The Campaign Builder's Guild

im trying to backwards engineer the mathematic formula used to get the angle of each point of an even (note: as in, equal; not as in even numbers) star. please note that this only needs to work for stars with an odd number of points, as far as i am personally concerned. since you guys are easily the smartest group of people i know on the net, i figured i would ask you for help.

with no further ado, the current version of the formula:

360/a=b
180-(b*c)=d

a = # of points to the star
b = # of degrees between the points
1 < c < ((a+1)/2)
d = degrees to make the angle of each point


does this make sense? can somebody actually try mapping this idea out, and seeing where it might need work? also, if any part of it can be simplified, can you point that out?

thanks in advance :)

edit- the formula has been simplified slightly, with a little bit of an explaination of the variables. note that 2 is always a valid option for variable c.

Assuming i understand your queary correctly, there is no actual way to work tha put for all stars, since the angles are not fixed.

imagine a seven pointed star with short points, and a seven pointed star with very long spikes. The angles inside the star with the short points would be far larger than the angles inside the points of the long pointed spiky star. It's a bit difficult to describe geometry without pictures, but let me know if i understand your question and you understand my point (heh, point! good one...).

you mean you can have both an obtuse septagram AND an acute septagram? whoa.... that kinda puts a dent in the issue.

that said, it doesnt have to be able to produce ALL septagrams, just a septagram, with equal points. or a pentagram, if thats what it was desired to be useed for. or an eleven-pointed star (heptagram, if im not mistaken?). or even....

you get the idea, i hope :D


edit- ok, i can see how septagrams can be both obtuse and acute, so im revising the last line of the formula by replacing the 2 in 2b with c, where:

1 < c < (a+1)/2

2 is always a possible option for c, so im keeping the original version of the formula up, with the 2.

edit- re-labeled the variables mentioned in this post, to keep consistant.

The obtuse angles are the same as the angles in the regular polygram with the corresponding number of sides:

Quote from:  180 degrees)

I'm not sure of this, I [ireally[/i] wish i'd tested it at work, where i had software that could autodraw an icosogon. :P

Is this even what you're asking? or are you wanting to like locate the points of a polygon from an origin or something crazy like that? If you're doing protractor & ruler, this is what you need.

If you've got a compass and straightedge, there's easier methods not involving measurement ;)

The first formula was true, but only for hexagons. :)

i think this is the obtuse polygram, supa?




It's been a long time since geometry class.

a1 = pi(1-2/n)
a2 = pi/n
a4 = 2*pi/n
a5 = pi(1-4/n)

a1 = 180 (1-2/n)
a2 = 360/n
a4 = 180/n
a5 = 180 (1-4/n)


I hope i got the question right. hell, i had fun, anyway...

whoa.... a tad over my head.

here, let me see if i can try to explain the theory behind the formula in the first post, using a pentagram as an example.


first off, with a pentagram, we have already decided a=5. at this point, i would like to define c, since it is much easier with a pentagram:

1<c<((a+1)/2) = 1<c<((5+1)/2) = 1<c<3, which means c=2.

we have now figured out how many points in the star (a), and how many points to "pass" (for lack of a more technical term) to get a polygram, as opposed to a polygon (c).

for b, we use the following equation:

360/a=b. since a=5, this is 360/5=b=64.

we have now, essentially, dissected a circle evenly into 5 (a) pieces. the circle is represented in brown, and the lines between each section are in blue.

at this point, we use the knowledge that the sum of all three angles in any triangle is 180 degrees. skipping over 2 (c) points, we get the triangle with the yellow hypotenus (sp?); we already know the angle opposing the hypotenus is b*c, or 2b, so the formula to figure out the angles of the other two corners would be:

(180-c*b)/2=d

but wait! the angle of the point would be 2d, so we can take out the step where we divide by two in that equation. this leaves:

180-c*b=d


the equations, step-by-step, would then be:

360/a=b
1<c<((a+1)/2)
180-c*b=d


at least, i think thats how that would work? i feel confident about it, but i also know that no amount of confidence can argue mathematics....

you're going from the center, okay. I was starting from the outside. pictures for the win.

which would explain exactly why all of what you said went over my head....

now does what im saying make sense?

and if it does, more importantly, is it mathematically accurate? if i had a protractor and ruled straightedge, i would test it out for myself, but...

dude 360/5 is 72.
that confused me for 30 frigg'n minutes. :(
Your answer is either giving double or exactly what it should. (is the answer 18 or 36 degrees? I can't tell)

If your formula is doing what you want it to do for pentagrams, then it matches my formula. (Your d is my a4)

180*(1-4/a) = 180 - c * 360/a
when c = 2

I didn't solve for anything but c = 2. (i really couldn't follow you without a picture and just found the value of every angle in an n-sided pentagram i could ;))

Quote from: brainfacedude 360/5 is 72.
that confused me for 30 frigg'n minutes. :(
Your answer is either giving double or exactly what it should. (is the answer 18 or 36 degrees? I can't tell)

If your formula is doing what you want it to do for pentagrams, then it matches my formula. (Your d is my a4)

180*(1-4/a) = 180 - c * 360/a
when c = 2

I didn't solve for anything but c = 2. (i really couldn't follow you without a picture and just found the value of every angle in an n-sided pentagram i could ;))

i honestly have no clue whatsoever how i was getting 64. honest.

assuming my formula is accurate, the angle on each point of a pentagram (as marked by the gray dots in my picture) would be 36 degrees. keep in mind, the only possible value for c in a pentagram is 2. polygrams with more points, such as a septagram, would have more possible values for c.

incidently, according to my formula:

when a=5, d=36

when a=7 and c=2, d=77.142857~

when a=7 and c=3, d=25.714285~

if somebody could use a protractor/straightedge, or mapping program to either confirm or deny these numbers....

it works on a 23-gon at c=2,c=3,c=4, and c=11. i'm not testing any further than that. ;)