[RESOLVED] Dividing a circle

[Pine_Apple]

Hi there!
I’m having big trouble trying to design some stimuli for a visual experiment. Mainly because programming (and maths) is not my strong point, plus I’m not even sure if VB can do what I need…

Anyways, I’ve managed to draw a centred circle with specific parameters and a fixation point in the middle, but what I now need is to divide it into 6 equal parts by passing a line through each equally spaced point and the centre. While I know how to do that with a protractor on a piece of paper, I have no clue how to do it in code And it pains me to think this is just the simplest manipulation I need to do in order to draw this stimulus. I’m trying to take it one step at a time and figure out as much as I can on my own, but unfortunately time is against me . I’d appreciate any help, suggestions, advice… Thank you all!

PS: I’m using VB 6.0 and only have basic beginner knowledge.

[Pine_Apple]

Ha! It took me a while, but I brushed up on my trigonometry and I figured it out myself (with little help from Google of course ). I consider this resolved now

[Pine_Apple]

Rrrright... I spoke too soon. I don't mean to be annoying, but I have another problem now. I've created most of the stimuli as I wanted, but there seems to be something wrong. I've uploaded a picture in combination to the code. The small circles within the 'doughnut' shape should be equally spaced from each other and should be centred on the outline of the green circle. However, as it appears obvious from the image - this is not the case. The bottom and top two are not centred where they should be. Also, they are not really equally spaced
Is there something wrong with the code? Am I overlooking something really obvious
My code:
Private Sub Form_Paint()
Const pi As Double = 3.142
'draw the circle on which the stimuli will be centred
Me.Circle (Me.ScaleWidth / 2, Me.ScaleHeight / 2), 5.74, vbGreen

'find points on the circle at specified radians
'(6 equally spaced locations)
'and centre a small circle at each point
Dim x, y As Integer
x = (Cos(2 * pi) * 5.74) + Me.ScaleWidth / 2
y = (Sin(2 * pi) * 5.74) + Me.ScaleHeight / 2
Form1.Circle (x, y), 0.64, vbGreen
Dim x2, y2 As Integer
x2 = (Cos(pi / 3) * 5.74) + Me.ScaleWidth / 2
y2 = (Sin(pi / 3) * 5.74) + Me.ScaleHeight / 2
Form1.Circle (x2, y2), 0.64, vbRed
Dim x3, y3 As Integer
x3 = (Cos(5 * pi / 3) * 5.74) + Me.ScaleWidth / 2
y3 = (Sin(5 * pi / 3) * 5.74) + Me.ScaleHeight / 2
Form1.Circle (x3, y3), 0.64, vbBlue
Dim x4, y4 As Integer
x4 = (Cos(4 * pi / 3) * 5.74) + Me.ScaleWidth / 2
y4 = (Sin(4 * pi / 3) * 5.74) + Me.ScaleHeight / 2
Form1.Circle (x4, y4), 0.64, vbYellow
Dim x5, y5, x6, y6 As Integer
x5 = (Cos(pi) * 5.74) + Me.ScaleWidth / 2
y5 = (Sin(pi) * 5.74) + Me.ScaleHeight / 2
Form1.Circle (x5, y5), 0.64, vbCyan
x6 = (Cos(2 * pi / 3) * 5.74) + Me.ScaleWidth / 2
y6 = (Sin(2 * pi / 3) * 5.74) + Me.ScaleHeight / 2
Form1.Circle (x6, y6), 0.64, vbMagenta

'Draw a point at the centre of each circle
Me.DrawWidth = 3
Me.PSet (Me.ScaleWidth / 2, Me.ScaleHeight / 2)
Me.PSet (x2, y2)
Me.PSet (x5, y5)
Me.PSet (x6, y6)
Me.PSet (x, y)
Me.PSet (x3, y3)
Me.PSet (x4, y4)
End Sub

The result:


PS: The scale is set to centimetres.

[Pine_Apple]

Ok, here I am some time later to continue my potentially annoying monologue... I realised my mistake is in declaring variables as Integers when clearly they are NOT whole numbers... duh! When I switch to Double it all works fine

I'm really sorry and kind of ashamed , I promise next time I'll give it at least 24 hours before I post something :P

[ThomasJohnsen]

If you don't want to bother with the trig-functions, the coordinates for the regular unit-hexagon with center (0, 0) are:
{ (1, 0), (0.5, sqrt(3/4)), (-0.5, sqrt(3/4)), (-1, 0), (-0.5, -sqrt(3/4)), (0.5, -sqrt(3/4)) }

[Pine_Apple]

Ah, yes! That would be quite helpful because although I've managed to centre the circles, they're still not equally spaced
However... I'm not sure how to draw a hexagon... I mean, what kind of argument do you use to input these coordinates? I'd be great if there was a Me.Hexagon option, but I sense it's bound to be more to it and I don't seem to be able to figure it out

Oh! And of course, thank you very much for the suggestion!

[opus]

......they're still not equally spaced

As I see it, the coordinates you are using are correct, you are already using the same coordinates as Pine_Apple suggested, although you do compute them using Cos and Sin.
Your point x;y uses Cos(2Pi) and Sin(2Pi) which are 1 and 0
your point x1;y1 uses Cos(Pi/3) and Sin(Pi/3) which are 0.5 and Sqrt(3/4) or better 0.86666..
etc...
So, why do you think your points (which form a hexagon, although "you can't draw it") are not equally spaced?

[Pine_Apple]

So, why do you think your points (which form a hexagon, although "you can't draw it") are not equally spaced?

Yeah... it makes no sense for them to not be spaced equally cause the coordinates are correct, I'm sure of that! However, they aren't! In the end I actually measured them manually by placing a protractor on the monitor (oh yes, I went there!) and it turned out the first angle was 62 instead of 60, and then next proportion was 58 So everything was off by approximately 2 degrees. No idea why... so I 'cheated' by centering the small circles at 59, 121, 239 and 301 degrees. This almost corrected for it. It seems fine now...

[ThomasJohnsen]

As I see it, the only place that the error can arise is due to your approximation of Pi with 3.142.
Pi represented as a double is actually 3,14159265358979.

[Pine_Apple]

Yeah, I thought so too! But it didn't make any difference There must be something fundamentally wrong with the scales because I also just noticed that the centimetre measurements are not accurate either... e.g. I set diameters of 1.28 and and 7.02, instead it generates them as 1.6 and 8.7 (approximately). Well... I don't think I can do anything about that but bite the bullet and somehow try to work around it

Thanks for the help and tips!

Edition: Oups.. I messed up those values a bit, that's not exactly how they turned out, but it doesn't matter cause the fact is that the centimetre measurements are not accurate.

[Pine_Apple]

Rrrright! As embarrassing as this is... I have finally realised what the issue is... The resolution of the screen!!! It's fixed now and everything corresponds to the correct measurements as it should have on the first place! What a relief!!