Drawing a clock in vb6 - Xtreme Visual Basic Talk

02-18-2009, 06:48 AM

picogsm

Freshman

Join Date: Mar 2005

Location: Uk

Posts: 43

Drawing a clock in vb6

hi,

Whats wrong with this code?

I want to draw 360 degrees of lines radiating from the centre of the picture box, but it does not work.

All I see is a small segment (maybe 2 degree worth) in the upper left quadrant, no matter what I do...

Code:

Dim x1 As Integer
Dim y1 As Integer

Dim x2 As Double
Dim y2 As Double

Dim i As Double

Dim j As Double

x1 = Picture1.ScaleWidth / 2
y1 = Picture1.ScaleHeight / 2

For i = 0 To 360 Step 20

j = (i / 180) * 3.14159

x2 = Cos(j) * 2
y2 = Sin(j) * 2

Picture1.Line (x1, y1)-(x2, y2), vbRed

Next i

Any ideas?

02-18-2009, 09:52 AM

OnErr0r

Obsessive OPtimizer

Administrator
* Guru *

Join Date: Jun 2002

Location: Debug Window

Posts: 13,685

I'll let you figure out what is wrong with yours.

Here is my working code:

Code:

Option Explicit

Const RAD_SECOND As Double = 0.10471975511966 ' PI / 60

' interval = 10
Private Sub Timer1_Timer()
  DrawSecondHand Second(Now)
End Sub

Private Sub DrawSecondHand(ByVal lSecond As Long)
  Dim x As Long
  Dim y As Long
  Dim lCenterX As Long
  Dim lCenterY As Long
  Dim lRadius As Long
  Static lLastSecond As Long
  
  ' Flicker protection
  If lSecond <> lLastSecond Then
    Caption = lSecond
    Cls
    
    ' Determine centers
    lCenterX = ScaleWidth \ 2
    lCenterY = ScaleHeight \ 2
    lRadius = lCenterX
    
    ' Use minimum
    If lCenterY < lCenterX Then lRadius = lCenterY
    
    Circle (lCenterX, lCenterY), lRadius
    
    x = Sin(lSecond * RAD_SECOND) * lRadius
    y = Cos(lSecond * RAD_SECOND) * lRadius
    Line (lCenterX, lCenterY)-(lCenterX + x, lCenterY - y)
    lLastSecond = lSecond
  End If
End Sub

__________________
Quis custodiet ipsos custodues.

02-19-2009, 04:54 AM

picogsm

Freshman

Join Date: Mar 2005

Location: Uk

Posts: 43

Hi, thanks for the hints!

Problem is, I do not really want to create a clock (its just to help explain).

I would like to draw a line at any of the 360 degrees in a circle.

Had a play more with the code:

Code:

Dim x1 As Long
Dim y1 As Long

Dim x2 As Long
Dim y2 As Long

Dim ra As Long

Dim i As Long

Dim j As Long


x1 = Picture1.ScaleWidth / 2
y1 = Picture1.ScaleHeight / 2

If x1 > y1 Then ra = y1 Else ra = x1

Picture1.Circle (x1, y1), ra

For i = 0 To 360

j = (i * 3.14159) / 180  'My original calc was wrong!!!

x2 = Cos(j) * ra
y2 = Sin(j) * ra

Picture1.Line (x1, y1)-(x1 + x2, y1 - y2), vbRed

Next i

Now what this does is draw a circle which is segmented at around 60 degrees, no matter what step value i put into the for..next loop.

Its really confusing

02-19-2009, 07:15 AM

Iceplug

MetaCenturion

Retired Moderator
* Guru *

Join Date: Aug 2001

Location: California, USA

Posts: 16,583

Well, if you are using Longs, you aren't going to get any detail between numbers such as 59 and 60. However, if you use a Single/Double/Currency, you can use Step 0.5 or lower and get a more detailed circle.

__________________

Iceplug, USN
Quadrill 1 Quadrill 2 (full) Quadrill 3 JumpCross .NET Website is ALIVE! - DL Platform Tour for VB.NET! Posting Guidelines Hint: Specify your location in your user cp profile if you want compassion!

02-19-2009, 07:31 AM

picogsm

Freshman

Join Date: Mar 2005

Location: Uk

Posts: 43

Guys, between the two of you it works.

many thanks for taking the time to help

02-19-2009, 09:16 AM

passel

Sinecure Expert

Super Moderator
* Guru *

Join Date: Jun 2003

Location: Upstate New York, usa

Posts: 7,714

Hopefully, you realize that it wasn't a problem with the step, but the fact that you were using a Long Type (all Integer, no fractional part) to hold your converted Angle in Radians, which is why it jumped by about 60 degrees.
1 Radian = 57.295779... Degrees
2 Radians = 114.591559... Degrees
3 Radians = 171.887338.... Degrees
etc...

__________________
There Is An Island Of Opportunity In The Middle of Every Difficulty.
Miss That, Though, And You're Pretty Much Doomed.

02-19-2009, 05:24 PM

OnErr0r

Obsessive OPtimizer

Administrator
* Guru *

Join Date: Jun 2002

Location: Debug Window

Posts: 13,685

Also note that my code only used floating point operations (Sin/Cos) as intermediates. After being multiplied with the radius the lack of a decimal really isn't that important, since pixel operations are integers (even though the line method takes singles).

In the same way you could have used all longs, omitted j and simply done this:

Code:

x2 = Cos((i * 3.14159) / 180) * ra
y2 = Sin((i * 3.14159) / 180) * ra

__________________
Quis custodiet ipsos custodues.

02-20-2009, 01:45 AM

picogsm

Freshman

Join Date: Mar 2005

Location: Uk

Posts: 43

Quote:

Originally Posted by passel

Yes, I see it now. I also used a Print statment to debug the coordinates being created and saw them all with no decimal places!

Really wish I had taken some formal programming course when younger, don't know what I would do without websites like this

Thanks again