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Re: parabolic trough edge's length
From: williamsdavid65@gmail.com (dow)
On Jul 9, 10:51 pm, Morris Dovey <mrdo...@iedu.com> wrote:
On 7/9/2010 8:50 PM, Steve D wrote:
Are you talking the length along a parabolic curve? If you are, I know
there is an exact math solution to this one. It involves a curve
integral. I'm sure of it since I calculated it when I was in college. Of
course I can't remember how to do it any more, but if you find an
appropriate egghead I'm sure you can get the formula. If I was more
ambitious I'd try to re-fire those old neurons (it's been more than 20
years).
Sigh - me too, except that I took the course 45 years ago, and my old
textbook was ruined in a basement flooding 30 years ago.
It should be a fairly simple definite integral, but I'm down to my last
two neurons - one is for breathing and the other... I forget what the
other one does. :)
Between the time I finished the program and the time I posted here I
visited Amazon and ordered a used copy of my old textbook (Abraham
Schwartz's "Calculus and Analytic Geometry"). Perhaps it'll stick better
the second time around.
My iterative approach is to approximate the curve with straight line
segments and sum their lengths. At each iteration I increase the number
of segments by a factor of ten, and when the difference between
successive approximations is less than or equal to the error threshold,
the program outputs the sum and quits. On this old IBM PC the last
non-zero difference (before running out of bits) was on the order of
10^-13 - which is probably close enough for most DIY work.
I remember using the calculation to make something resembling a
parabolic dish. Knowing the length along the parabolic curve I cut paper
into wedges that varied in width based on the length along the parabolic
curve. I taped the wedges together and got a dish.
I suspect that project would have exceeded my attention span, but I did
write a Z-80 program to control three lead screws to spin dishes from
aluminum disks back in C- and K-band days.
--
Morris Dovey
I think the integral is:
<integral> sqrt(1 + 4Ax^2 +4ABx + B^2) dx
between whatever limits you want. Maybe I could have integrated it
analytically sometime back in the middle of the last century, but
nowadays I'd do like you did, and just program a computer to grind out
a numerical answer.
The paraboloid problem was a bit different. If f is the focal length,
then we can define k = 4f^2. The integral becomes:
<integral> r.sqrt(k+r^2).(k-r^2) dr
where r is the radius of the rim of the dish. The lower limit of the
integral is zero, but the upper limit is what we want to find in order
to make the value of the definite integral zero. Usually, we know the
limits and want to find the value of the integral. In this problem, we
know the value of the integral and one of the limits, and we want to
find the other limit. Writing a program that would would produce an
answer correct to ten significant digits in less than a human lifetime
was a bit of a challenge!
Fun stuff....
dow- Hide quoted text -
- Show quoted text -
I tried writing program to solve your problem. In 13 seconds, it got
an answer correct to only about seven digits. So I guess my 15-year
old 386 is way slower than whatever you're using.
Here's the program I used to solve the paraboloid problem, in QBasic:
DEFDBL A-Z
f = 1 ' focal length
dr = f / 100000 ' step size
d2 = dr / 2
k = 4 * f * f
t = 0
r = d2
DO
r2 = r * r
q = r * (k - r2) * SQR(k + r2)
t = t + q
r = r + dr
LOOP WHILE t > 0
r = r - d2 - dr * t / q
d = r * r / (4 * f)
a = ATN(r / (d - f)) * 45 / ATN(1)
PRINT "Focal length:"; f; "unit(s)"
PRINT "Depth of dish:"; d; "units"
PRINT "Radius of rim:"; r ; "units"
PRINT "Angular radius of rim, seen from focus:"; a; "degrees"
END
With the given values for f and dr, it gets results correct to ten
significant digits, but on my machine takes about ten minutes to do
it How long does it take on your computer?
dow
