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Torque...it's Power


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#201 manolis

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Posted 18 April 2012 - 04:56

You're assuming equal RPM range for both engines, which wasn't stated.


Dmitiy_Guller,

No, in my posts I am not assuming that both engines have equal RPM range.

What I write is:
“Provided you mean that:
1. The rev limit of the first engine is higher than the revs of the peak torque of the 2nd engine (which means the 1st engine peak power is higher then the peak power of the 2nd engine) . . .”, which is completely different

On the other hand, and unless I am wrong,
what you (and others) are assuming is that the 1st engine has a rev limit exactly at the revs of the torque bump of the 2nd engine.
No such assumption is stated by the problem.
With the torque bump of the 2nd engine happening at the middle revs of the 2nd engine, your assumption is a very specific and “convenient” one, which limits the problem making easier its solution; but the “problem” is no longer the original interesting general one.

In a similar way, you can assume that the 1st car has flat tires. The problem does not state specifically that the tires of the 1st car are OK. This way the solution becomes even easier; but is it worthy?

This is why I posted the link for the RoadLoad program that can help somebody to understand how the transmission ratios of a gearbox are selected, the effect of different transmission ratios on the performance of the vehicle, the effect of the way the clutch is used on the performance of the vehicle, how the transmission ratios are optimized, etc. A far more interesting problem.

Here is another problem that fits to the “Torque …it’s Power” title of this thread:

When it is calculated the inertia torque in any reciprocating piston engine, the one way is to calculate the inertia forces resulting from the motion of the parts, then to calculate the resulting torques and then to make the plot of the inertia torque versus the crankshaft angle (as in the balance.exe DOS program at http://www.pattakon....attakonEduc.htm ).
The other way is "the energy approach", i.e. to calculate the energy of the moving parts, and from the energy to take the inertia torque.
With the second way it is easily understood why the 4-cylinder even-firing 4-stroke has such strong inertia torque (besides the crankshaft and the engine block, it loads heavily the gearbox and the rear tire of a motorcycle, as well), and why the 4-cylinder 4-stroke with the cross-plane crankshaft is so smooth (like a turbine) in comparison.

The "energy approach" is a great tool for solving technical problems.

Thanks
Manolis Pattakos

Edited by manolis, 18 April 2012 - 05:00.


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#202 pugfan

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Posted 18 April 2012 - 06:13

Not sure what has you perplexed but I hope the red text above fixes the problem.


The statement that energy has left the system and yet no nett work has been done on the system seems a little contradictory to me...

#203 gruntguru

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Posted 18 April 2012 - 11:00

There are a number of ways this can happen.
1. "Work" is mechanical energy. Energy can leave a system in the form of heat with no work occurring.
2. "Nett work done on a system" means the total of all work entering the system minus the total of all work leaving the system. For vehicles in motion but not changing their energy level, the "Nett work done on the vehicle" is zero. This is despite the fact that the propulsion system is performing work on the system - the reason being that parasitic losses are removing work from the system at the same rate.

#204 pugfan

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Posted 18 April 2012 - 23:57

There are a number of ways this can happen.
1. "Work" is mechanical energy. Energy can leave a system in the form of heat with no work occurring.
2. "Nett work done on a system" means the total of all work entering the system minus the total of all work leaving the system. For vehicles in motion but not changing their energy level, the "Nett work done on the vehicle" is zero. This is despite the fact that the propulsion system is performing work on the system - the reason being that parasitic losses are removing work from the system at the same rate.


I still disagree but it's largey a matter of semantics so not really worth further discussion.


#205 gruntguru

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Posted 19 April 2012 - 01:07

:wave:

#206 bigleagueslider

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Posted 19 April 2012 - 03:49

Manolis
Slider stated that both engines have the same peak power and the same peak torque. This makes the problem much easier to solve. Engine 2 can be operated constantly at its power peak and so the car fitted with engine 2 will have superior performance at all speeds except at the speed where engine one is at peak power. For that instant only, the performance of the two engines will be identical.


gruntguru,

Thanks for clarifying what I meant to say in my post! Like many engineers, my ability to express my thoughts in writing are limited at best.

Fortunately, both you and manolis grasped the basic concept of engine torque and power that I was attempting to get across. That point being a transmission can make up for differences in the power & torque characteristics of IC engines. And the characteristics of the transmission is just one of many factors that must considered when comparing the relative merits of engine power and torque curves for aa particular application.

I was just trying to stimulate further discussion on the topic.
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#207 johnny yuma

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Posted 19 April 2012 - 05:39

gruntguru,

Thanks for clarifying what I meant to say in my post! Like many engineers, my ability to express my thoughts in writing are limited at best.

Fortunately, both you and manolis grasped the basic concept of engine torque and power that I was attempting to get across. That point being a transmission can make up for differences in the power & torque characteristics of IC engines. And the characteristics of the transmission is just one of many factors that must considered when comparing the relative merits of engine power and torque curves for aa particular application.

I was just trying to stimulate further discussion on the topic.
slider

If you want more torque (and who doesn't) it's a no-brainer to be in the lowest gear possible that the engine can tolerate.If you have a formula 1 engine,you can run really low ratios right through because
your engine will rev to 19000 rpm.The engine doesn't actually make much more torque than a top
of the range 2 litre car from your local dealer. It makes a lot more engine power because it is always revving so high with the torque band sitting very high in the rev range, (there are three times more spark plug firings-and thus force impulses- per minute than a street engine)and as well torque is multiplied by the extreme gearing l.

Edited by johnny yuma, 19 April 2012 - 05:55.


#208 manolis

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Posted 19 April 2012 - 10:11

If you want more torque (and who doesn't) it's a no-brainer to be in the lowest gear possible that the engine can tolerate.If you have a formula 1 engine,you can run really low ratios right through because
your engine will rev to 19000 rpm.The engine doesn't actually make much more torque than a top
of the range 2 litre car from your local dealer. It makes a lot more engine power because it is always revving so high with the torque band sitting very high in the rev range, (there are three times more spark plug firings-and thus force impulses- per minute than a street engine)and as well torque is multiplied by the extreme gearing l.


Torque and Power problem

Suppose the F1 car of "johnny yuma" above, has the following torque vs rpm, power vs rpm and speed on 1st gear vs rpm characteristic curves:

Posted Image

The total “car and driver” mass is 500 Kg.
The tires are ideal (like gear wheels on rack gears, i.e. no slipping is allowed) with zero inertia.
The inertia of the moving parts of the engine is zero, too.
The air resistance is zero, too.
The clutch is of the “mechanical friction” type with ideal reliability.

How the driver can accelerate from 0 Km/h to 100 Km/h in the minimum possible time ?
What is this minimum possible time ?

Thanks
Manolis Pattakos

#209 gruntguru

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Posted 20 April 2012 - 00:45

Slip the clutch holding the revs at the torque peak (12,000).
0.86 sec 0-100.

#210 Canuck

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Posted 20 April 2012 - 01:13

Wait...what? Why not at the power peak?

#211 John Brundage

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Posted 20 April 2012 - 02:29

Wouldn't you dump the clutch at the peak hp rpm?






















#212 gruntguru

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Posted 20 April 2012 - 03:10

Canuck. Has to be at the torque peak because the gear ratio is specified. If the ratio was free, I would gear it for 100 kph @ 18,000 rpm and slip the clutch at peak torque and lock the clutch from the torque peak to the power peak.

John. He specified zero engine inertia so dropping the clutch would instantly bring the revs to zero and the acceleration from there would follow the torque curve.

Edited by gruntguru, 20 April 2012 - 03:31.


#213 johnny yuma

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Posted 20 April 2012 - 03:14

Wouldn't you dump the clutch at the peak hp rpm?

You want to get a stationary car moving,12000 rpm is where the motor has greatest potential to do work ,why would you use the power peak ?
You risk bogging down and stopping the gasflow more than if you used the torque peak rpm..Note also gruntguru says slip the clutch,not dump it.

#214 gruntguru

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Posted 20 April 2012 - 06:42

You want to get a stationary car moving,12000 rpm is where the motor has greatest potential to do work ,why would you use the power peak ?
You risk bogging down and stopping the gasflow more than if you used the torque peak rpm..Note also gruntguru says slip the clutch,not dump it.

I am sure John was suggesting a clutch dump at 12,000. This could be the correct strategy if the engine had some inertia.

"Bogging down" and "gasflow" are phenomena associated with real engines (which this is not).

#215 manolis

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Posted 21 April 2012 - 06:57

Gruntguru's answer is correct.
However, it seems only few realize why.

Let’s modify a little the initial problem.

Now the road is 45 degrees uphill.
The Torque and Power vs rpm plots remain the original ones (post 209)
All gears, except the top gear of the gearbox, are not working (broken), with the "Speed vs rpm" plot for the top gear of the gearbox being:

Posted Image

The car is stopped.
Can the car start without increasing the engine rpm over 11,000 ?
Can the car start with its engine revving at 18,000 rpm? (at 18,000 rpm is where the engine makes its peak power).
Can the car start without decreasing the engine rpm below 15,000 ?
What is the best 0-100 Km/h acceleration?

If the car was moving initially with 630 Km/h and 18000 rpm on the uphill, can it keep its speed? At what speed it stops decelerating?

If the two problems seem difficult to you, the “article” at http://www.pattakon.com/educ/gear.txt (written several years ago) explains, at the paragraphs “CLUTCH OPERATION” and “EVEN MORE HELP FOR CLUTCH”, how the clutch actually operates.

Thanks
Manolis Pattakos

#216 gruntguru

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Posted 23 April 2012 - 03:10

No takers!
- 45 deg hill requires 3468 N of thrust to maintain velocity.
- Thrust available at 11,000 RPM is approx 3428 N so NO.
- Thrust available at 18,000 RPM is approx 3170 N so NO.
- Thrust available at 15,000 RPM is approx 3513 N so YES.
- RPM (above 12,000 RPM) where thrust equals 3468 N is approx 15,400 RPM.

Edited by gruntguru, 23 April 2012 - 03:39.


#217 manolis

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Posted 25 April 2012 - 13:29

Thanks again Gruntguru.

In the post 209 it was given:
the torque vs rpm plot,
the speed vs rpm plot,
and the power vs rpm plot.
The power is not necessary because it results from the torque and the rpm:

Power (in W) = 2* pi * Torque(in Ntm) * rpm / 60 = 0.1047 * Torque (in Ntm) * rpm..........(1)

The speed is linearly proportional to the rpm, so speed= gr1*rpm, where gr1 is the total transmission ratio from the engine to the rear tire, multiplied by the circumference of the rear tire. I.e: Speed ( in Km/h ) = gr1 * rpm..........(2)
with gr1 being 0.008333 m (at 18000 rpm an fully engaged clutch the car moves with 150Km/h) .

For the Speed in m/sec the formula is:
Speed ( in m/sec ) = gr1 * rpm / 3.6..........(3)

The work (or energy) W is related to the force F and the distance s at the direction of the force:
W= F * s

Similarly the Power is related to the force the tire applies to the road and the speed of the car:
Power (in W ) = Force ( Nt ) * speed ( m/sec ),
which according the (1) relation gives:
Force ( Nt ) * speed ( m/sec ) = 0.1047 * Torque (in Ntm ) * rpm
and according (3) :
Force ( Nt ) * gr1 * rpm / 3.6 = 0.1047 * Torque (in Ntm ) * rpm

Which gives:
Force ( Nt ) = 0.1047 * Torque (in Ntm) * 3.6 / gr1 = 45.2 * Torque (in Ntm)

With 36 Kpm at 12000 rpm, the force is 16000 Nt.
The form of the Force versus rpm plot is the same with the Torque vs rpm plot:

Posted Image

When the car is on the dynamometer, its tyres push (force) a high inertia drum to rotate. The torque applied to the drum (which is the force the tyres apply to the drum, multiplied by the radius of the drum) is calculated by the angular acceleration of the drum and the momentum of inertia of the drum. This torque is the torque provided by the engine multiplied by the total transmission ratio.

The force in the plot above is the maximum available force – with 1st gear in the gearbox – that accelerates the car. With proper use of the clutch and the engine at 12000 rpm (wherein the peak torque is provided) this force maximizes (16000 Nt).

With 500 Kg mass and 16000 Nt accelerating the car, the acceleration is

a=16000Nt / 500Kg = 32 m/sec^2 (or 3.3g)

100 Km/h is 100Km*(1000m/Km)/ (h / (3600sec/h)), which is 27.8 m/sec

with acceleration 32 m/sec^2, the necessary time for 0-100Km/h is:

T= (27.8m/sec) / (32m/sec^2)= 0.86 sec

If the engine was revving not at the peak torque, but at the peak power revs (18000 rpm), the maximum available accelerating force drops 17%, and the time for 0-100 Km/h increases to 1.04 sec.



The second problem, post 216, with the top gear in gearbox:

Now the Force vs rpm plot is:

Posted Image

The horizontal orange/red line is the force opposite to the motion caused by the weight of the car and the 45 degrees uphill road: 500 Kg*9.81m/sec^2*sin(45 deg) = 3500 Nt.

The maximum available force at 18000 rpm is only 3200 Nt, i.e. the car cannot start moving.

At 12000 rpm the available force is 3800 Nt. This means that keeping the revs at 12000 rpm the car starts and accelerates by a force 3800-3500= 300Nt.

The acceleration is 300Nt/500Kg=0.6m/sec^2.

The necessary time for 0-100 Km/h is T=(27.8m/sec) / (0.6 m/sec^2) = 46 sec.

I hope a couple of useful things about the power, the torque and their relation were shown.

Thanks
Manolis Pattakos

Edited by manolis, 26 April 2012 - 03:44.


#218 bigleagueslider

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Posted 28 April 2012 - 04:03

If you want more torque (and who doesn't) it's a no-brainer to be in the lowest gear possible that the engine can tolerate.If you have a formula 1 engine,you can run really low ratios right through because
your engine will rev to 19000 rpm.The engine doesn't actually make much more torque than a top
of the range 2 litre car from your local dealer. It makes a lot more engine power because it is always revving so high with the torque band sitting very high in the rev range, (there are three times more spark plug firings-and thus force impulses- per minute than a street engine)and as well torque is multiplied by the extreme gearing l.


johnny yuma,

The relationship between engine polar inertia, engine torque, and the power needed to accelerate quickly from a stop in an F1 car is a good illustration of the principles involved. An F1 engine makes high power at high rpms, but also has very low peak torque and a narrow peaky torque curve that falls off rapidly below about 10,000 rpm. The engine also has a very small polar inertia, since it has very light components and no flywheel. Before launch assist became available, it was quite common to see even highly skilled professional race drivers stall the engine when leaving the pits. They would rev the engine to 12,000 or 14,000 rpm and dump the clutch. And the engine would immediately stall since it didn't have enough combustion and inertia torque to keep the engine speed high enough to prevent stalling.

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#219 saudoso

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Posted 29 April 2012 - 00:01

johnny yuma,

The relationship between engine polar inertia, engine torque, and the power needed to accelerate quickly from a stop in an F1 car is a good illustration of the principles involved. An F1 engine makes high power at high rpms, but also has very low peak torque and a narrow peaky torque curve that falls off rapidly below about 10,000 rpm. The engine also has a very small polar inertia, since it has very light components and no flywheel. Before launch assist became available, it was quite common to see even highly skilled professional race drivers stall the engine when leaving the pits. They would rev the engine to 12,000 or 14,000 rpm and dump the clutch. And the engine would immediately stall since it didn't have enough combustion and inertia torque to keep the engine speed high enough to prevent stalling.

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Lauch assist is completelly ilegal today, race start or pit stop. That's the half the point of the SECU. They do use the hand clutch and their right foot to get it running.

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#220 rms

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Posted 29 April 2012 - 23:02

Lauch assist is completelly ilegal today, race start or pit stop. That's the half the point of the SECU. They do use the hand clutch and their right foot to get it running.


And when does the "anti stall" kick in ?????




#221 johnny yuma

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Posted 30 April 2012 - 01:48

So getting back to post #1,Sam Elliot speaks truth for most people.If your engine has torque,it will have power.It used to be said Australian drivers don't like to rev their engines.Climbing a hill in top gear was something to be proud of.With most cars now being Auto, and modern engines quiet and smooth, we often would not know what gear we climbed a hill with ...and today's 2 litre 4s produce more torque than 3 litre 6 cylinder cars of the 60s.

#222 rms

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Posted 30 April 2012 - 02:38

...and today's 2 litre 4s produce more torque than 3 litre 6 cylinder cars of the 60s.


Can you back that up with some actual figures ??? or is it wishfull thinking ?

Mitsubishi FTO Mivec 2ltr 24 Valve V6 -- 147 lb/ft, 20.4 Kg/Mtr
173 Holden Red 2840 cc 6 cyl 12 valve pushrod --- 168 lb/ft, 22.8Kg/Mtr



#223 gruntguru

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Posted 30 April 2012 - 03:03

Yes, 50% improvement in BMEP - hasn't happened since . . . . . hmmm

#224 bigleagueslider

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Posted 30 April 2012 - 04:08

There has been a big improvement in the average BMEP of production engines over the past couple of decades. But it's almost entirely due to improvements in the fuel and ignition controls. With closed loop digital fuel and ignition controls, the SI engine can run closer to its detonation limit.

If you consider production diesel engines, their BMEP rates have improved substantially over the past 50 years. Maybe not quite 50% though.

#225 johnny yuma

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Posted 30 April 2012 - 04:44


The last thing I read before posting my comment was a newspaper report of the new 4 cylinder 2 litre engine the Falcon will soon have as an option.Quoted at 179kw,353n/m torque.Of course it is a turbo (petrol).

The 179 (3 litre)Holden 6 put out 237n/m on an 8.8:1 compression in 1963. As compression ratios climb higher into the 11s with better engine management, non-turbo 2 litre cars may approach this figure but not quite yet.The 2004 Honda CRV 2 litre managed 220n/m,but is there some benchmarking issue as with Kw in the late 70s ?

Would we not consider turbocharged engines pretty mainstream now for comparison purposes?Certainly turbocharging is an old technology, but much better behaved now..it was not an option in the 60s for a sensibly priced car,but is now ubiquitous.

#226 saudoso

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Posted 30 April 2012 - 20:10

And when does the "anti stall" kick in ?????



Anti stall is not launch assit, it will screw your start:

Kamui Kobayashi: “It was a tricky race for me. At the start when I released the clutch the car suddenly was in anti-stall mode, and that was why I lost five positions straight away.



#227 D-Type

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Posted 03 May 2012 - 21:47

Coming back to torque and power, I can never understand why people have problems with this issue and don't have problems with acceleration, speed and velocity, or with momentum and kinetic energy

#228 Magoo

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Posted 03 May 2012 - 22:08

Coming back to torque and power, I can never understand why people have problems with this issue and don't have problems with acceleration, speed and velocity, or with momentum and kinetic energy



Most folks had no problem with Aristotelian motion. It was the prevailing mechanical system for two millennia and when Galileo disproved it they put him in prison. There, problem solved. I daresay a number of people wouldn't know the difference today.

#229 Kelpiecross

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Posted 04 May 2012 - 02:57

Coming back to torque and power, I can never understand why people have problems with this issue and don't have problems with acceleration, speed and velocity, or with momentum and kinetic energy


I have a problem with momentum and kinetic energy - why one is conserved in a collision and one isn't - and situations when to apply each.

#230 Greg Locock

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Posted 04 May 2012 - 03:50

I have a problem with momentum and kinetic energy - why one is conserved in a collision and one isn't - and situations when to apply each.

Typically you need to apply each all the time, in interesting problems. Sometimes you have to include kinetic energy as an inequality, for instance in a simple car collision the total KE of the vehicles after the crash will be less than it was before the crash. Momentum is always conserved so you always use that. Have you got any specific examples?



#231 bigleagueslider

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Posted 04 May 2012 - 04:19

In an impact momentum is transferred while energy is converted. For example, much of the kinetic energy is converted to strain in the crumpled metal chassis.

Of course, I only have a high school level physics education, so my explanation may be wrong. :)

#232 johnny yuma

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Posted 04 May 2012 - 05:11

In an impact momentum is transferred while energy is converted. For example, much of the kinetic energy is converted to strain in the crumpled metal chassis.

Of course, I only have a high school level physics education, so my explanation may be wrong. :)

A car crashing exactly head on into a solid cliff-face will" bounce" back a bit.Most of it's momentum is expended in damage work ,but a little remains briefly for a new direction before friction stops it.

A billiard ball hit square against the opposing cush preserves most of it's momentum in a new direction,as virtually no damage is caused to slow it down.

I too have only high school physics... but have crashed cars and played billiards !!!

#233 gruntguru

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Posted 04 May 2012 - 07:20

A car crashing exactly head on into a solid cliff-face will" bounce" back a bit.Most of it's momentum is expended in damage work ,but a little remains briefly for a new direction before friction stops it.

A billiard ball hit square against the opposing cush preserves most of it's momentum in a new direction,as virtually no damage is caused to slow it down.

I too have only high school physics... but have crashed cars and played billiards !!!

In both your examples you are referring to energy, not momentum. In the first, most of the energy is converted to heat via deformation of metal. All of the momentum is conserved, so the car stops, but the Earth gains all the car's lost momentum.

In the second example, the energy of the ball (a scalar) remains the same after hitting the cushion (assuming an elastic collision), wheras the momentum of the ball (a vector) becomes negative and the table (along with the earth) acquires double the initial momentum of the ball.

#234 Dmitriy_Guller

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Posted 06 May 2012 - 02:48

Coming back to torque and power, I can never understand why people have problems with this issue and don't have problems with acceleration, speed and velocity, or with momentum and kinetic energy

Some people do as well, often in the course of arguing power vs. torque.

Edited by Dmitriy_Guller, 06 May 2012 - 02:49.


#235 Kelpiecross

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Posted 07 May 2012 - 04:25

Typically you need to apply each all the time, in interesting problems. Sometimes you have to include kinetic energy as an inequality, for instance in a simple car collision the total KE of the vehicles after the crash will be less than it was before the crash. Momentum is always conserved so you always use that. Have you got any specific examples?


I doubt if I could actually put into words exactly what my problem is with momentum and KE. I have done quite a bit of physics involving KE and momentum etc. both at high school and in first year physics and engineering at uni - and I always liked it. I am not questioning any of the equations etc. - clearly they accurately describe the behaviour of objects in the real world.
The origin of momentum (or inertia) is not known and is not even really speculated about. Should we also wonder about the origin of KE?
KE and gravitational potential energy are quite happy to interchange with each other in an object rolling up and down hills, pendulums, etc. and KE is conserved in elastic collisions - but KE is not conserved in inelastic collisions - it just never seemed to me to be consistent that this should be so.

Despite what physicists claim, I think some of the most basic questions in science are still unanswered. Gravity - due to space being curved around the earth (or whatever)? Just what is curved? - space is just that - space/empty.

Electromagnetic waves travel through "empty" space? We are told - "Don't worry about - it's just a property of space that this happens." Hardly an explanation.

Momentum/inerta? According to Mach, due to the distant stars. Was he serious? (pissed maybe?)

#236 Dmitriy_Guller

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Posted 07 May 2012 - 17:24

Kinetic energy never did make sense to me either. Why should it be squared with speed, from intuition POV? Absent friction, why should it take more energy to speed up the object by 1 mph when it's traveling at 30 mph as opposed to when it's traveling at 20 mph? I've seen mathematical proof, so I know that it's true within the boundaries of classical mechanics, but that doesn't help me understand intuitively why it's true.

#237 Canuck

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Posted 07 May 2012 - 21:51

That precise thing bothers me too. How is it that a wee flight attendant can accelerate a beverage cart that outweighs her beyond it's cruising speed of 500+mph?

#238 Greg Locock

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Posted 07 May 2012 - 23:37

That precise thing bothers me too. How is it that a wee flight attendant can accelerate a beverage cart that outweighs her beyond it's cruising speed of 500+mph?

So long as you use the same reference frame for both displacement and velocity, it all works out.

Say she accelerates it by 3 m/s in 3 seconds and the trolley weighs 100 kg. V0=300 m/s

KE0==1/2*m*v^2=1/2*100*300^2 J

F=m.a=100*3/3=100 N

distance travelled in 3 seconds=v0*t+1/2*a*t^2=300*3+.5*1*9=904.5m

Therefore work done by accelerating force = 904.5*100 <so the hostie is adding 30 kW of power? No, she's doing 4.5*100/3 W, the aircraft does the rest>

KE3=1/2*100*303^2

so change in KE=work done by force QED

Getting into the right reference frame and being consistent with them is crucial for solving these problems.

Incidentally, yes the basic inertial frame of the universe is probably defined by star masses, one reason you know that there is a basic one is the effects of angular velocities, which only make sense if they are absolute. <I think>
<late edits in angle brackets>

Edited by Greg Locock, 08 May 2012 - 05:42.


#239 Dmitriy_Guller

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Posted 08 May 2012 - 05:28

Frames of reference can themselves be a bit of a mind bender. Again, it's nothing that can't be dealt with analytically, but getting the intuitive feel for it can be challenging.

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#240 Dmitriy_Guller

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Posted 08 May 2012 - 05:40

What if the flight attendant, through miraculous feat of strength, accelerates the cart to 100 mph, at which point it leaves the plane (let's say there was an opening in the front of the plane). With the cart now being out of the plane, does it have the kinetic energy of a cart going 600 mph? The plane only accelerated it from 0-500, and the attendant only accelerated it from 0-100. That doesn't up to the kinetic energy of going from 0-600 due to squared relationship.

#241 gruntguru

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Posted 08 May 2012 - 11:36

You missed the bit where Greg said "and the plane did the rest". What he mean't was, while the hostie was accelerating the cart, she added the usual amount of energy (as if she were on the ground and went from 0 to 3 m/s), when in fact the energy of the cart increased by far more than the amount she added ("and the plane did the rest").

#242 saudoso

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Posted 08 May 2012 - 11:38

While the flight attendant accelerates the cart, she pushes the airplane backwards. And the jet engines need to compensate for that keeping the plane at 500 mph.

So that's from the jet engines that the missing energy comes from.

EDIT:Grunt beat me to it.

Edited by saudoso, 08 May 2012 - 11:39.


#243 Canuck

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Posted 08 May 2012 - 13:36

How is the energy transferred from the aircraft to the cart? Through the flight attendant? Seems a bit taxing on the bone structure.

#244 Dmitriy_Guller

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Posted 08 May 2012 - 15:10

You missed the bit where Greg said "and the plane did the rest". What he mean't was, while the hostie was accelerating the cart, she added the usual amount of energy (as if she were on the ground and went from 0 to 3 m/s), when in fact the energy of the cart increased by far more than the amount she added ("and the plane did the rest").

So, if I understand it right, pushing the cart from rest on a plane isn't exactly like pushing it on the ground? Your body does the same kind of work in a physics definition of the word, but now also has to transfer energy from the floor of the plane?

#245 saudoso

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Posted 08 May 2012 - 16:36

Using hypothetical numbers: The lady is doing 10N on the cart and at a given time shes walking at 1m/s. So she's pushing the plane back by 10N also. The plane is flying around 250m/s.

The lady is developing 10N*m/s = 1 watt. But the jet engines have to deliver 2500N*m/s to keep the movement.

So the cart knetic energy increasing rate is 2510W in this exact moment. While it accelerates, if the atentant is able to keep the 10N, this rate will increase accordingly.

#246 Wolf

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Posted 08 May 2012 - 17:55

What if the flight attendant, through miraculous feat of strength, accelerates the cart to 100 mph, at which point it leaves the plane (let's say there was an opening in the front of the plane). With the cart now being out of the plane, does it have the kinetic energy of a cart going 600 mph? The plane only accelerated it from 0-500, and the attendant only accelerated it from 0-100. That doesn't up to the kinetic energy of going from 0-600 due to squared relationship.


Dmitriy, is this something like what you were asking?

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#247 Dmitriy_Guller

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Posted 08 May 2012 - 18:13

Yep, that was it, translated in numbers. The action-reaction causing jet engines to do extra work explanation makes sense to me.

#248 Dmitriy_Guller

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Posted 08 May 2012 - 18:22

Thinking of another example, if you mount a cannon to the front of the train, and fire it while moving. Clearly the cannon's recoil would either slow down the train, or require it to put some energy out to maintain its speed, in which case the locomotive would work together with the powder charge to put the energy in the cannon ball.

#249 Wolf

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Posted 08 May 2012 - 19:54

Yep, that was it, translated in numbers. The action-reaction causing jet engines to do extra work explanation makes sense to me.


Yep, but I was hoping that somebody will come up with formula to explain it... (the problem is that when using 'force x displacement' formula I got the wrong result and can't come to terms why I did, but, appropriately for this thread, got the correct result when I considered the power consumption)

As I wrote this reply it occurred to me where the error was, so we'll pretend I was posting it to test everybody else... :lol:

#250 GSpeedR

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Posted 08 May 2012 - 20:11

Yep, but I was hoping that somebody will come up with formula to explain it... (the problem is that when using 'force x displacement' formula I got the wrong result and can't come to terms why I did, but, appropriately for this thread, got the correct result when I considered the power consumption)

As I wrote this reply it occurred to me where the error was, so we'll pretend I was posting it to test everybody else... :lol:


Greg Locock already presented the concept. This won't turn out as nice as Wolf's equations do above (sorry):

Ei = Ef

KEi + PEi = KEf +PEf - Wext

PEi = PEf = 0

Force = m * a = m * vc/t [t wasn't defined but it will cancel]
X = v0*t + 1/2*a*t^2 = v0*t + 1/2*(vc/t)*t^2

Wext = Force*X = m*vc/t * (v0*t+1/2*vc*t) [note that t's cancel]

KEi = 1/2*m*v0^2 = 0.5*100*100^2 = 500kJ
KEf = 1/2*m*(v0+vc)^2 = 0.5*100*(110)^2 = 605kJ
Wext = m*vc*(v0+1/2*vc) = 100*10*(100+0.5*10) = 105kJ

KEi = KEf - Wext => 500 = 605 -105

In the situation Wolf presented we have changed reference frames and must account for the difference in work done in each frame. In the 'general' ground-based frame, the work done is shown in the equation above (105kJ). In the moving reference frame of the plane, the velocity of the plane is neglected and the displacement of the cart is X = 1/2*a*t^2 = 1/2*vc/t*t^2. Thus the work is m*vc/t * (1/2*vc*t) = 100*10*0.5*10 = 5kJ. The difference in work done between these two frames is 100kJ, which is the "unaccounted for" energy that Wolf was left with.

Edited by GSpeedR, 08 May 2012 - 20:24.