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Q&A for linguists, teachers, and students wanting to discuss the finer points of the Latin language

--calculate circle points

declare @Radius decimal(28,23) = 1400  --float
declare @spirals float = 10
declare @numberOfPositions float = 36 ---
declare @radiansPerPosition float = (360 / @numberOfPositions) * pi() / 180.0 --'radians = degrees × pi / 180°
declare @xStop float, @yStop float
create table #x (x float, y float)
declare @z float, @spiral int=0

while @spiral <= @spirals 
  begin
    set @spiral=@spiral+1
    set @z=0
    while @z <= @numberOfPositions - 1
      begin --calculate points
        set @xStop = Cos(@radiansPerPosition * @z) * @Radius
        set @yStop = Sin(@radiansPerPosition * @z) * @Radius
        insert into #x(x,y) values (@xStop,@yStop)  --write to table (double the height)
        set @z=@z+1
        set @radius = 1.0*@radius - ((1.0*@radius/@spirals)/@numberOfPositions)
      end
  end
	
print @z
select x,' ',y from #x 
union all select 2660.0,'   ',1333.0 --the keep circle away from edge
union all select -2660.0,'     ',-1333.0

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edited aug 3 18

created aug 3 18 ashleedawg