Trig equations

  • Thread starter philipc
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Main Question or Discussion Point

I'm looking for a long list of trig equations.
Mainly as of right now I'm looking for the equation to find the surface area of a sector of a sphere?
Any help would be great,
Thanks
Philip
 

Answers and Replies

  • #2
s = r*θ ; where θ is the angle of sector s.
 
  • #3
NateTG
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IIRC area of a sphere is 4πr2.

One way to calculate the area of a sector is to determine the fraction of the surface of the sphere that it is.

There are alternatives, but they either involve shape-specific forumlas, or calculus.
 
  • #4
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Ever tried measuring angles on a sphere? You need solid angles for spherical surfaces. The formula you gave is for circles.

The area of the sector depends on how you segment the surface. A few are given here

http://mathforum.org/dr.math/faq/formulas/faq.sphere.html

The one I think you were after is 'sector of a sphere' which is just a simple extension of the circular case.
 
  • #5
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I need to calculate how much light will hit an aperture from the stars above. The area from the aperture will in a since form a region on the outer sphere.
Calculus I can handle, but not sure how to set up the problem.
Thanks
Philip
 
  • #6
Area of a sphere:

(4 * pi * r^2) / θ

where θ is in radians.
 
  • #7
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Thanks everyone for your help.

Lonewolf,
Due to the fact I nearly made my second post at the exact time you made yours, I didn't see it. But the link you gave me was great thanks. You got me going on the right track with the solid angle, I later found to be called a "steradian", sorry it's been over 10 years since I've had high school Geometry, so I barely know left from right :)
So I've found Area to be Omega = A/r^2, and this helps solve my physics problem.
Again thanks everyone for your help.
Philip
 
  • #8
I hope you really found the answer,... or are you fooling yourself?

[tex]\omega[/tex] usually represents angular velocity in physics.

If you meant to look for angular displacement, which can be considered area of a sector of a circle if u integrated

[tex]\theta = l/r[/tex]
[tex]l = \theta * r[/tex]
 
  • #9
Hurkyl
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Solid angles are often represented by Ω.

Anyways, don't feel bad you didn't know much about this; I don't think I've encountered the term "solid angle" or "steradians" in any of my math courses; they're things I've only picked up on the side.
 
  • #10
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I came across solid angles in stellar astrophysics, incidentally, and later in EM. Wonder if they're of any use to mathematicians?
 
  • #11
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Good, I don't feel so stupid now [?] thought maybe I was just sleeping in class that day or something :)

So I might see this again in EM? I've heard it's a tuff class, how does it compare to Physics II, because it's taking at least 20+ hours a week for me to get through this stuff, and the stories I hear about EM, well I'm sure you've heard them too.
Philip
 
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  • #12
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I'm not sure how the courses compare across the atlantic. We don't 'major' in anything like you guys. We pick the degree we're going to take sometime in high school. When we get to university, there are some modules that are compulsory, while there are options related to the chosen degree.
20+ hours a week for a physics course sounds about right though. Depends what's involved in Physics 2. EM was one of the toughest last year. The good news is, once you get through it, things seem to become easier.
 
  • #13
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More Trig questions,
how does arccot(x) and slope of a ray hitting a parabolic surface go together?


Lonewolf,
My Physics II class is mostly Electricity and Magnetism, but now we are hitting up optics.
It's been very interesting, but a bunch of new stuff for me.

Philip
 
  • #14
Derivative of [tex]Cot^-1(x)[/tex] looks like what you described.
 
  • #15
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Try plotting a parabola against [tex]Cot^-1(x)[/tex] and compare the two graphs. Then think about how rays reflect on parabolic mirrors and such.

A lot of the new stuff you learn will be repeated throughout the degree. They're most likely to be 'breaking you in' during your first few courses. Just keep at it, and you'll be rewarded for it.
 

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