# Acceleration - Torque vs Power

308 replies to this topic

### Poll: Max accelleration at max torque or max power (65 member(s) have cast votes)

#### Assume we have a car traveling at speed and a gearing so we can select gear to have the engine running at a rpm where it's either produces max torque or max power. What gear would achieve maximum acceleration, the one that put engine rpm at max torque or at max power?

1. I am sure it is at max torque and know the formula to prove it (9 votes [13.85%])

Percentage of vote: 13.85%

2. I know it is at max torque as I read it in books (3 votes [4.62%])

Percentage of vote: 4.62%

3. I believe it is at max torque by reading forums like this (1 votes [1.54%])

Percentage of vote: 1.54%

4. I think it is at max torque (5 votes [7.69%])

Percentage of vote: 7.69%

5. I have no idea what rpm would generate max acceleration (4 votes [6.15%])

Percentage of vote: 6.15%

6. I think it is at max power (13 votes [20.00%])

Percentage of vote: 20.00%

7. I believe it is at max power by reading forums like this (1 votes [1.54%])

Percentage of vote: 1.54%

8. I know it is at max power as I read it in books (3 votes [4.62%])

Percentage of vote: 4.62%

9. I am sure it is at max power and know the formula to prove it (26 votes [40.00%])

Percentage of vote: 40.00%

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### #151 Dmitriy_Guller

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Posted 01 May 2013 - 02:49

But anyway, I'd assume you wouldn't be prescribing shifting around the peak torque RPM on the curve. And bear with me, flat top torque curve means a straight pointing up power curve, not a flat one.

Depends on where you look. No torque curve stay flat forever, it decays at some point. That region where it decays creates for unusually flat power curve, with peak power somewhere in the middle.

### #152 manolis

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Posted 01 May 2013 - 03:29

Best exposition yet of the underlying logic. And thanks for keeping the math(s) simple.

Desmo,
did you see the post #13?

"Look at the given problem from the energy viewpoint.

In the question: Where the energy provided by the engine goes?

The answer is: One part is consumed to overcome the overall resistance (mechanical, aerodynamic etc) of the vehicle to move with the given speed (so, this part is the same in either case), the rest part is added to the kinetic energy of the vehicle and increases vehicle’s speed.

So, the more the energy provided per second by the engine (i.e. the more the power), the more the increase of the speed of the vehicle per second (i.e. the higher the acceleration).

So, with the engine operating at its max power, the maximum acceleration is achieved.

Thanks
Manolis Pattakos"

### #153 johnny yuma

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Posted 01 May 2013 - 06:29

Dammit I feel like Homer Simpson going back to his poisonous 4 day old submarine sandwich...mmm how
could I leave you.
Can I ask what is being expressed exactly in engine specifications where for example Torque is 300 N/M
at 3000 rpm,Power is 150KW at 5000rpm. I have always thought it meant the best,or most energetic,
combustion chamber explosions,were occuring at 3000 rpm, while the engine was doing the most work at
5000rpm as lessening torque was still balanced by the fact more combustion chamber explosions per second
or minute or whatever time period were occuring, this being followed soon by a decay in engine function as torque took a huge dive.
You may,or may not,experience maximum acceleration RIGHT AT THAT POWER PEAK,but the trouble is IT'S OVER !
Isn't acceleration linked inextricably with time ? Are we looking for a Sweet Spot,or a Sweet Period ?

Edited by johnny yuma, 01 May 2013 - 06:32.

### #154 gruntguru

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Posted 01 May 2013 - 06:52

But anyway, I'd assume you wouldn't be prescribing shifting around the peak torque RPM on the curve..

Correct.

And bear with me, flat top torque curve means a straight pointing up power curve, not a flat one.

No, I did mean flat-top power curve - often all the way to rev limit. They also have a flat torque curve over a substantial rpm range, then falling in the constant power region.

Detroit examples.

### #155 gruntguru

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Posted 01 May 2013 - 07:03

Can I ask what is being expressed exactly in engine specifications where for example Torque is 300 N/M at 3000 rpm,Power is 150KW at 5000rpm. I have always thought it meant the best,or most energetic, combustion chamber explosions,were occuring at 3000 rpm, while the engine was doing the most work at 5000rpm as lessening torque was still balanced by the fact more combustion chamber explosions per second or minute or whatever time period were occuring, this being followed soon by a decay in engine function as torque took a huge dive.

All correct except the huge dive doesn't have to exist. Many engine designs see a much more gradual drop in power.

You may,or may not,experience maximum acceleration RIGHT AT THAT POWER PEAK, but the trouble is IT'S OVER !

At a given road speed the maximum acceleration occurs when you gear to put the engine "RIGHT AT THAT POWER PEAK". Sure you will get greater acceleration in that gear at the torque peak, but you are now at a LOWER ROAD SPEED.

Isn't acceleration linked inextricably with time ? Are we looking for a Sweet Spot,or a Sweet Period ?

Yes.
A sweet spot. (lots of them if the acceleration test is conducted over a finite time or distance eg 1/4 mile, 0-60 etc)

### #156 Canuck

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Posted 01 May 2013 - 15:37

At a given road speed the maximum acceleration occurs when you gear to put the engine "RIGHT AT THAT POWER PEAK". Sure you will get greater acceleration in that gear at the torque peak, but you are now at a LOWER ROAD SPEED.

This is where I keep falling down in the "max power" camp. The thrust curves show a max accel at peak torque so there's no question that peak torque produces maximum acceleration in any given gear.

At any given speed - assuming identical gear ratios, then yes, max power (because we're now multiplying torque by using a lower gear) however the answer to the question as posed:

Assume we have a car traveling at speed and a gearing so we can select gear to have the engine running at a rpm where it's either produces max torque or max power. What gear would achieve maximum acceleration, the one that put engine rpm at max torque or at max power?

is clearly to have the engine running at an rpm where it produces max torque. The question does not appear to require identical gearing though as written it is a bit undefined and open to interpretation.

### #157 rgsuspsa

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Posted 01 May 2013 - 20:11

This is where I keep falling down in the "max power" camp. The thrust curves show a max accel at peak torque so there's no question that peak torque produces maximum acceleration in any given gear.

At any given speed - assuming identical gear ratios, then yes, max power (because we're now multiplying torque by using a lower gear) however the answer to the question as posed:

is clearly to have the engine running at an rpm where it produces max torque. The question does not appear to require identical gearing though as written it is a bit undefined and open to interpretation.

Acceleration of a fixed mass is solely a function of the rate at which kinetic energy is added to the mass, and the rate at which energy is added is defined as power. The type of mechanism supplying power to the mass is irrelevant. The power supplier may be an internal combustion engine with rotation of a flywheel, rotating electric motor, linear electric motor (by definition having zero torque because there is no rotary motion involved), wind power supplied to a sail, also no torque involved, etc. Without recourse to differential calculus and its associated mathematical expressions the subject of this thread is correctly answered by GruntGuru's earlier
post No. 141 regarding kinetic energy.

Ron Sparks

Edited by rgsuspsa, 01 May 2013 - 21:41.

### #158 saudoso

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Posted 01 May 2013 - 20:24

Assume we have a car traveling at speed and a gearing so we can select gear to have the engine running at a rpm where it's either produces max torque or max power. What gear would achieve maximum acceleration, the one that put engine rpm at max torque or at max power?

This is where I keep falling down in the "max power" camp. The thrust curves show a max accel at peak torque so there's no question that peak torque produces maximum acceleration in any given gear.

At any given speed - assuming identical gear ratios, then yes, max power (because we're now multiplying torque by using a lower gear) however the answer to the question as posed:

is clearly to have the engine running at an rpm where it produces max torque. The question does not appear to require identical gearing though as written it is a bit undefined and open to interpretation.

My dear northern friend, plase read again the question and your own statements.

Edited by saudoso, 01 May 2013 - 20:26.

### #159 Canuck

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Posted 01 May 2013 - 21:47

Edited.
If starting ground speed is fixed then max power rpm will allow a lower gear. That will always be quicker.

Edited by Canuck, 01 May 2013 - 22:26.

### #160 johnny yuma

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Posted 02 May 2013 - 01:12

Edited.
If starting ground speed is fixed then max power rpm will allow a lower gear. That will always be quicker.

That sums up the quandary neatly.However I still can't see a way to answer the multiple chice question by ticking one
of the boxes available without some sort of qualifying statement involving time.

### #161 NTSOS

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Posted 02 May 2013 - 01:43

Using a 2013 GT500 as an example:

Peak HP = 609 HP @ 6400 RPM and 500 lb-ft (same RPM)

Peak Torque = 609 lb-ft @ 4300 RPM and 502 HP (same RPM)

Maximum acceleration is not the same as maximum rate of acceleration. Maximum "overall" acceleration occurs when the vehicle is geared for and shifted at the power peak. Maximum "rate" of acceleration occurs at the torque peak in any gear. The vehicle is still accelerating after the torque peak, simply at a declining rate.....why then would you want to gear for the torque peak and @ 502 HP and throw away the additional 107 HP that is available at the power peak?

PS (that is of course the additional 107 HP that is generated by the 500 lb-ft of torque at the rpm of peak power)

Edited by NTSOS, 02 May 2013 - 14:06.

### #162 gruntguru

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Posted 02 May 2013 - 04:45

That sums up the quandary neatly.However I still can't see a way to answer the multiple chice question by ticking one of the boxes available without some sort of qualifying statement involving time.

The question as stated implies instantaneous acceleration. So for each test you take a high speed photo of the driver's head in profile. Which test shows the drivers head pressing furthest into the headrest?

BTW. Did you know the window you type your posts into has word-wrap?

### #163 Kelpiecross

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Posted 02 May 2013 - 12:13

The question as stated implies instantaneous acceleration. So for each test you take a high speed photo of the driver's head in profile. Which test shows the drivers head pressing furthest into the headrest?

BTW. Did you know the window you type your posts into has word-wrap?

What's "word-wrap"?

### #164 saudoso

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Posted 02 May 2013 - 12:50

johny finishes his lines with [enter] as he types, making his text all broken for people with different screens. Word-wrap means the text box will automatically break lines to fit the presentation rectangle no matter the rectangle and the font size.

### #165 NTSOS

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Posted 02 May 2013 - 23:08

Big engines and flat torque curves are great to haul big weights and to drive around confortably. Cross the RFID toll gate in top gear and just floor it back to cruising speed.

But what would be the optimal shifting point for maximum acceleration with a 952 ci Detroit engine?

Interesting conversation today with the owner of Berry Brothers Towing & Transport Company that uses DD16 engines. He just returned from a tour of the engine assembly plant and he told me that Detroit is owned by Daimler....I didn't know that. The guy was very impressed with the amount of money that Daimler has invested in R&D and automated engine assembly machinery to circumvent human errors induced by repetitive/boring tasks. I also didn't know that the secondary turbo is a compound turbo and it drives the flywheel using a mechanical/hydraulic interface. Anyway, he has owned the company for 40 years and he states that there is no current engine that is comparable to the DD16's technology, performance and durability and he has owned them all....he said a Cummins would be second best, but a distant second. He still drives occasionally so I jumped at the opportunity to ask him how he actually drives it. For max fuel economy he shifts at 1300 RPM, for max acceleration he shifts at 1800 RPM and he regularly pulls 80,000 + pound loads over the Altamont Pass in northern CA.

### #166 gruntguru

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Posted 02 May 2013 - 23:39

Here's why. DD16 Power and turque curves.

### #167 saudoso

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Posted 02 May 2013 - 23:45

Peak power at 1800rpm.

### #168 gruntguru

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Posted 02 May 2013 - 23:52

Yes. So if you shift at 1800 (600hp) and the revs drop to 1600, you still have 585hp. So he should be shifting when the power drops to 585 - that would be about 1830 rpm - someone better tell him he's short shifting!

### #169 NTSOS

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Posted 03 May 2013 - 00:14

Yes. So if you shift at 1800 (600hp) and the revs drop to 1600, you still have 585hp. So he should be shifting when the power drops to 585 - that would be about 1830 rpm - someone better tell him he's short shifting!

Gulp......I'm not exactly sure if I would want to tell him that.....any volunteers?

Edited by NTSOS, 03 May 2013 - 17:45.

### #170 Kelpiecross

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Posted 03 May 2013 - 04:17

What's "word-wrap"?

Thank you for that. JY's posts appear normal on my screen.

### #171 MatsNorway

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Posted 03 May 2013 - 13:49

Yes. So if you shift at 1800 (600hp) and the revs drop to 1600, you still have 585hp. So he should be shifting when the power drops to 585 - that would be about 1830 rpm - someone better tell him he's short shifting!

hehe. but its actually "risky" missing out even as little as 20rpm up to 1850 (575hp) that is a bigger miss than hes current shift point. It all depends on the time you got to time it. So it depends on the rate acceleration.

Does F1 teams do risk assesment on shift points?? that would indicate a silly level of performance hunting.

"time lost when you hit the rpm limiter has given us the statistics to say that you should try to shift at 17,5k in first, second and third from there you should eek it out to 17,8k"

### #172 desmo

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Posted 03 May 2013 - 20:16

Is reaction time really best approximated as a constant? Even driver to driver?

### #173 saudoso

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Posted 03 May 2013 - 20:33

So they're always going as close as possible to redline without getting caught by the rev limiter due to driver lag, right?

### #174 Dmitriy_Guller

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Posted 04 May 2013 - 00:13

I'm not sure what you mean with "risk assessment", but what you describe in your last sentence is pretty common procedure, not only in F1.
You will move your shift points, and any "half decent" shift light/display system, will allow you to select different shift rpm´s for different gears (speed).
The main reason for this, is that your aerodynamic resistance increases, with the square of speed, which means, that if we assume a constant shift time (interrupt of power/acceleration)
the car slows more in a higher gear (shift 5-6) (at higher velocity) then it does in a lower gear (shift 2-3).
Now if, like F1, you have no interruption in power during upshifts, then this is less of an issue.

But, you have still the "reaction time" of the driver, which is a constant too, to deal with. I think, this was your point, and yes it is taken into consideration.
How long does it take, the driver after seeing the shift light to up shift, the gradient with which rpm's increase in any gear, will define the amount of "overshoot" you get.
You try to account for this, by setting the shift lights accordingly.
I'm sure, that even the top teams in Formula Ford take this into account.

Another reason you would have different shift points in each gear is that gear ratios, to my knowledge, are not usually set to be in fixed proportion from one to the next. Instead these proportions approach 1 as you go up through the gears, and thus the faster you go, the closer to power peak you stay.

### #175 gruntguru

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Posted 04 May 2013 - 08:16

When a series is artificially rev-limited (e.g. F1, V8 supercars) the tendency as engine development progresses is for peak power to occur at or very close to the rev-limiter. In this case max acceleration is achieved by shifting at the rev limit which of course is difficult to do precisely and quite risky since power drops to zero immediately above the rev limit. It is important for the engine developer to produce an engine with a wide enough power band to accommodate the gear ratio spreads and the speed-range of the worst corner encountered (to avoid gearshifting mid-corner).

So if for example the requirement is for the engine to operate within 20% of rev-limit without significant power loss, the design will necessarily sacrifice a little peak power at the limiter to obtain a higher average power over that 20% band.

### #176 bigleagueslider

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Posted 07 May 2013 - 06:33

The question as stated implies instantaneous acceleration.

There is no such thing as "instantaneous" acceleration, since by definition acceleration is the change in velocity with respect to time. An instantaneous event would require an infinitely small time period, and would thus produce infinitely high acceleration rates. Of course, you can have instantaneous moments, forces or velocities.

As for the shift events in a race car drivetrain, there will always be some interruption in power delivery during any shift. But it may be imperceptible if the shift event is rapid and the inertias in the system are large enough.

### #177 smitten

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Posted 07 May 2013 - 07:49

There is no such thing as "instantaneous" acceleration, since by definition acceleration is the change in velocity with respect to time. An instantaneous event would require an infinitely small time period, and would thus produce infinitely high acceleration rates.

Many mathematicians would disagree with you on a conceptual level. The calculation (derivative) is as the time approaches zero, not when it is zero.

### #178 Tony Matthews

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Posted 07 May 2013 - 08:29

As for the shift events in a race car drivetrain, there will always be some interruption in power delivery during any shift. But it may be imperceptible if the shift event is rapid and the inertias in the system are large enough.

Here we go again.

### #179 Dmitriy_Guller

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Posted 07 May 2013 - 14:04

There is no such thing as "instantaneous" acceleration, since by definition acceleration is the change in velocity with respect to time. An instantaneous event would require an infinitely small time period, and would thus produce infinitely high acceleration rates. Of course, you can have instantaneous moments, forces or velocities.

How exactly do you get an infinitely high acceleration rate as your time period gets infinitely small?

### #180 munks

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Posted 07 May 2013 - 17:29

There is no such thing as "instantaneous" acceleration, since by definition acceleration is the change in velocity with respect to time.

This is one of the silliest things you've ever written, sorry. What if I said there is no such thing as "instantaneous" velocity, since by definition velocity is the change in position with respect to time?

### #181 munks

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Posted 07 May 2013 - 17:32

Also, you've totally contradicted yourself by saying there *is* such a thing as instantaneous force. In case you haven't noticed, F=ma and the mass can often be considered a constant.

### #182 rgsuspsa

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Posted 07 May 2013 - 19:38

There is no such thing as "instantaneous" acceleration, since by definition acceleration is the change in velocity with respect to time. An instantaneous event would require an infinitely small time period, and would thus produce infinitely high acceleration rates. Of course, you can have instantaneous moments, forces or velocities.

As for the shift events in a race car drivetrain, there will always be some interruption in power delivery during any shift. But it may be imperceptible if the shift event is rapid and the inertias in the system are large enough.

Rate of change of velocity as a function of time as time approaches an infinitesimally small increment, but not zero, is in fact, "Instantaneous Acceleration". Fundamental laws of physics and mathematics are necessary analytical skills in order to correctly apply Newton's Laws of Motion to physical objects. Verbal language and rationalizations derived from it are inadequate means of analysis and understanding of physical phenomena.

Ron Sparks

Edited by rgsuspsa, 07 May 2013 - 19:40.

### #183 gruntguru

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Posted 07 May 2013 - 23:05

More relevent is one of Newton's other little discoveries - calculus.

a=dv/dt= the limit of (delta v)/(delta t) as delta t approaches zero.

Note that as delta t approaches zero, so does delta v so "a" does not go to infinity but represents the slope or gradient of the v-t curve at the instant in question.

### #184 bigleagueslider

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Posted 15 May 2013 - 07:34

More relevent is one of Newton's other little discoveries - calculus.

a=dv/dt= the limit of (delta v)/(delta t) as delta t approaches zero.

Note that as delta t approaches zero, so does delta v so "a" does not go to infinity but represents the slope or gradient of the v-t curve at the instant in question.

Obviously a=dv/dt . So for a given dv, as dt approaches zero, wouldn't "a" approach infinity? What you are describing seems to be a case of fixed acceleration rate, rather than the variable acceleration rate case being considered. Maybe we just have a different notion of what the term "instantaneous" implies. I consider the term to mean an infinitely short period of time, while you seem to consider it to mean a specific point of time within a given event.

The point of my original post was that the notion of "instantaneous acceleration" was impractical due to the fact that you cannot quantify acceleration without having some associated finite period of time. On the other hand, characteristics such as velocity, force, or moment can be considered in "instantaneous" terms, since no change in time is required to quantify them.

I'd also appreciate it if you cut me some slack regarding my math and physics. I only have a high school level education in calculus and physics.

### #185 Dmitriy_Guller

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Posted 16 May 2013 - 03:46

Obviously a=dv/dt . So for a given dv, as dt approaches zero, wouldn't "a" approach infinity?

For a given dv, dt isn't going to be approaching anything. It would just be the dt corresponding to the "given" dv. You're missing the fact that dv and dt are inextricably linked, so you can't just hold one of them fixed and vary the other. That's exactly the mistake that torque people make, by fixing the gear and and varying the RPM, ignoring the fact that at any given moment you can't vary the RPM without changing the gear.

Edited by Dmitriy_Guller, 16 May 2013 - 03:47.

### #186 Greg Locock

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Posted 17 May 2013 - 04:31

I'd also appreciate it if you cut me some slack regarding my math and physics. I only have a high school level education in calculus and physics.

Sure, but cut us some slack when we say your objections have been answered many times over. Acceleration, and inertia, have interesting properties mechanically compared with displacements, velocities, springs, and dampers (because mass and acc result in single ended forces) but mathematically d^2x/dt^2 is just one of a bunch of derivatives from dx/dt through to d^100000x/dt^100000 or more, it is not very special.

### #187 munks

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Posted 17 May 2013 - 14:46

Sure, but cut us some slack when we say your objections have been answered many times over. Acceleration, and inertia, have interesting properties mechanically compared with displacements, velocities, springs, and dampers (because mass and acc result in single ended forces) but mathematically d^2x/dt^2 is just one of a bunch of derivatives from dx/dt through to d^100000x/dt^100000 or more, it is not very special.

Now *I* am going to ask for some slack, even though I had some college-level calculus. I never thought to ask yet I couldn't figure it out on my own: In these examples, why does the power go after the delta on the top, but after the value on the bottom?

Edited by munks, 17 May 2013 - 14:46.

### #188 blkirk

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Posted 17 May 2013 - 17:03

Now *I* am going to ask for some slack, even though I had some college-level calculus. I never thought to ask yet I couldn't figure it out on my own: In these examples, why does the power go after the delta on the top, but after the value on the bottom?

My understanding is that the more correct (pure, pedantic, math-snob) way of writing it would be (d/dt)^2 x. Translated to spoken words, it would read "the second derivative with respect to time of x". Or at least that's what my professors in college did.

### #189 saudoso

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Posted 17 May 2013 - 18:22

Or x = f(t), v = f'(t) and a = f"(t).

Edited by saudoso, 17 May 2013 - 18:22.

### #190 saudoso

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Posted 17 May 2013 - 18:23

This stuff is starting to come back to my brain and it's starting to hurt.

### #191 GodHimself

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Posted 17 May 2013 - 21:25

Or x = f(t), v = f'(t) and a = f"(t).

Or x = f(t), v = ẋ and a = ẍ

On a side note, the discussion about derivatives looks rather funny.

### #192 saudoso

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Posted 17 May 2013 - 23:47

I'll second munks question, the d^2x/dt^2 notation looks odd to me, but then again we didn't use computers when I was at university but for numerical analysis with pascal programing.

blkirk's notaion looks more familiar to me.

### #193 GodHimself

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Posted 18 May 2013 - 00:52

It isn't odd, it's perfectly logical and equivalent to blkirk's notation.

Operators (very roughly speaking, functions that take other function as a parameter and produce yet another function) are ubiquitous in physics, both classical and quantum mechanics. A derivative is an operator - a differential operator - commonly denoted (in one-dimensional case) by d / dx. Now, we can find second derivative by applying the operator twice, denoted by d/dx (d/dx) and likewise n-th derivative by a d/dx (d/dx (d/dx (...))), which then can be abbreviated as (d / dx)^n = d^n / (dx)^n (notice the parentheses in denominator). The numerator part should be obvious, since it's simply a delta to the n-th power. The dx part however is an infinitesimal change of x and as such should be treated as a single, indivisible symbol (or to put it differently, in the context of this operation, neither d, nor x from the denominator have any mathematical meaning). Therefore we can skip the parentheses, which gives us the final formula of d^n/dx^n.

tl;dr
Both (d/dx) ^ n and d^n / dx^n notations, are correct.

Edited by GodHimself, 18 May 2013 - 01:05.

### #194 NTSOS

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Posted 18 May 2013 - 16:23

Yes sir.....thinking mathematically vs physically.....so, how did Einstein think?

Interesting article!

How Did Einstein Think?

Juanito

Edited by NTSOS, 18 May 2013 - 17:22.

### #195 saudoso

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Posted 18 May 2013 - 18:40

Einstein took a hint from aliens, future people or from god himself.

That's not thinking outside the box, it's thinking outside the [then] know universe.

### #196 Greg Locock

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Posted 18 May 2013 - 23:19

The other way of writing it is to replace d/dt by the D operator. And of course in the old days we just put a dot over the x. Counting dots is a bit scary.

The D operator approach is the easiest typographically, and for setting up partial fractions and other horrible things i can't remember to do with poles and sinks in electronic filter design and fluids.

### #197 GodHimself

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Posted 19 May 2013 - 01:22

Fascinating read, that Einstein article, thank you NTSOS. And yep saudoso, that was me. I like to drop a hint from time to time.

GodHimself

### #198 NTSOS

NTSOS
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• Joined: February 05

Posted 19 May 2013 - 15:25

Fascinating read, that Einstein article, thank you NTSOS. And yep saudoso, that was me. I like to drop a hint from time to time.

GodHimself

### #199 pizzalover

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• Joined: February 12

Posted 06 June 2013 - 19:10

From Wikipedia http://en.wikipedia....roll_(Internet)

In Chinese, trolling is referred to as bái mù (Chinese: 白目; literally "white eye"), which can be straightforwardly explained as "eyes without pupils", in the sense that whilst the pupil of the eye is used for vision, the white section of the eye cannot see, and trolling involves blindly talking nonsense over the internet, having total disregard to sensitivities or being oblivious to the situation at hand, akin to having eyes without pupils. An alternative term is bái làn (Chinese: 白爛; literally "white rot"), which describes a post completely nonsensical and full of folly made to upset others, and derives from a Taiwanese slang term for the male genitalia, where genitalia that is pale white in colour represents that someone is young, and thus foolish. Both terms originate from Taiwan, and are also used in Hong Kong and mainland China. Another term, xiǎo bái (Chinese: 小白; literally "little white") is a derogatory term that refers to both bái mù and bái làn that is used on anonymous posting internet forums. Another common term for a troll used in mainland China is pēn zi (Chinese: 噴子; literally "sprayer, spurter").

In Japanese, tsuri (釣り?) means "fishing" and refers to intentionally misleading posts whose only purpose is to get the readers to react, i.e. get trolled. arashi (荒らし?) means "laying waste" and can also be used to refer to simple spamming.

In Icelandic, þurs (a thurs) or tröll (a troll) may refer to trolls, the verbs þursa (to troll) or þursast (to be trolling, to troll about) may be used.

In Korean, nak-si (낚시) means "fishing", and is used to refer to Internet trolling attempts, as well as purposefully misleading post titles. A person who recognizes the troll after having responded (or, in case of a post title nak-si, having read the actual post) would often refer to himself as a caught fish.[citation needed]

In Portuguese, more commonly in its Brazilian variant, troll (produced [ˈtɾɔw] in most of Brazil as spelling pronunciation) is the usual term to denote internet trolls (examples of common derivate terms are trollismo or trollagem, "trolling", and the verb trollar, "to troll", which entered popular use), but an older expression, used by those which want to avoid anglicisms or slangs, is complexo do pombo enxadrista to denote trolling behavior, and pombos enxadristas (literally, "chessplayer pigeons") or simply pombos are the terms used to name the trolls. The terms are explained by an adage or popular saying: "Arguing with fulano (i.e. John Doe) is the same as playing chess with a pigeon: the pigeon defecates on the table, drop the pieces and simply fly, claiming victory."

In Thai, the term "krean" (เกรียน) has been adopted to address Internet trolls. The term literally refers to a closely cropped hairstyle worn by most school boys in Thailand, thus equating Internet trolls to school boys. The term "tob krean" (ตบเกรียน), or "slapping a cropped head", refers to the act of posting intellectual replies to refute and cause the messages of Internet trolls to be perceived as unintelligent.[citation needed]

Urban myth. The term Troll had nothing to do with trawling. If it did, the proper name would not be Troll which is a tenuous link at best.

Troll come from the three Billy Goats Gruff fairy tale and refers to Trolls lurking unseen under bridges. Users of chat rooms over 20 years ago who did not make a contribution, but were only present for voyeuristic reasons were called Trolls for this reason.

Troll then become a term for any undesirable internet user of comments pages, chat rooms having become practically defunct.

### #200 MatsNorway

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• Joined: December 09

Posted 06 June 2013 - 19:30

I suggest watching the movie trollhunter if you want proper trolls. None of that tiny stuff. Proper Norwegian ones.

My recommendation is to not read up or watch the front cover. As it spoils the movie.

Cover spoiler if you must.
http://ia.media-imdb...,0,214,317_.jpg

^^

Edited by MatsNorway, 06 June 2013 - 19:37.