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position along a circle

Math and Physics Programming Algorithm

Started by Tanzan March 02, 2018 01:40 PM

5 comments, last by Alberth 6 years, 4 months ago

Tanzan

113

Author

March 02, 2018 01:40 PM

Hello all,

So i want to place objects along a (half) circular line with the formula (x-a)^2 + (y-b)^2 = r^2

object pixel size are 36px.

so if i have 10 objects then r would become (10.36):2 = 180.

finding Y for X =0 (0 degrees)becomes sqrt(r^2 - 0^2) Y becomes 180

finding Y for X =5 (90 degrees) becomes sqrt(r^2 - (5*36)^2) Y becomes 0

this makes perfect sense...

but if i do this in a for loop then the sequence for x will be (0,36,72,108,144,180) this means that the 6th object gets the 90 degrees Y value....and not the 5th.

(for i= 0 etc)

x = i*36...

I know its probably very 'dumb'question but i just don't see it anymore

Thx in advance...

Alberth

10,216

March 02, 2018 06:41 PM

Use sine and cosine instead


import math
r = ...
for i in range(10):
    x = r * cos(math.pi * i / 10)
    y = r * sin(math.pi * i / 10)

# 0 radian <-> 0 degrees
# pi/2 radian <-> 90 degrees
# pi radian <-> 180 degrees

# For proper centering, you may want to add 0.5 to i

Tanzan

113

Author

March 03, 2018 05:35 AM

Thx alot Alberth! ;-)

Still wondering though.....

Alberth

10,216

March 03, 2018 06:07 AM

I am not quite sure what you mean, actually. What is "6th"?

You are only speaking about X index values (which seems to start at 0), and you use such index values for i in the "for" statement as well. Then suddenly you have a 6th object??

The only thing I can think of is that you somewhere switch to a different indexing system, namely 1st, 2nd, 3rd, etc. Alternatively, you have switched to a languages that starts counting at index 1 instead of 0, like Matlab or Lua.

If so, then yes, when you start counting at 0 versus when you start counting at 1 makes exactly 1 unit difference everywhere

Tanzan

113

Author

March 03, 2018 07:06 AM

Sometimes i'm almost human ;-)

what i did was:


for(int i=0;i<aantal;i++) {
    x = i * 36;
    y = (int) Math.sqrt(Math.pow(radius, 2) - Math.pow(x, 2));
}

so the 5th object at index 4 got (4*36) obv. instead of (5*36) grrr

thx again Alberth.

Alberth

10,216

March 03, 2018 04:39 PM

Just always use the index value to refer to an item

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position along a circle

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