Torque and RPM relation

[xpensive]

and once again http://s22.photobucket.com/user/STRAD2/ ... e.gif.html . six pages

Oh my I didn't want to hurt you, I'm just a jealous guy

[Greg Locock]

jz11 while you are at it please write to these people and let them know that they are doing it wrong.

http://fetweb.ju.edu.jo/ME/courses/labs ... Torque.pdf

Have you ever actually used an engine dyno?

[jz11]

all I tried to say - in order to calculate (this is what I mean by indirect measurement) torque for continuously rotating object, you need to let it do work and observe the amount of work done, measure it and derive (calculate) torque from that measurement, and it is exactly how they do it in that manual of yours, doesn't matter that there are feetpounds or nm on the scale, they still make it work against the load and observe the amount of work done and record the current rpm for that particular reading <- if that is called - direct measurement, I apologize for wasting everyones time

another thing I said - torque alone doesn't mean anything, in order for it to gain meaning in context of work done, that as I understood is the subject of this topic, you need to know the rotation speed - the mistake in the very first post in the thread

and no, I haven't used lone engine dyno, I have used only car dynos, same thing, but I'm trying to avoid any confusion, since the question was specific about which type of dyno I may or may have not used

[gruntguru]

jz11. Not sure where you got lost but your ideas are far from reality. Actual torque - the amount of twisting force - in a shaft can be directly measured a number of ways. This is true for stationary shafts and rotating shafts.

Torque is the rotary analogue of force. If you place a brick on the floor and drag it along with a fish scale you can measure the horizontal force applied to the brick - whether it is moving or not. If you measure the distance the brick has moved along the line of the force and multiply by the force, you have calculated the "work done" or energy transferred to the brick. If you measure the velocity and multiply by the force you have the "power" - the rate of doing work or energy transfer.

The torque case is no different. If you have say a vertical axis millstone sitting on its base, attach a radial arm (crank) with a horizontal fish scale attached at 90 deg to the end of the crank. You then have a horse pull on the fish scale. The force measured by the scale times the length of the crank (measured from fish scale to axis of rotation) is the torque generated by the horse - whether the millstone is turning or not. The work done is torque times angle of rotation. Power is torque times speed of rotation.

If the horse track has a large enough diameter it may seem almost straight. You can measure distance and speed. Combining these with the fish scale force measurement, you can calculate work and power in the linear sense. Surprise surprise they come out the same as the rotational work and power - even though the torque from the same horse can vary considerably by changing the length of the crank.

[strad]

I give, you are right the moon is made of green cheese

[xpensive]

I give, you are right the moon is made of green cheese

Don't, you still have all the "motorsport journos" on your side, most of them still confusing torque with low-end power?

[xxChrisxx]

Xpensixe.

It's generally very clear that when talking about torque they are referring in gear pulling power from low engine speed.

No, it's not technically correct so separate low engine speed performance into 'torques' and high rpm performance into 'horsepowers'.

But then again saying that torque is completely irrelevant isn't technically correct either. Infact as we know P=Tw.

Life is full of colloquialisms.


Do you get as mental when people use the unit kg for weight and not the more technically correct kgf? Probably not.

[xpensive]

Xpensixe.

It's generally very clear that when talking about torque they are referring in gear pulling power from low engine speed.

No, it's not technically correct so separate low engine speed performance into 'torques' and high rpm performance into 'horsepowers'.

But then again saying that torque is completely irrelevant isn't technically correct either. Infact as we know P=Tw.

Life is full of colloquialisms.

Do you get as mental when people use the unit kg for weight and not the more technically correct kgf? Probably not.

There is no unit, at least not within SI, named "kgf", Mass is kg and Force is N.

I hate to brake it to you, but what Jeremy Clarkson in Top Gear feels when he pushes the accelerator at low revs is not
torque in any shape or form, it's the propulsion force between tire and surface, as in Force equals Power over Speed.

The force has a relation to the engine's torque, but then you need to know the gearbox ratio, final drive and wheel radius,
why just using Force, Power and Speed is so much simpler.

[Brian.G]

Wheel radius plays a part also in terms of 'pulling power'

Brian,

[xxChrisxx]

The force has a relation to the engine's torque, but then you need to know the gearbox ratio, final drive and wheel radius,
why just using Force, Power and Speed is so much simpler.

I'm going to set out a scenario and ask you a question:

We want to work out the in gear acceleration for a vehicle with 5 gears (already pre-set, but we don't know what the ratios are). Giving it's actual acceleration profile throughout it's speed range. Assuming no slip, or traction limit.

Are you sure you don't need the overall ratio between engine speed and road speed, for you to use power force and speed.


Think about your answer carefully.

[godlameroso]

It's a proven fact that even though torque diminishes at a certain point in the rev range, horsepower continues to increase up to a point. This is because of the relationship between speed and force to total energy. For any given mass, doubling the speed roughly quadruples the energy. However, doubling the mass only doubles the energy for any given speed. The energy being the expansion of burned gasses in the combustion chamber, converted to crankshaft torque. Which means that system inertia becomes instrument in determining power output.

All horsepower measures is the ability of engine speed to increase when subjected to a particular load. Practically speaking the faster all the parts move, the more energy there is in the system. That's why crashing at 160kph is a lot worse than crashing at 60kph. As long as the combustion pressure can overcome the inertia in the system, power can keep increasing at a logarithmic rate as engine speed increases. Two equal sized engines, the one that makes the most torque at the highest engine speed will always be the fastest, provided the engines are indestructible.

In the end, you are using torque at the crank to overcome the inertia in the system. When a car is stationary, the engine must overcome the friction caused by the entire system, so you use a very high gear ratio to give you the most torque, so that system inertia can be overcome. Once moving, the system inertia provides some energy, so less is needed to overcome inertia and more is used to accelerate engine speed, to a point. This point is when friction/load cannot be overcome by engine torque, be it because the engine cannot produce enough torque at that speed to overcome said friction.

[xpensive]

The force has a relation to the engine's torque, but then you need to know the gearbox ratio, final drive and wheel radius,
why just using Force, Power and Speed is so much simpler.

...
Think about your answer carefully.

Save it for the Quiz thread Chris.

[xxChrisxx]

The force has a relation to the engine's torque, but then you need to know the gearbox ratio, final drive and wheel radius,
why just using Force, Power and Speed is so much simpler.

...
Think about your answer carefully.

Save it for the Quiz thread Chris.

It's not a quiz. It's just utter bullshit that you are dodging this.

There are two reasons as far as can see.

1. You've worked out what I was getting at.
2. You don't have a clue and don't want to get tripped up.

Either way it'll be interesting.

So what will it be:

We want to work out the in gear acceleration for a vehicle with 5 gears (already pre-set, but we don't know what the ratios are). Giving it's actual acceleration profile throughout it's speed range. Assuming no slip, or traction limit.

Are you sure you don't need the overall ratio between engine speed and road speed, for you to use power force and speed.

[xpensive]

Not at all, I'm just not in the mood to question Newtonian mechanics, not with someone who only wants to make a point.

[xxChrisxx]

Not at all, I'm just not in the mood to question Newtonian mechanics, not with someone who only wants to make a point.

It's not a trick, I will work through it at some point anyway. I just want to know your honest answer. No discussion involved.

There are 2 outcomes:
You are right, and you can feel warm and fuzzy that you are right.
You are wrong, and you'll learn something.

You know the question . All it needs is a yes or no.

Think of it as a game! How exciting.


EDIT: It's also a bit ironic that you now don't want to 'discuss' it. Even though you've been pretty damn eager for the last 7 pages.