(.01)

 

So using the formula which you gave without proof we have:

 

95,000 = T2p/(.01)

 

Therefore:

 

95,000 (.01) = T2p

95,000 (1/100) = T2p

950 = T2p

 

 

Therefore:

 

T = 151.197...

 

 

So T has units of energy per radian.

 

Torque has units of energy.

 

Here is a good article on torque.

 

In particular, note where they say the following:

 

A torque of 1 N·m applied through a full revolution will require an energy of exactly 2π joules. Mathematically,

 

E = T X θ

 

where

 

E is the energy

 

θ is the angle moved, in radians.

 

So, i believe the answer is 151 joules.

 

I'm going to go think about it for a few.

___________________________________________________________

 

Well if i use that formula, together with Dr. Swanson's, the answer is 151 Joules.

 

They have:

 

E = T times theta(in radians)

 

So that torque would have units of energy per radian.

 

I still have to think about this a bit though.

 

Something about spinning...

 

I need to develop a new kind of algebra to handle this.

 

It has to do with the truth value of statements about spinning frames of reference.

 

Well that's what I am thinking about anyways.

 

This goes back to Geistkiesel's post on the Sagnac effect, and a problem I was working on there.

 

It has to do with rotation being absolute, not relative.

 

Something about the coriolis force is coming to mind.

 

Think about a spinning top, or a spinning wheel axis, or what have you.

 

There are a couple of things you must do, in order to mathematically analyze that which spins.

 

First you have to identify a spin axis.

 

Then you must attach that axis to the thing which spins.

 

Now, in the rest frame of the object, the change in angle is zero.

 

So change in angle, necessarily involves what I am now thinking about. The non-relativistic character of that which spins.

 

Think about that merry go round ride, where your back is against a circular wall, then the whole thing starts spinning, then the floor drops out, and you don't fall.

 

I've been on one before, long ago, and I still remember something quite vividly...

 

I moved my head towards the center of the spinning ride, just to see what would happen.

 

Instantly, I got nauseous.

 

I moved my head back to where it was, and it went away.

 

The fact that I was spinning, is absolute.

 

You cannot say that I was at rest, and the rest of the universe was spinning.

 

That's incorrect.

 

From a purely kinematical point of view, you can, but not a dynamical one.

 

And so in that rest frame, (the spinning one), my change in angle was zero.

 

Of course that frame was a non-inertial one.

 

What I am concerned about, is the mathematical analysis only.

 

In Dr. Swanson's answer, the angle theta is introduced, but little is said about what frame to define it in.

 

And then of course we use units of radians, when in fact 2 times pi is a pure number.

 

Rotation