Drag in accelerated reference

1. Jul 2, 2010

TimeHorse

I am a bit confused by the equations for Fdrag (the force drag applies to an object) and Pdrag (the power required to overcome drag).

$$F_{drag} = \frac{1}{2} C_{d} \times \rho \times A \times \nu^{2}$$

Where:
$$C_{d}$$ is the Drag Coeffecient (a dimensionless, empirically-derived constant).
$$\rho$$ is the density of the fluid (air or liquid)
A is the cross-sectional area incident incident on the object by the moving fluid (or moving body in a fluid)
$$\nu$$ is the speed of the object relative to the fluid.

This is fine. It's interesting that it has the same form as Kinetic Energy, with it's $$\frac{1}{2} \nu^{2}$$ term but since it was derived using dimensional analysis and the Buckingham π theorem, it seems that's just a coincidence.

Where I think things break down is when you go from Force to Power. Normally, one can compute Power from force and velocity by observing:

$$P = \vec{F} \cdot \vec{\nu}$$

So most texts derive:

$$P_{drag} = \frac{1}{2} C_{d} \times \rho \times A \times \nu^{3}$$

I assume this is born out by experiment as true so I can't argue it. But at the very least, I take it that assumes $$\nu$$ is constant. What if $$\nu$$ wasn't constant? Wouldn't you then have to, effectively, take the integral:

$$P = \frac{dW}{dt} = \frac{\int{F ds}}{dt}$$

So would the power required to overcome drag in an accelerated reference frame be given by:

$$P_{drag} = \frac{1}{6} C_{d} \times \rho \times A \times \nu^{3}$$

Can somebody please explain why power is given by the former and not the later equation? Thanks.

2. Jul 2, 2010

hikaru1221

This is where you got it wrong: $$dW = \int {Fds}$$

3. Jul 2, 2010

TimeHorse

Okay, but if:

$$W = \int{F ds}$$

How does the discrete dW relate to force? Why isn't Power given by:

$$P = \int{F d\nu}$$?

4. Jul 2, 2010

hikaru1221

No :(
Work is defined in this way: $$dW=Fds$$ (so $$W=\int {Fds}$$ )

And power is defined as: $$P=\frac{dW}{dt}$$

From the two definitions: $$P=\frac{Fds}{dt}=Fv$$
(I'm sorry for being too lazy to write those in vector form, but you should understand F and ds are vectors, not scalar quantities )

5. Jul 2, 2010

TimeHorse

No need to apologize, you are being quite helpful here. So what about mass and acceleration?

$$F = m \times a$$

$$E = \frac{1}{2} m \times v^{2}$$

So if you use $$P = \frac{dW}{dt}$$ you get $$P = \frac{1}{2} m \times \frac{dv^{2}}{dt}$$ but with $$P = \vec{F} \cdot \vec{v}$$, you get $$P = m \times v \times a$$, a velocity times an acceleration. Dimensionally, both are correct, but wouldn't the change in Kinetic Energy over the Change in Time better describe the Power required than the acceleration for some velocity times mass?

Again, thanks for helping clarifying things here. Just trying to understand why we sometimes use calculus and sometimes don't.

6. Jul 2, 2010

hikaru1221

There is a little (but deadly) mistake:
$$P=\frac{1}{2}m\frac{d(v^2)}{dt}$$
not $$P=\frac{1}{2}m\frac{(dv)^2}{dt}$$
From that, you can easily deduce this: $$P=mv\frac{dv}{dt}=mva$$ which is the same result.
Clear?

7. Jul 2, 2010

TimeHorse

Okay, yes, I meant the former and see your derivation is sound. So let me put this in more practical terms. Say I have an object whose velocity I measure in snapshots of 1 second interval. I want to compute an estimate of the power exerted by that object in order to go from one sample speed to the next. As it is, I could do:

$$P = m \times v_{n} \times (v_{n+1} - v_{n}) \times \omega$$

or

$$P = \frac{1}{2} m \times (v_{n+1}^{2} - v_{n}^{2}) \times \omega$$

for my estimate, where $$\omega$$ is my sample rate. Which is more correct? Clearly the former would return a Power of 0 when accelerating from 0 so I believe the later is more correct. But you've proven me wrong before.

8. Jul 2, 2010

hikaru1221

Are the average speed and the instantaneous speed the same? Obviously not. So are the average power and the instantaneous power the same? Obviously not too. The formula I derived in post #6 is for instantaneous power (which is why I could take the derivative of $$v^2$$ w.r.t time t). For average power, which you want to find in the above problem, you must use the original definition:
$$<P>=\frac{\Delta W}{\Delta t}$$
That means the latter one is correct in this case

By the way, why did you set $$v_n$$ as the 2nd factor in the former? The formula P=mva doesn't specify that, so it can be $$v_{n+1}$$, $$0.5(v_n+v_{n+1})$$, $$v_n - 2v_{n+1}$$, or anything whose dimension is speed, right? Be careful! You must derive a new formula from the original one

9. Jul 2, 2010

TimeHorse

Thank you kindly, hikaru! Glad to know my knowledge of basic Physics isn't completely shot. So going back to the original problem of Drag, clearly the instantaneous drag on an object is given by the standard formula:

$$P_{drag, instant} = \frac{1}{2} C_{d} \times \rho \times A \times \nu^{3}$$

But again going back to my velocity samples, assuming I know $$\rho$$, A and $$C_{d}$$, which are fixed regardless of velocity and acceleration, what is the average power used to overcome drag by my moving object? Is it still given by this formula, or should I apply calculus here as well?

BTW, point well taken on the arbitrary decision of which $$v_{n}$$ to use. Thanks.

10. Jul 2, 2010

hikaru1221

Calculus is essential in this case for a precise result But sometimes approximation is a better choice.

11. Jul 5, 2010

TimeHorse

Okay, so I think I got it now. The average power required to overcome drag is therefore given by the equation:

$$P_{drag,average} = \frac{1}{6} C_{d} \times \rho \times A \times \left( \frac{\Delta s}{\Delta t} \right)^{3}$$

Does that sound right? And doesn't that just become:

$$P_{drag,average} = \frac{1}{6} C_{d} \times \rho \times A \times \Delta \nu^{3}$$

Regardless of the time interval of the sample?

Last edited: Jul 5, 2010
12. Jul 5, 2010

K^2

ds/dt IS NOT dv. It's just v

$$P = \frac{dW}{dt} = \frac{d \int F ds}{dt} = \frac{F ds}{dt} = F \frac{ds}{dt} = Fv$$

This works because given h(t)=dH/ds, dH/dt = (dH/ds)(ds/dt) = h(t)(ds/dt). Chain rule applied to integrals.

Always be very careful of what you are integrating over and what you are differentiating with respect to.

13. Jul 5, 2010

TimeHorse

Yes, yes, I know how the Power-Force relation is derived. But I'm talking specifically drag where v is not constant and is averaged over an interval of the form:

1 sec | 4 m/s
2 sec | 6 m/s
3 sec | 6 m/s
4 sec | 5 m/s

etc.

Obviously there is some acceleration component between each sample and what I want to compute is what the average power required to overcome drag when the object is moving at the given speed at the given instant. Yes, I could just assume the instant drag at each interval is applied equally throughout the interval, but in practice this interval is applied in a gradated scale from one sample to the next. I want a better estimate than just estimating the instant drag for each velocity.

I tried Wolfram Alpha for the drag equation and found the following:

http://www.wolframalpha.com/input/?i=d+%28int+%28c+%2F+2+a+%28d+x+%2F+%28d+t%29%29^2+rho%29+dx%29+%2F+%28d+t%29

Which is where I derived my delta expressions from. But what is the right approach then? $$\frac{v_{n}^{3} + v_{n + 1}^{3}}{2}$$? What is the correct equation for the average power in this situation?

14. Jul 6, 2010

hikaru1221

Sorry, that's not right
Just as K^2 said, $$\Delta s/\Delta t \neq \Delta v$$, even if you mean to deal with the average values. It should be: $$\Delta s/\Delta t = <v> = v_{average}$$ When everything goes to the limit, i.e. $$\Delta t \rightarrow 0$$, you will have $$\Delta$$ becomes d, and $$\Delta s/\Delta t \rightarrow ds/dt$$ obviously different from dv or $$\Delta v$$, right?

But there is another flaw: $$P_{drag,average}$$ doesn't equal to $$\frac{1}{6} C_{d} \times \rho \times A \times \Delta \nu^{3}$$. I'll show you why.

I'm not familiar with using WolframAlpha, but I guess that there must be something wrong in the input so that WolframAlpha misunderstood it.

Here is the exact integral you must compute:
$$<P>=\frac{\Delta W}{\Delta t}=\frac{1}{t_2-t_1}\int_{t_1} ^{t_2} Kv^3dt$$

where: $$K=\frac{1}{2}C_d\rho A$$

As you can see, to compute the integral, you must know v as a function of t. To derive it, you should come back to the Newton's 2nd law:
$$mg-F_{drag}=mdv/dt$$.

You may foresee that the result will definitely be very crazy, and I'm not sure whether we can compute that crazy integral So I think it's just not so simple as the result you got from WolframAlpha.

Is this problem asking you to derive an exact formula of the average power, or are you conducting some experiment where the need for average power arises?

Last edited: Jul 6, 2010
15. Jul 6, 2010

TimeHorse

Thank you Hikaru and $$K^{2}$$ for trying to help!

So what I'm trying to do is take the EPA test suit known as USSD or LA4 and get a more accurate estimate of the power used by a car in that test suit. Or, more specifically, to find the average power for each interval and multiply back by the period length to get the total Energy used.

http://www.epa.gov/nvfel/methods/uddscol.txt [Broken]

I want to do this to more accurately model the power / energy / range of an EV based on its calculated range extrapolated from these test results.

It seems to me that most components are fairly straight-forward: acceleration using changes in Kinetic Energy, slope through changes in Potential Energy, Rolling Resistance, which is proportional to the Normal Force and velocity in a linear function and of course Drag which is a vector addition of the car speed and the direction and strength of the wind.

Fortunately, AFAICT, the changes in Potential Energy and in Wind Speed are 0 (there are no hills and no wind) in these EPA tests so those components drop off in the calculation. Obviously, there is some constant power component to run the car console and perhaps the head lights. There is some linear term in the rolling resistance and a square term in the changes in velocity when accelerating. Finally, as I do understand this, there is a cubic value which is related to the drag experienced. Thus, I am fairly confident that for the most part a car can be modeled by a cubic power function. There may be other higher-order polynomial terms, especially at lower velocities, but I'd expect this to be the over-all case.

Again, I'm not concerned with the Kinetic or linear terms since a linear curve can be very well estimated by the average of the velocity and since the sample rate is a constant, it's typically good enough to just use the sampled velocity in those equations rather than:

$$\nu = \frac{\nu_{n} + \nu_{n + 1}}{2}$$

Since for the most part, $$\frac{\nu_{n}}{2}$$ will be used in its sample and the previous one, which would have the net effect of $$\nu_{n}$$ (except for the border conditions).

It's the $$\nu^{3}$$ term that causes me concern. Certainly $$\left(\nu_{n} + \nu_{n+1}\right)^{3} \neq \nu_{n}^{3} + \nu_{n+1}^{3}$$, and the same goes for the difference, although at least with cubics the sign is preserved .

So, let me consider this discrete problem given your very very helpful advice.

$$\frac{1}{t_{n + 1} - t_{n}}\int_{t_{n}} ^{t_{n + 1}} K\nu(t)^{3}dt$$

If $$\nu$$ is assumed to be linear from $$\nu_{n}$$ to $$\nu_{n + 1}$$, then the function $$\nu(t) = \nu_{n} + a_{n} t$$ where $$a_{n} = \frac{\nu_{n + 1} - \nu_{n}}{t_{n + 1} - t_{n}} = \left(\nu_{n + 1} - \nu_{n}\right)\omega$$. Going back to the original integral:

$$\omega\int_{0} ^{\frac{1}{\omega}} K\left(\nu_{n}^{3} + 3\nu_{n}^{2}a_{n}t + 3\nu_{n}a_{n}^{2}t^{2} + a_{n}^{3}t^{3}\right)dt$$

Here, I've replaced the definite integral range with the time for one cycle since the calculation for each sample should look the same independent of it's position in the cycle. This is of course much easier to solve:

$$\omega K\left(\nu_{n}^{3}t + \frac{3}{2}\nu_{n}^{2}a_{n}t^{2} + \nu_{n}a_{n}^{2}t^{3} + \frac{1}{4}a_{n}^{3}t^{4}\right)\right|_{0}^{\frac{1}{\omega}}$$

$$= \omega K\left(\frac{\nu_{n}^{3}}{\omega} + \frac{\frac{3}{2}\nu_{n}^{2}a_{n}}{\omega^{2}} + \frac{\nu_{n}a_{n}^{2}}{\omega^{3}} + \frac{\frac{1}{4}a_{n}^{3}}{\omega^{4}}\right)$$

$$= K\left(\nu_{n}^{3} + \frac{\frac{3}{2}\nu_{n}^{2}a_{n}}{\omega} + \frac{\nu_{n}a_{n}^{2}}{\omega^{2}} + \frac{\frac{1}{4}a_{n}^{3}}{\omega^{3}}\right)$$

And replacing $$a_{n}$$:

$$= K\left(\nu_{n}^{3} + \frac{3}{2}\nu_{n}^{2}\left(\nu_{n + 1} - \nu_{n}\right) + \nu_{n}\left(\nu_{n + 1} - \nu_{n}\right)^{2} + \frac{1}{4}\left(\nu_{n + 1} - \nu_{n}\right)^{3}\right)$$

And collecting like terms:

$$= K\left(-\frac{1}{2}\nu_{n}^{3} + \frac{3}{2}\nu_{n}^{2}\nu_{n + 1} + \nu_{n}\left(\nu_{n + 1}^{2} - 2\nu_{n + 1}\nu_{n} + \nu_{n}^{2}\right) + \frac{1}{4}\left(\nu_{n + 1}^{3} - 3\nu_{n + 1}^{2}\nu_{n} + 3\nu_{n + 1}\nu_{n}^{2} - \nu_{n}^{3}\right)\right)$$

$$= K\left(-\frac{1}{2}\nu_{n}^{3} + \frac{3}{2}\nu_{n}^{2}\nu_{n + 1} + \nu_{n + 1}^{2}\nu_{n} + -2\nu_{n + 1}\nu_{n}^{2} + \nu_{n}^{3} + \frac{1}{4}\nu_{n + 1}^{3} + -\frac{3}{4}\nu_{n + 1}^{2}\nu_{n} + \frac{3}{4}\nu_{n + 1}\nu_{n}^{2} + -\frac{1}{4}\nu_{n}^{3}\right)$$

Rearranging:

$$= K\left(-\frac{1}{2}\nu_{n}^{3} + -\frac{1}{4}\nu_{n}^{3} + \nu_{n}^{3} + \frac{3}{2}\nu_{n}^{2}\nu_{n + 1} + -2\nu_{n}^{2}\nu_{n + 1} + \frac{3}{4}\nu_{n}^{2}\nu_{n + 1} + \nu_{n}\nu_{n + 1}^{2} + -\frac{3}{4}\nu_{n}\nu_{n + 1}^{2} + \frac{1}{4}\nu_{n + 1}^{3}\right)$$

$$= K\left(\frac{1}{4}\nu_{n}^{3} + \frac{1}{4}\nu_{n}^{2}\nu_{n + 1} + \frac{1}{4}\nu_{n}\nu_{n + 1}^{2} + \frac{1}{4}\nu_{n + 1}^{3}\right)$$

$$= \frac{1}{4}K\left(\nu_{n}^{3} +\nu_{n}^{2}\nu_{n + 1} + \nu_{n}\nu_{n + 1}^{2} + \nu_{n + 1}^{3}\right)$$

And doing some polynomial arithmetic:

$$= \frac{1}{4}K\left(\nu_{n} +\nu_{n + 1}\right)\left(\nu_{n}^{2} + \nu_{n + 1}^{2}\right)$$

Which is just the average velocity times the average square of the velocity:

$$= K\frac{\nu_{n} +\nu_{n + 1}}{2}\frac{\nu_{n}^{2} + \nu_{n + 1}^{2}}{2}$$

And of course, this reduces to:

$$= K\nu_{n}^{3}$$

when $$\nu_{n} = \nu_{n + 1}$$, which is as we'd expect.

Now am I on the right track?

(Now to check my work so make sure there are no TeX mistakes...)

Last edited by a moderator: May 4, 2017
16. Jul 6, 2010

TimeHorse

And of course, this means more generally:

And now it all seems to make sense! Can you very smart Physicists confirm my conclusion? I appreciate y'all letting me know if I finally got it right! :)

Last edited: Jul 6, 2010
17. Jul 7, 2010

hikaru1221

I don't understand much about the experiment, as I cannot get access to that site, so no comment on the experiment. The reason I asked you whether you were conducting an experiment or not is to find a way to do some approximations here from the data. But now, you assume the speed is linear from v_n to v_n+1, so I assume that you have checked its accuracy.

I haven't checked your work yet, but the way you carefully typed it ensures its correctness. It's just math, right? So good luck with your experiment

By the way, I saw your reply to The Laws of Nature, but his/her post was deleted. I have some comment on his/her line:
Consider mass m in 1-D motion: x=kt^4.
_ Speed: v=4kt^3.
_ Average speed from 0 to T: <v> = delta(x)/T = kT^3.
_ Average force from 0 to T: <F> = delta(p)/T = m.delta(v)/T = 4mkT^2.
_ Work done from 0 to T: W= delta(E_kinetic) = 8mk^2.T^6. So average power: <P> = W/T = 8mk^2.T^5.
Is <P> equal to <F>.<v>? Only in some particular cases does that statement hold its validity.

@The Laws of Nature: I'm sorry about commenting on the post you deleted. No offense though. Just to make it clear to the OP.

18. Jul 7, 2010

TimeHorse

It's a very simple text based spreadsheet, time in seconds in column A, velocity in mph in column B. This is given 1. Given 2, we know the car evaluated under this test procedure is estimated to have a total range of R miles. From this information, calculate this car's power function.

Obviously, you can't compute the Power function with just that information, but if you know a bit about the car: it's mass, it's Drag Coefficient, it's Coefficient of Rolling Resistance, its Frontal Area, the amount of energy recaptured by Regenerative Breaking, the Mass Density of the Atmosphere during the test and the average power used by the console, head lights, A/C unit and Heater, you can get a pretty good idea of what the car can do under other estimated conditions.

And if you don't know all of these other constants about the car, you could estimate them with enough similar test results to narrow down the variables.

As this car is not yet available to consumers and as an EV it's range is critical to the consumer, this is the only way we currently have to estimate the car's abilities.

Well, the math came out to a very simple and intuitive result, which is satisfying, so either it's right or something is grossly wrong!

Sorry about that. I was just trying to re-iterate what people were telling me that I was having so much trouble understanding. But I agree, that statement only holds true if acceleration is constant. The thing with this result is we don't know what the car is doing between samples. It could be 2 mph at 50 seconds, 150 mph at 50.5 seconds and 4 mph at 51 seconds. We don't have samples in the EPA LA4 test at finer granularity than 1 Hz. Since we know velocity is not constant, and hitting 150 mph between seconds 50 and 51 is highly unlikely, the best estimate for what is going on is that the acceleration from 2 to 4 mph is linear from seconds 50 to 51. It probably isn't purely as such, but it seems a very good approximation, and certainly better than taking the cube of each sample to get the Drag Power. I suppose you could fit the 1371 or so samples to a 1371-degree polynomial as well to get an even better fit, though since some of the test points are 0 velocity it doesn't seem a polynomial fit would be quite appropriate. In the end, approximating discrete linear acceleration seems the right approach.

Thanks again for your help hikaru and sorry again about the confusion!

19. Jul 7, 2010

hikaru1221

As for linear changes, you can always expect nice, simple results
Just some thoughts on your approximation: Your reasoning is good, as it's unlikely to make such great jump from 2mph to 150mph in just 0.5s. For a normal, popular vehicle, you may expect that during a short period (e.g 1s), the acceleration is constant, because a great change usually implies a particular phenomenon (e.g. car crash!). However when the sampling rate is low (e.g. 5s), it might be erroneous to say it's linear between 2mph and 4mph, but it's still okay to say so between 50mph to 52mph, as the difference between the speeds in that case << the speeds particularly.
Anyway, your problem is solved. So good luck

20. Jul 8, 2010

TimeHorse

Not to belabor a dead (time) horse here but I was curious what would arise if we instead assumed the car was traveling quadratically between samples. I'm not asking for help this time, just doing a thought experiment that I though I'd share with the group. This should also illustrate hikaru point about the average power not always equal to the average force times the average velocity.

Assume this time velocity is given in terms of 3 samples, $$\nu_{n}$$, $$\nu_{n + 1}$$, and $$\nu_{n - 1}$$.

Thus:

$$\nu(t) = \left(\frac{\nu_{n - 1} + \nu_{n + 1}}{2} - \nu_{n}\right) \omega^{2} t^{2} + \frac{\nu_{n + 1} - \nu_{n - 1}}{2} \omega t + \nu_{n}$$

Thus, Power is given by:

$$\omega K \int^{\frac{+1}{2 \omega}}_{\frac{-1}{2 \omega}} \nu(t)^{3} dt$$

I decided to cheat this time and use sage:

$$\frac{1}{17920} K \left(14072 \nu_n^3 + 1628 \nu_n^2 \nu_{n + 1} + 1628 \nu_n^2 \nu_{n - 1} + 1090 \nu_n \nu_{n + 1}^2 - 1628 \nu_n \nu_{n + 1} \nu_{n - 1} + 1090 \nu_n \nu_{n - 1}^2 + 89 \nu_{n + 1}^3 - 69 \nu_{n + 1}^2 \nu_{n - 1} - 69 \nu_{n + 1} \nu_{n - 1}^2 + 89 \nu_{n - 1}^3 \right)$$

------------------------------

Ignore this; the manual integration of $$\nu(t)^{2}$$.

$$\omega K \int^{\frac{+1}{2 \omega}}_{\frac{-1}{2 \omega}} \left(\frac{\nu_{n - 1} + \nu_{n + 1}}{2} - \nu_{n}\right)^{2} \omega^{4} t^{4} + \left(\frac{\nu_{n - 1} + \nu_{n + 1}}{2} - \nu_{n}\right) \left(\nu_{n + 1} - \nu_{n - 1}\right) \omega^{3} t^{3} + \left(2 \nu_{n} \left(\frac{\nu_{n - 1} + \nu_{n + 1}}{2} - \nu_{n}\right) + \left(\frac{\nu_{n + 1} - \nu_{n - 1}}{2}\right)^{2}\right) \omega^{2} t^{2} + \nu_{n} \left(\nu_{n + 1} - \nu_{n - 1}\right) \omega t + \nu_{n}^{2} dt$$

$$\omega K \left( \frac{1}{5} \left(\frac{\nu_{n - 1} + \nu_{n + 1}}{2} - \nu_{n}\right)^{2} \omega^{4} t^{5} + \frac{1}{4} \left(\frac{\nu_{n - 1} + \nu_{n + 1}}{2} - \nu_{n}\right) \left(\nu_{n + 1} - \nu_{n - 1}\right) \omega^{3} t^{4} + \frac{1}{3} \left(2 \nu_{n} \left(\frac{\nu_{n - 1} + \nu_{n + 1}}{2} - \nu_{n}\right) + \left(\frac{\nu_{n + 1} - \nu_{n - 1}}{2}\right)^{2}\right) \omega^{2} t^{3} + \frac{1}{2} \nu_{n} \left(\nu_{n + 1} - \nu_{n - 1}\right) \omega t^{2} + \nu_{n}^{2} t \right) \right|^{\frac{+1}{2 \omega}}_{\frac{-1}{2 \omega}}$$

$$\omega K \left( \frac{1}{5} \left(\frac{\nu_{n - 1} + \nu_{n + 1}}{2} - \nu_{n}\right)^{2} \omega^{4} \left(\frac{1}{32 \omega^{5}} - \frac{-1}{32 \omega^{5}}\right) + \frac{1}{4} \left(\frac{\nu_{n - 1} + \nu_{n + 1}}{2} - \nu_{n}\right) \left(\nu_{n + 1} - \nu_{n - 1}\right) \omega^{3} \left(\frac{1}{16 \omega^{4}} - \frac{1}{16 \omega^{4}}\right) + \frac{1}{3} \left(2 \nu_{n} \left(\frac{\nu_{n - 1} + \nu_{n + 1}}{2} - \nu_{n}\right) + \left(\frac{\nu_{n + 1} - \nu_{n - 1}}{2}\right)^{2}\right) \omega^{2} \left(\frac{1}{8 \omega^{3}} - \frac{-1}{8 \omega^{3}}\right) + \frac{1}{2} \nu_{n} \left(\nu_{n + 1} - \nu_{n - 1}\right) \omega \left(\frac{1}{4 \omega^{2}} - \frac{1}{4 \omega^{2}}\right) + \nu_{n}^{2} \left(\frac{1}{2 \omega} - \frac{-1}{2 \omega}\right) \right)$$

$$\omega K \left( \frac{1}{5} \left(\frac{\nu_{n - 1} + \nu_{n + 1}}{2} - \nu_{n}\right)^{2} \frac{1}{16 \omega} + \frac{1}{4} \left(\frac{\nu_{n - 1} + \nu_{n + 1}}{2} - \nu_{n}\right) \left(\nu_{n + 1} - \nu_{n - 1}\right) \left(0\right) + \frac{1}{3} \left(2 \nu_{n} \left(\frac{\nu_{n - 1} + \nu_{n + 1}}{2} - \nu_{n}\right) + \left(\frac{\nu_{n + 1} - \nu_{n - 1}}{2}\right)^{2}\right) \frac{1}{4 \omega} + \frac{1}{2} \nu_{n} \left(\nu_{n + 1} - \nu_{n - 1}\right) \left(0\right) + \nu_{n}^{2} \frac{1}{\omega}\right)$$

Dropping 0 Terms:

$$\omega K \left( \frac{1}{5} \left(\frac{\nu_{n - 1} + \nu_{n + 1}}{2} - \nu_{n}\right)^{2} \frac{1}{16 \omega} + \frac{1}{3} \left(2 \nu_{n} \left(\frac{\nu_{n - 1} + \nu_{n + 1}}{2} - \nu_{n}\right) + \left(\frac{\nu_{n + 1} - \nu_{n - 1}}{2}\right)^{2}\right) \frac{1}{4 \omega} + \nu_{n}^{2} \frac{1}{\omega}\right)$$

$$K \left( \frac{1}{80} \left(\frac{\nu_{n - 1} + \nu_{n + 1}}{2} - \nu_{n}\right)^{2} + \frac{1}{12} \left(2 \nu_{n} \left(\frac{\nu_{n - 1} + \nu_{n + 1}}{2} - \nu_{n}\right) + \left(\frac{\nu_{n + 1} - \nu_{n - 1}}{2}\right)^{2}\right) + \nu_{n}^{2} \right)$$

$$\frac{K}{960} \left( 3 \left(\nu_{n - 1} + \nu_{n + 1} - 2 \nu_{n}\right)^{2} + 80 \nu_{n} \left(\nu_{n - 1} + \nu_{n + 1} - 2 \nu_{n}\right) + 20 \left(\nu_{n + 1} - \nu_{n - 1}\right)^{2} + 960 \nu_{n}^{2} \right)$$

$$\frac{K}{960} \left( 3 \left(\nu_{n - 1}^{2} + \nu_{n + 1}^{2} + 4 \nu_{n}^{2} + 2 \nu_{n - 1} \nu_{n + 1} - 4 \nu_{n} \nu_{n + 1} - 4 \nu_{n} \nu_{n - 1}\right) + 80 \nu_{n} \nu_{n - 1} + 80 \nu_{n} \nu_{n + 1} - 160 \nu_{n}^{2} + 20 \left(\nu_{n + 1}^{2} + \nu_{n - 1}^{2} - 2 \nu_{n + 1} \nu_{n - 1}\right) + 960 \nu_{n}^{2} \right)$$

$$\frac{K}{960} \left( 3 \nu_{n - 1}^{2} + 3 \nu_{n + 1}^{2} + 12 \nu_{n}^{2} + 6 \nu_{n - 1} \nu_{n + 1} - 12 \nu_{n} \nu_{n + 1} - 12 \nu_{n} \nu_{n - 1} + 80 \nu_{n} \nu_{n - 1} + 80 \nu_{n} \nu_{n + 1} - 160 \nu_{n}^{2} + 20 \nu_{n + 1}^{2} + 20 \nu_{n - 1}^{2} - 40 \nu_{n + 1} \nu_{n - 1}+ 960 \nu_{n}^{2} \right)$$

$$\frac{K}{960} \left( 23 \nu_{n - 1}^{2} + 23 \nu_{n + 1}^{2} + 812 \nu_{n}^{2} - 34 \nu_{n - 1} \nu_{n + 1} + 68 \nu_{n} \nu_{n + 1} + 68 \nu_{n} \nu_{n - 1}\right)$$

Since 23 is a prime number, I don't think this can further be reduced, but if $$\nu_{n - 1} = \nu_{n} = \nu_{n + 1}$$, the expression reduces to $$K \nu_{n}^{2}$$, which is not right, it should be cubed, so there is a flaw in this. BTW, if it was being squared, the math does match my calculations in Sage.

Last edited: Jul 9, 2010
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