Recent content by BvU

Coming back to and when you report Where I'm afraid I have to remark that this doesn't satisfy the 'at least 6-8 correct digits' :rolleyes: ... (and the ##\mathcal O## is also on edge -- I personally would take 0.5E-7 or even 0.5E-8 ... - - - - When I finally got something to work (see...

I don't follow. Does this explain how you calculate the error ? When I compare ##y## after a step of size ##h## (your 5e-4) with ##y## after two steps ##h/2##, I get close to machine precision ! ##\ ##

I think you are quoting the OP, not Filip ...

And the distance vs time plot is ...?

No. You got ψ1(a) = 0 and ψ2(a) = ## (a-d)^2-c ##. Equating ψ1 and ψ2 at x=a means you have ψ1(a) = ψ2(a) Then: what about continuity of the first derivative ? ##\ ##

Is this a new exercise or still about the bouncing ball ?

Beg to differ. With step size 5e-4 (as in post #18) we get a first ##y<0## at step 532 with ##y=##-3.16e-4. ##\dot y\approx## -2.1 so the Newton step size is -1.5e-4 . Such a step (number 533) approximately doubles the amount of significant digits and takes us to ##y=##-7e-8. I had...

:welcome: ! Did you do a little research ? Google 'parabolic optical reflector design calculator'and NASA has a nice webpage. You of course don't want a dish but a kind of parabolic gutter. Formulas in one dimension stay the same. And: make a sketch ! Do you want the beam to be parallel...

Does your textbook / do your lecture notes have anything about steam turbines ? ##\ ##

I'm having sooooo much fun. (*) -0.353496957103320 ##\qquad## 0.433854331752568 ##\qquad## 0.913422978195822 :biggrin: Step sizes ten times bigger (5e-4) -0.353496957103362 ##\qquad## 0.433854331752487 ##\qquad## 0.913422978195706 Another...

:welcome: ##\qquad## ! Great ! I didn't have any formal NA courses but picked up quite a bit while going along. That assignment itself probably isn't a secret. Cold you post the full text verbatim ? What is fixed (given), what is required, what is optional, etc. Well, with a step size of...

Coming back to that global error. Here is an RK4.F90 including a simple test: ##\ddot y = -y, \ y(0)=1, \dot y = 0## from x=0 to 10. The answer is a cosine. Ran it with 50 steps, 100, 200, 400. Error ##\equiv y(x)-\cos(x)## plotted. Clearly 50 and 100 steps is not good enough (although all...

Further comments: The error estimate is for the local error. For the global error you have to factor in the number of steps. I still wonder about your algorithm to locate ##y=0## points -- in connection with the notion that your real concern is the global error. I wonder if it isn't equally...

Here we have 1 degree of longitude = 288200 ft. That's at 38 deg latitude. So for latitude 47 you have ##\cos 47/\cos 38## times 288200 feet per degree of longitude. Someone should check this :wink: And don't forget to set the calculator to 'degrees' ! ##\ ##

:smile: that would be latitude 47 or thereabouts. Latitude matters to convert longitude differences to furlongs! ##\ ##