Algorithms

Re: Finding a point on a circle [modified]

20-Jul-07 12:36

This one has me frustrated... What I need is a function that will give me the x and y coordinates of a point on a circle (defined by it's x and y coordinates and its width and height) when given a number in degrees. For example...

public Point GetPointOnCircle(int circleX, int circleY, int circleWidth, int circleHeight, int degrees) {
	return new Point(x, y);
}

I've tried to figure it out on my own but it's proven to be a bit more complicated than I thought it would be. Could anyone help me out?

P.S. I'm using C#

-----
*** Never give me an answer having anything to do with Visual Studio. I don't have this because I have two computers, one being my dad's mac, which is connected to the internet, the other being my pc, which is, sadly, not connected to the internet. The setup for the Visual C# program I think is called a "bootstrap" program, and it needs to connect to the internet to install the program. THEREFORE I cannot install this program onto my pc.***

Luc Pattyn20-Jul-07 16:23

20-Jul-07 16:23

Hi Max,

for a real circle with origin (0,0) and radius 1 the coordinates of the point at angle alpha
are given by (cos(alpha), sin(alpha))

Sanity checks: alpha=0 ==> (1,0) alpha=90 ==> (0,1) alpha=180 ==> (-1,0)

multiply by r to enlarge the circle, and add (cx,cy) to translate it.

BTW I expect Math.Sin/Cos to accept radians, not degrees; so you need to convert that
probably based on Math.Pi

for an ellipse it is similar but slightly more complex, ~~dont know by heart~~.

it is (cx+rx*cos(alpha), cy+ry*sin(alpha))
where rx and ry are the horizontal and vertical radii (that's for an orthogonal ellipse,
one that has its major axes parallel with the coordinate system's axes)

Sanity check: x must and will be in range (cx-rx,cx+rx) and y in range (cy-ry,cy+ry)

Smile | :)

-- modified at 0:30 Saturday 21st July, 2007

Luc Pattyn

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max2929720-Jul-07 20:09

20-Jul-07 20:09

Thanks a lot! That's pretty much exactly what I needed.

Luc Pattyn21-Jul-07 0:49

21-Jul-07 0:49

You're welcome.

I am surprised though you asked this, it basically is the definition of cos() and sin(),
hence part of trigonometry 101, isnt it ?

Smile | :)

Luc Pattyn

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Paul Conrad21-Jul-07 5:24

21-Jul-07 5:24

Luc Pattyn wrote:
I am surprised though you asked this, it basically is the definition of cos() and sin(),
hence part of trigonometry 101

I was curious as to why? Maybe he was just a little rusty on the Trig...

"The clue train passed his station without stopping." - John Simmons / outlaw programmer

Luc Pattyn21-Jul-07 6:06

21-Jul-07 6:06

Well, I dont expect any of the kids his age in my neighbourhood to ask this kind of question;
either they are not interested in programming, math and science, so they dont need it
and dont care to ask; or they are interested and hence they remember the definition and
the use of cos() and sin() they learned at school, and they are keen enough to figure
things out by themselves...
In the end, that's what schools are for: provide the basic information and the drive
to investigate and discover, isnt it ?

But then (this does not apply to Max, he asks lots of mostly good questions in diverse areas,
my comment was intended as an extra stimulus for him), I continuously am amazed by the number of
people lacking such drive : they ask basic questions on CP, often using the right keywords,
but "forget" to search the answers themselves; some of them, when they get a reply refering
to CP/MSDN/Google, see it fit to reply with "Can you provide a link ?".

Smile | :)

Luc Pattyn

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Paul Conrad21-Jul-07 6:11

21-Jul-07 6:11

I didn't realize the OP was only 14 years old until I looked at his profile. Trigonometry the Easy Way[^] is a great book I had when I was in high school back in the late '80s. I still have it and it is the second edition. It helped me with my grades.

"The clue train passed his station without stopping." - John Simmons / outlaw programmer

max2929721-Jul-07 16:31

21-Jul-07 16:31

ha! Yeah, I'm kind of going into high school next year so...

Paul Conrad21-Jul-07 16:53

21-Jul-07 16:53

max29297 wrote:
I'm kind of going into high school next year so...

Good luck to you, and any questions about your maths studies are welcome here.

"Any sort of work in VB6 is bound to provide several WTF moments." - Christian Graus

max2929721-Jul-07 17:52

21-Jul-07 17:52

Thanks!

Paul Conrad21-Jul-07 18:01

21-Jul-07 18:01

No problem. I'd look at the Douglas Downing book that I posted about earlier. He has an Algebra one, and Calculus one which is really good.

"Any sort of work in VB6 is bound to provide several WTF moments." - Christian Graus

David Crow26-Jul-07 10:53

David Crow

26-Jul-07 10:53

How do you "kind of" go into high school? Confused | :confused:

"A good athlete is the result of a good and worthy opponent." - David Crow

"To have a respect for ourselves guides our morals; to have deference for others governs our manners." - Laurence Sterne

Color Ranges

Malcolm Smart20-Jul-07 5:29

20-Jul-07 5:29

Hi
I am monitoring CPU performance and disaplying it in a colour ('cos we don't care about the poor sighted where I work!).

0% = Green.
100% = Red.

and anywhere in between should be a mixed shade of these two. I tried setting the mid point as yellow, so:-

RGB

0% 00,255,00
50% 255,255,00
100% 255,00,00

formula for this was

n < 50 : r = n * 5, g = 255 , b = 0;
n >=50 : r = 255 , g = (255-((n-50) * 5)) , b = 0;

near enough, and I could handle the maths, but the results looked ****. The green was going lighter as it headed towards yellow, not darker (redder?). The type of affect i am after is best shown in Microsoft Word, Gradient fill, where you select two colours, the first green, the second red.

(Now if Luc answers this, he might as well have written the whole class for me!!)

Cheers

"More functions should disregard input values and just return 12. It would make life easier." - comment posted on WTF

Luc Pattyn20-Jul-07 5:45

20-Jul-07 5:45

You rang for me, my lord ?

What you want is a gradient in HSB space; take S=100% B=100% and let hue vary from
yourGreen.Hue (probably 33%) to yourRed.Hue (probably 0)

I leave the coding to you tho.

Smile | :)

Luc Pattyn

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Malcolm Smart20-Jul-07 10:37

20-Jul-07 10:37

HSB space is all new to me - I sense another nights googling and reading ahead.

Cheers Luc

"More functions should disregard input values and just return 12. It would make life easier." - comment posted on WTF

Paul Conrad20-Jul-07 10:40

20-Jul-07 10:40

Malcolm,

Would this article help? Colorspace[^], it's right here on CP...

"Any sort of work in VB6 is bound to provide several WTF moments." - Christian Graus

Malcolm Smart20-Jul-07 11:22

20-Jul-07 11:22

It certainly would. I've read it before, some time ago, and as usual, didn't bookmark as I didn't need it at the time. Amazing how a little request from the boss about cpu monitoring has turned into learning about color manipulation.

"More functions should disregard input values and just return 12. It would make life easier." - comment posted on WTF

Paul Conrad20-Jul-07 11:27

20-Jul-07 11:27

Malcolm Smart wrote:
didn't bookmark

I bookmark alot of articles around here. Nice to bookmark but out of fear of losing the bookmarks, I added an entry into my CP blog on my profile page to act as a bookmark to an extent.

"The clue train passed his station without stopping." - John Simmons / outlaw programmer

Luc Pattyn20-Jul-07 15:42

20-Jul-07 15:42

Hi Malcolm,

sorry, forgot to mention HSBtoRGB is not part of .NET;
formulas and code are available in this nice article (my preferred one for june).[^]

Smile | :)

Luc Pattyn

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CPallini20-Jul-07 11:33

CPallini

20-Jul-07 11:33

I think (on empirical grounds) that you can try to use the following formula:

g(n)= (int) ( 255 * cos( n * pi / 200) + .5)
r(n)= (int) ( 255 * sin( n * pi / 200) + .5)

where:

double pi = 3.14159..........
 0 <= n <= 100

The starting point for the above formulas was the fact that at the middle point of the Microsoft Word's gradiend there is no pure yellow, but a darker one.
So I made the following hypothesis (maybe are wrong, maybe that are, neverthless, useful):

(1) You have to pass from Red to Green keeping constant intensity.
(2) Intensity is proportional to the square root of the summed up squares of the rgb components, namely:

i(r,g,b) = sqrt( r*r + g*g + b*b)

I don't know if the above formulas give a good gradient approximation, I have not yet made a test...
Hope that helps.

Smile | :)

If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.

Luc Pattyn21-Jul-07 3:49

21-Jul-07 3:49

Hi CPallini,

There is lots of discussion about brightness, luminance, intensity, etc,
most often the formulas used are linear ones, or they rely on MAX(r,g,b) and/or MIN(r,g,b).
Your sqrt is an interesting twist.

Here are five samples of your formulas

n =    0%  25%  50%  75%  100%

g(n) 100%  92%  70%  38%    0%
r(n)   0%  38%  70%  92%  100%

and they match very closely with what MS Word is offering with their "Fill Gradient between
Two Colors".

Well done !

BTW: curious how you would solve it if you are not as lucky as to find one of (r,g,b)
at a constant value tho.

Smile | :)

Luc Pattyn

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Malcolm Smart21-Jul-07 3:58

21-Jul-07 3:58

Thanks all - I can feel my first CP article coming on now. Not specifically to do with this control, but the whole concept of doing something, hitting problems, the investigation, the questions asked, teh answers, the application of the answers to the problem etc.

Thanks again.

"More functions should disregard input values and just return 12. It would make life easier." - comment posted on WTF

Luc Pattyn21-Jul-07 4:03