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## #1 2005-10-25 04:37:28

John E. Franklin
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Registered: 2005-08-29
Posts: 3,562

What is the angle in the center of a sphere making 1.000 steradians?

If the sphere has a radius of one meter.
Then you draw a circle on the sphere, and inside this
circle, the area is one meter². (But this area is not flat, it
is curved on the sphere.)

I don't know how to solve this angle.    I guess it is somewhere
between 40 and 150 degrees.    The area of the sphere is
4 pi times that of the steradian's circle.

igloo myrtilles fourmis

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## #2 2005-10-25 05:27:05

mathsyperson
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Registered: 2005-06-22
Posts: 4,900

Wow. Tough question. I can't think of how to go about solving that, but if anyone else wants to have a go, I've got a nugget of knowledge that might help.

If you have a sphere and cut a bit off it, then if the radius of the chopped bit is r and the height of the chopped bit is h, then the volume of the chopped bit is pi*h*(3r²+h²)/6. I haven't a clue whether or not that helps.

Why did the vector cross the road?
It wanted to be normal.

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## #3 2005-10-25 09:21:07

MathsIsFun
Registered: 2005-01-21
Posts: 7,535

Would the cone angle be 1 radian (180/pi degrees) ?

Not answered, but nice illustration here: http://www.squ1.com/light/quantities.html

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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## #4 2005-10-27 06:39:57

John E. Franklin
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Registered: 2005-08-29
Posts: 3,562

I'm working on an approach to this problem by first working out the area of a sphere by using slices of cones.  If I get that to work out to 4 pi r² for the surface area of a sphere, then I'll use the same procedure to go just part of the way forming the steradian.  Wish me luck.  I'll let you know how it goes.  So far, I intend on doing a computer approximation since I am not good at Calculus, and would need help even getting the calculus equation.

igloo myrtilles fourmis

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## #5 2005-10-27 15:54:04

ganesh
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Registered: 2005-06-28
Posts: 14,542

A unit of measure equal to the solid angle subtended at the center of a sphere by an area on the surface of the sphere that is equal to the radius squared: The total solid angle of a sphere is 4π (4 pi)steradians.

The unit of solid angle adopted under the System International d'Unites.

Character is who you are when no one is looking.

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## #6 2005-12-09 08:33:24

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,562

Irspow, could I learn your surface of rotation stuff for this steradian calculation??
That would be really great!

igloo myrtilles fourmis

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## #7 2005-12-09 10:26:39

irspow
Member
Registered: 2005-11-24
Posts: 455

I don't know if a surface of revolution will help in this problem as it calculates a volume.  Anyway it works like this.

Make a function that models the outer edge of an object that is three dimentional for a cartesian coordinate system.  After you have the function it is simply a matter of integration.

If y=f(x) the Volume would be ∫2pixf(x) dx] evaluated from 0 to the radius.  (for a solid object)  Perhaps an example is in order.

Take a cone.  If you look at the right half of it the side will match a decreasing line with the y intercept being the apex.

The equation for that line will be y = H - (H/R)x, this is true for all cones.  This line intercepts the x axis at x = R.

Now multiply your line function by (2 pi x) and then integrate from zero to R.  You will get:

2pix(H - (H/R)x) = 2piH ∫x - x^2 / R dx  (remember to factor out constants?)

2piH [ x^2 / 2 - x^3 / 3R ] from 0 to R

2piH (R^2 / 2 - R^3 / 3R)

2piH ( 3R^2 - 2R^2 ) / 6

This all equals piHR^2 / 3

This is the common formula for the area of a cone.  This is the same way that the volume of a sphere was calculated using the curve of one quadrant and multiplying the integration by two to add the bottom half.  You can use this method for any object with a little imagination.

Note that we used y = f(x) because we were rotating around the y axis.  If you rotate around the x axis you will need a function like x = f(y).

You can also calculate surface area if at the end of your computation you divide by the thickness or radius of the object in question.  If you want a volume of an object that is not solid you simply subtract the integration of the inner boundary from that of the outer boundary.

I hope this helps and would love if you post any uncommon volumes that you figure out in the future.

edit:  I am sure that you understand that an integration can be interpreted as the area below the curve or line.  What we have done above is that we have integrated that area for every value of the radius also.  That is why we multiply the function by 2pix (circumference formula) or 360 degrees.

Last edited by irspow (2005-12-09 10:33:26)

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## #8 2005-12-09 12:01:29

irspow
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Registered: 2005-11-24
Posts: 455

Now I have never heard of these steradians before, but after a little research it appears that:

sr = A/4pi

And more importantly:  4pi sr = 129600/pi degrees

So in our case 1 sr = 32400/pi^2 degrees ≈3282.8

These are the relationships from the experts, although I am still having difficulty grasping the jump from regular units to square units.

The reason that the number is so large is because steradians are literally square radians.

So they say that a sphere contains 4pi sr (square radians) = 4pi rad^2

Since each rad = 180/pi degrees:

4pi(180/pi)^2

The only logic that I can grasp at here is that this number represents the interior angle of the apex at the center of the sphere multiplied by the degrees of revolution.

If my logic holds then the angle at the apex would be 3282.8/360 = 9.1189..revolved 360.

I'll see if this makes sense in a while.

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## #9 2005-12-09 16:12:20

MathsIsFun
Registered: 2005-01-21
Posts: 7,535

(Put a ":" in front of "pi" and you get π, see http://www.mathsisfun.com/forum/viewtop … 623#p17623)

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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## #10 2005-12-09 17:30:45

irspow
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Registered: 2005-11-24
Posts: 455

Thanks MathIsFun, I learn something new everyday here.

Now I can write 32400 ÷ π² instead of 32400/pi^2

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## #11 2005-12-10 13:19:00

irspow
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Registered: 2005-11-24
Posts: 455

John, I was trying to verify my suggestion that the interior angle was indeed correct at 90/π²°.  This was because if you multiply this by 360° you will get the amount of square degrees in a steradian.

As it turns out I am having a very difficult time integrating the formula to prove it.  I have to integrate ∫2πf(y)√1+(dx/dy)²  dy.

I tried integration by parts and trigonomic substitution to no avail.

The integration is the general form for a surface of revolution about the y axis.  I think that that is what you need for this proof.

f(y) = √1-y²  is what needs to be applied to the formula above.  Without going much further at this point the upper and lower bounds are no problem here but do depend on how the integration is solved.

So if anyone can solve the integration above for the general case, we could all finally put an end to the mystery of how exactly square degrees are related to steradians within three dimensional space.  Because once the integration above is completed and evaluated over the correct range it should simply equal 1.

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## #12 2005-12-11 11:19:08

John E. Franklin
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Registered: 2005-08-29
Posts: 3,562

irspow wrote:

∫2πf(y)√1+(dx/dy)²  dy.

Tell me if this is the same thing?  This is way over my head unless I learn a lot.

which by the way was created with
[ m a t h ] \int  2\pi f(y) \sqrt{1 + (dx/dy)^2} dy  [/ m a t h ]

irspow wrote:

The equation for that line will be y = H - (H/R)x, this is true for all cones.  This line intercepts the x axis at x = R.

Now multiply your line function by (2 pi x) and then integrate from zero to R.  You will get:

2pix(H - (H/R)x) = 2piH ∫x - x^2 / R dx  (remember to factor out constants?)

I finally figured out that you are making a series of hollow pipes that get
shorter and shorter as you approach R.
Thanks for teaching me so well!!!!!!
I am very grateful to everyone for spending so much time helping me.

Last edited by John E. Franklin (2005-12-11 12:07:42)

igloo myrtilles fourmis

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## #13 2005-12-11 12:24:42

irspow
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Registered: 2005-11-24
Posts: 455

Yes John, that is precisely what I was trying to write.  As far as my efforts to figure this thing out this is what I have come up with.

The problem that I was having with the integration was silly after realizing that I was making substitutions too early complicating the matter more than was necessary.

f(y) = √1-y²

When I first saw this I immediately start using the trigonomic substitution which made this completely unworkable, at least for me.

Leaving f(y) alone until I had the integrand simplified as much as possible was the solution for me.  Sadly it took a lot of crumpled pieces of paper and a lot of time pouring over the integration tables to figure this out.

I solved it easily after that point.  Just setting u = y². (smacking myself!)  Then I was reproached as the surface area for the angle that I proposed earlier was only .019883871 units²!

So then I solved the integral backwards to find the true angle within the sphere at the apex.  Basically I just solved for when the integral equalled one. (That was what was stated in your original post)

This solution is only half of the interior angle because of the way the surface formula works.  The angle that solved the integral was about 32.77°.  So the whole angle would be about 65.541°.  It turns out my theory of 360° multiplied by the interior angle was further off than your original guess from the beginning.

I could not find a relationship between this interior angle and the square degrees, and I tried very hard to believe me.  That 65.541° is reliable though because that integral is used to calculate just the type of thing that we are doing here.

By the way you should note that if you remove the 2πf(y) from the integrand it will now solve for the length of the arc!  You can use either f(x) or f(y) depending on which is more convenient for the length of arc.

I hope all of this jabbering that I've done on this thread has helped you somewhat, I know that I have learned a thing or two since I first ventured into this topic.

Oh!  Is there somewhere that I can learn about that [math] syntax?

Last edited by irspow (2005-12-11 12:25:29)

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## #14 2005-12-11 15:39:38

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,562

I can't believe the progress you made!!!!!!
That's amazing.
(Google Latex math for graphical math symbols, notice _ for subscript and ^ for superscript, \frac{numerator}{denominator}, etc.  I just learned some today.
You've helped me a lot, and I also have been trying to do what you have done,
except I only got as far as the following, except I had the x and y axis
backwards from you:
Curved surface area of steradian = 1 =

Note that I am missing the 2 for 2
.
This is because I used the formula for the surface area of a cone and you
divide by two.
Cone Area Piece=dSlant 1/2 2 π y   (I'm rotating around the x-axis, so y is R base of cone)
y' is y prime or derivative in above; I'm very sloppy still; I need reform.
Where

and
, which I got
from a table, and then I gave up and read your post.
I must have made a mistake on the 2, but I thought I'd let you know just
in case.  I tried figuring out the rotation myself and must have failed by a
factor of two somehow.
...
I just examined what I did, and it turns out I was lucky to get as close as I
did because I should have used slices of cones, not the whole cone.
And slices of the cones would involve something I can't figure out yet.
It would be a slice of a cone at the base, incrementally thin.  Somehow the
1/2 goes away, don't ask me how.  But the surface area of a cone can be
shown "unfolded" as a bunch of triangles, and the area of a triangle has the
1/2 in it, that's a simple explanation I saw a few weeks ago.
...
But you are way ahead of me.
How come your formula didn't get really messy with the derivative
of the circle??  Your substition?  I'll have to try to learn all this when I can.
Bye.

Last edited by John E. Franklin (2005-12-11 15:59:21)

igloo myrtilles fourmis

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## #15 2005-12-11 16:18:30

irspow
Member
Registered: 2005-11-24
Posts: 455

I saw that you posted in another thread about the arc length integral, very good. (useful too) Try to remember that one it comes in handy ofter.  I actually solved the integral that I was having so much trouble with.

As funny as it sounds, in the end it was nothing more than 2π (√u) evaluated from cos²0° to cos²32.77053659° which provided the answer of 1 square unit.  The u came from a substitution for y² after much simplification.

Anyway, I still haven't found a relationship between this interior angle (65.54107318° which is twice the half angle used for the surface of revolution) and the 3282.80635°² in one steradian.  I think it is because there is no relationship to be found.  By definition a 4π steradians = 4πr².

Sorry that is the best that I can do on this one for now.

Last edited by irspow (2005-12-11 16:21:05)

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