On Fri, 21 Jun 2019 15:01:35 +0100, Paul wrote:

Commander Kinsey wrote:
On Thu, 20 Jun 2019 01:15:46 +0100, Eric Stevens
wrote:

On Wed, 19 Jun 2019 16:48:01 +0100, "Commander Kinsey"
wrote:

--- overdue snip ---

You think torque is relevant when considering engines, I don't.
Simple enough for you?

Let me help you understand the error of your ways.

Consider a car about to climb a hill. The car has a mass of 1500kg

The gradient is such that a the force applied in the direction of the
slope to propel the car against gravity and rolling resistance is 20%
of the car's weight

= 0.20 x 9.807 x 1500 = 2942 Newtons.

The propulsive force is applied by the driving wheels against the
road. The wheels are 600mm in diameter which means that the total
torque which has to be supplied to the wheels by the differential

= 2942 x 0.300 = 882.6 Newton.Metres.

The gears of the final drive have a reduction ratio 3.14:1 which
meaans that the driveshaft input torque to the final drive is:

882.6
------ = 281.08 Newton.Metres.
3.14

The car will not climb the hill if it cannot supply the driveshaft
with at least this torque. This leaves no margin for acceleration.

Question: How much power does the engine need to enable the car to
climb the hill?

Look how complicated your question is. Now try to work it out
using a simple equation of balancing energy.

I think you are fudging. If I am misjudging you, please use my example
to explain it your way.

I can't be bothered looking at your overly complicated explanation.
Your mind must be really twisted to try to calculate things in such a
longwinded way.

What you are really saying is that you don't properly understand the
problem.

Way back in this thread Carlos correctly wrote:

Actually, it is the torque which changes with the gear change. The
power output is the same - except that the power curve is not
linear.

And you replied:
The power output is not the same. Double the revs gives you double
the power. Changing gear changes the revs and therefore the power.

In writing that you completely failed to recognize that double the
revs means double the gear reduction and twice the torque applied to
the driving wheels.

Since then, in spite of numerous hints, you have failed to acknowledge
the role of torque in all of this.

The point is that power does not enter anywhere in my examplar
calculation. Nor could it without any mention of speed. You can have
all the power in the world available to get you up the hill but it
will be totally useless unless you have the torque at the driving
wheels sufficient to overcome gravity.

Its torque that gets you up the hill, not power. Power only affects
the speed at which you might climb the hill, assuming you have
sufficient torque to climb it at all. When you change down its to get
more torque. You will only get more power if you maintain vehicle
speed. If the down change forces you to halve your speed you are
getting the same power as before but now you are getting more torque.


You have said:
" Torque means precisely **** all, because depending on the revs,
you can have a completely different output. ..... And all I need
to know is the horsepower available. I want to know if it equates
to enough to lift the mass of the car up the hill at the speed I
want."

You look at the HP graph, you find the maximum power you can get out of
the engine, and you decide that would be enough to lift the car up the
hill at say 20mph. Is there a suitable gear for this? No? But one
will make it go 15mph at maximum output, so use that.

Um, it's a defining equation.

The controlling term is Torque.

End... of... story.

You can try and weasel around it all you like.

If you hold a parchment in this discipline, you're
taught to think in terms of defining equations.

*******

Just a thought experiment should be enough to convince you.

1) A car makes the same horsepower, each time you stomp the gas.

2) Put the car on a 7 percent slope, with the car on
flat ground, about to go up the hill.

3a) Put the car in fifth gear. Let out the clutch.
What happens ? The car stalls. Not sufficient
torque was generated, to counter the other terms
in the equation. The car refuses to roll forward.

3b) Put the car in first gear. Let out the clutch.
What happens ? The car goes up the hill.

In (3a) and (3b), in both cases the same horsepower was
available at the engine shaft... but after the transmission
is done with it, there is more torque by shifting to
first gear.

Same horsepower, but you're trying to make the car go many times further up the hill in the same time, due to the different gear ratio. Also as the revs are lower, the horsepower is probably also a lot less. I guess it's just two ways of calculating the same thing. I prefer to think in terms of energy and power.

Also notice, that if we manage to get the car into
fifth gear, shifting as we drive up that hill,
eventually the force equation turns against us,
and the car starts to slow down before it gets
to the top. When this happens, you gear down into
fourth. As fifth is "too tall a gear". This is
still the torque at work!

Now, being a physics graduate, what do you do ?

Why, a "units check". Take the units for each term
in the equation, and make sure that the output is
a force (or whatever).

Once we've figured out the net force, we can use
Fnet=MA or whatever.

The discipline that comes from doing a thousand
problem sets (without copying the work of a fellow
student), should be cutting in about now...