• ### Question

• hi guys! i just coded dynamic shadows in SB! looks really awesome! but its not done..
heres the import code: VMM615

move the lightsource with the arrow keys.

its still got some bugs, but i'll fix them.

i made that using the turtle cuz im too stupid to use trigonometry... i learned in school how to use it, but i forgot it... can someone tell me how to do this with trigonometry instead of using the turtle? it would run much better then..

i just need to find the point, where the shadow starts on the floor. for example: a "sunray" is coming from the lightsource and is going to the an edge of the Box. And going in that angle to the floor. and i need to find that point, where the "sunray" is touching the floor, so it can create a triangle with one edge on that point, one on one of the lower corners of the box and one on the top corner of the box.

i hope you guys understand what i need..
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Friday, August 28, 2009 9:40 PM

• Try the following import CXM437 as an idea method, but the screen clearing may be an issue for any game.  Its nothing other than the geometry shown above, just treating the various cases carefully, sun above, below, etc.
• Marked as answer by Monday, October 19, 2009 9:29 PM
Monday, October 12, 2009 6:21 PM

### All replies

• heres the new version with 2 shadows, so it looks realistic!
CPD121
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Friday, August 28, 2009 10:24 PM

• You don't need any trigonometric formula here. You simply need to use similarity.

B = b * (H/h)
C = a * (H/h)

You also must note there's two shadows if the sun is just over the house.

If the sun is below the top of the house, the shadow should be so :

In this case, you'll need a triangle + a rectangle.
Fremy - Developer in VB.NET, C# and JScript ... - Feel free to try my extension
• Proposed as answer by Friday, August 28, 2009 10:25 PM
• Marked as answer by Tuesday, September 15, 2009 8:43 PM
• Unmarked as answer by Sunday, October 11, 2009 10:52 PM
Friday, August 28, 2009 10:25 PM
• i posted an updated version with 2 shadows..

ill look through your stuff now! thx man!
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Friday, August 28, 2009 10:30 PM

• You don't need any trigonometric formula here. You simply need to use similarity.

B = b * (H/h)
C = a * (H/h)

You also must note there's two shadows if the sun is just over the house.

If the sun is below the top of the house, the shadow should be so :

In this case, you'll need a triangle + a rectangle.
Fremy - Developer in VB.NET, C# and JScript ... - Feel free to try my extension
dude, i dont get it.. how am i supposed to code this with that formula?? what do i need C and B for??

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Friday, August 28, 2009 10:52 PM
• lol, sorry! i found it out!
btw, this isnt the whole formula:
B = b * (H/h)
C = a * (H/h)

its:
B = math.squareroot(b * (H/h))
C = math.squareroot(a * (H/h))

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Friday, August 28, 2009 11:21 PM
• I think Fremy's similar triangle formulae are right.  Another way to work it out is with vectors.

The sun is at (Sx,Sy ) and a box corner is at (Bx,By ) and the shadow hits the ground at (Dx,Dy ).

A straight line is defined by y = A*x+B

The sun and the box top both lie on this line, so Sy = A*Sx+B and By = A*Bx+B

We can solve these for A and B by subtracting the second from the first.

Sy-By = A*(Sx-Bx) => A = (Sy-By)/(Sx-Bx) and therefore B = Sy-A*Sx

When we know A and B (from above) , we can find Dx, since we know this point must be on the same line and have Dy = gh, hence:

Dy = gh = A*Dx+B => Dx = (gh-B)/A

Putting this together:

Dx = (gh-(Sy-A*Sx))/A = (gh-Sy)/A+Sx => Dx= (gh-Sy)*(Sx-Bx)/(Sy-By)+Sx and Dy = gh

So if the sun is at (Sx,Sy) and a box corner is at (Bx,By), then the shadow will hit the ground at (Sx+(gh-Sy)*(Sx-Bx)/(Sy-By),gh ), assuming the sun is above the box.

We need to check two corners (Bx1,By) and (Bx2,By) and that Sy < By (the sun is above the box, remembering y increases down).  Also we only get a left shadow when Sx > Bx1 and a right shadow when Sx < Bx2.

PS , using Fremy's picture: and the right side shadow:

Dx = Sx+B

h = By-Sy
H = gy-Sy
b = Bx-Sx

B = b*(H/h) = (Bx-Sx)*(gy-Sy)/(By-Sy)

Hence Dx= Sx+(Bx-Sx)*(gy-Sy)/(By-Sy) The same answer
Friday, August 28, 2009 11:38 PM
• B = math.squareroot(b * (H/h))
C = math.squareroot(a * (H/h))
Not it isn't. You don't need the sqare root.

Fremy - Developer in VB.NET, C# and JScript ... - Feel free to try my extension
Saturday, August 29, 2009 7:53 AM
• yeah, i just found out, lol
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Saturday, August 29, 2009 11:05 AM
• yay! i finally got it working!

man! it took me so long till i found out that the picture for the formula doesnt really fit to the formula you gave me..

H is as big as h in that picture... but h is actually just from the top of the rectangle, to the lightsource... and H is the Heigt... (from the bottom of the rectangle to the lightsource...

works really awesome so far!

heres the import code if you wanna check it out:
XHZ643

btw, i found some amazing 2d dynamic shadow engines! (too bad its not made in SB, but its still awesome)

here you go:
http://www.gamedev.net/reference/articles/article2032.asp
and here at the bottom:
http://forums.xna.com/forums/p/34665/219412.aspx

i think thats extremely awesome! looks so real and stuff!
but that "shadow maze" ,on image Image 9 in the first link, isnt possible to be done in small basic? or is it? (but it would be propably as slow as a slideshow...)
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Saturday, August 29, 2009 11:05 AM
• yay! i finally got it working!

man! it took me so long till i found out that the picture for the formula doesnt really fit to the formula you gave me..

H is as big as h in that picture... but h is actually just from the top of the rectangle, to the lightsource... and H is the Heigt... (from the bottom of the rectangle to the lightsource...

works really awesome so far!

heres the import code if you wanna check it out:
XHZ643

btw, i found some amazing 2d dynamic shadow engines! (too bad its not made in SB, but its still awesome)

here you go:
http://www.gamedev.net/reference/articles/article2032.asp
and here at the bottom:
http://forums.xna.com/forums/p/34665/219412.aspx

i think thats extremely awesome! looks so real and stuff!
but that "shadow maze" ,on image Image 9 in the first link, isnt possible to be done in small basic? or is it? (but it would be propably as slow as a slideshow...)
Live for nothing, OR CODE FOR SOMETHING! (Happy now vijaye?^^)

Two remaining problems :
* It flashes (maybe you should use GraphicsWindow.DrawTriangle instead of AddTriangle, since you need the trianges only one time)
* If the sun is just on the top of the house, your program crash because you try to divide by 0. This is a special case because there's no triangle to show, just a rectangle.
Fremy - Developer in VB.NET, C# and JScript ... - Feel free to try my extension
Saturday, August 29, 2009 11:16 AM
• See XHZ643-0 to have both problems fixed.

Please note there's still problems in special conditions (sun below the house, on the house...) you could fix :-)
Fremy - Developer in VB.NET, C# and JScript ... - Feel free to try my extension
Saturday, August 29, 2009 11:39 AM

should be:

Saturday, August 29, 2009 11:51 AM

should be:

Fixed in XHZ643-2 - Nice catch BTW!

Fremy - Developer in VB.NET, C# and JScript ... - Feel free to try my extension
Saturday, August 29, 2009 12:19 PM
• i know that! i wanted it like that, i was just testing something out!

im working on a new version now. and im gonna chech out your ppls stuff now.

btw, im now trying to do Top down view shadows now, not from the side...

and

filltriangle will have the same effect like addtriangle, but different. if you use filltriangle, the triangle is always under every other object. but if you use addtriangle, you can make other objects be in the shadow.. but having objects hidden in the shadow of an object looks really cool, but if you always see the top of every object, its more logic... im gonna do the version with filltriangle.
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Saturday, August 29, 2009 2:49 PM
• heres a 2 shadow version:
RTF108

but somehow it freezes for a moment sometimes, idk why! (it isnt because its a 2 shadow version. i also have a 1 shadow version, and its the same there..)
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Saturday, August 29, 2009 3:57 PM
• DAMN! i cant GET IT TO WORK!!!!!!!!!! The shadow works really great when the lightsource is ABOVE the box.. but not when its under it... idk how i have to change the variables to work how it should...

heres the code:
PKD858

plz! someone help me!
(btw, the shadows are over the box, because the box is transparent. when the lightsource is above the box, it works great, so please dont change anything there...)
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Saturday, August 29, 2009 7:53 PM
• better use this one to fix:
BPQ575
the other one i posted before was horrible..
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Saturday, August 29, 2009 10:50 PM

• You don't need any trigonometric formula here. You simply need to use similarity.

B = b * (H/h)
C = a * (H/h)

You also must note there's two shadows if the sun is just over the house.

If the sun is below the top of the house, the shadow should be so :

In this case, you'll need a triangle + a rectangle.
Fremy - Developer in VB.NET, C# and JScript ... - Feel free to try my extension
another question. what would the formula look like when the Lightsource is under the box?? like the upper picture flipped upside down?
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Sunday, August 30, 2009 3:37 PM
• I think Fremy's similar triangle formulae are right.  Another way to work it out is with vectors.

The sun is at (Sx,Sy ) and a box corner is at (Bx,By ) and the shadow hits the ground at (Dx,Dy ).

A straight line is defined by y = A*x+B

The sun and the box top both lie on this line, so Sy = A*Sx+B and By = A*Bx+B

We can solve these for A and B by subtracting the second from the first.

Sy-By = A*(Sx-Bx) => A = (Sy-By)/(Sx-Bx) and therefore B = Sy-A*Sx

When we know A and B (from above) , we can find Dx, since we know this point must be on the same line and have Dy = gh, hence:

Dy = gh = A*Dx+B => Dx = (gh-B)/A

Putting this together:

Dx = (gh-(Sy-A*Sx))/A = (gh-Sy)/A+Sx => Dx= (gh-Sy)*(Sx-Bx)/(Sy-By)+Sx and Dy = gh

So if the sun is at (Sx,Sy) and a box corner is at (Bx,By), then the shadow will hit the ground at (Sx+(gh-Sy)*(Sx-Bx)/(Sy-By),gh ), assuming the sun is above the box.

We need to check two corners (Bx1,By) and (Bx2,By) and that Sy < By (the sun is above the box, remembering y increases down).  Also we only get a left shadow when Sx > Bx1 and a right shadow when Sx < Bx2.

PS , using Fremy's picture: and the right side shadow:

Dx = Sx+B

h = By-Sy
H = gy-Sy
b = Bx-Sx

B = b*(H/h) = (Bx-Sx)*(gy-Sy)/(By-Sy)

Hence Dx= Sx+(Bx-Sx)*(gy-Sy)/(By-Sy) The same answer

im sorry, but this was to complicated for me, that i could use it... (also because of the high english...)

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Sunday, August 30, 2009 3:39 PM
• I think Fremy's similar triangle formulae are right.  Another way to work it out is with vectors.

The sun is at (Sx,Sy ) and a box corner is at (Bx,By ) and the shadow hits the ground at (Dx,Dy ).

A straight line is defined by y = A*x+B

The sun and the box top both lie on this line, so Sy = A*Sx+B and By = A*Bx+B

We can solve these for A and B by subtracting the second from the first.

Sy-By = A*(Sx-Bx) => A = (Sy-By)/(Sx-Bx) and therefore B = Sy-A*Sx

When we know A and B (from above) , we can find Dx, since we know this point must be on the same line and have Dy = gh, hence:

Dy = gh = A*Dx+B => Dx = (gh-B)/A

Putting this together:

Dx = (gh-(Sy-A*Sx))/A = (gh-Sy)/A+Sx => Dx= (gh-Sy)*(Sx-Bx)/(Sy-By)+Sx and Dy = gh

So if the sun is at (Sx,Sy) and a box corner is at (Bx,By), then the shadow will hit the ground at (Sx+(gh-Sy)*(Sx-Bx)/(Sy-By),gh ), assuming the sun is above the box.

We need to check two corners (Bx1,By) and (Bx2,By) and that Sy < By (the sun is above the box, remembering y increases down).  Also we only get a left shadow when Sx > Bx1 and a right shadow when Sx < Bx2.

PS , using Fremy's picture: and the right side shadow:

Dx = Sx+B

h = By-Sy
H = gy-Sy
b = Bx-Sx

B = b*(H/h) = (Bx-Sx)*(gy-Sy)/(By-Sy)

Hence Dx= Sx+(Bx-Sx)*(gy-Sy)/(By-Sy) The same answer
Whats A and B? and g!?!?

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Tuesday, September 1, 2009 9:27 AM
• A and B here are coeffcients in the equation of a line y = Ax+B, gh (mistyped as gy sometimes) is the Graphics Window height, i.e. ground level of the bottom of the box, hence for example H = gh-Sy, or gh = H+Sy.
Tuesday, September 1, 2009 6:35 PM
• A and B here are coeffcients in the equation of a line y = Ax+B, gh (mistyped as gy sometimes) is the Graphics Window height, i.e. ground level of the bottom of the box, hence for example H = gh-Sy, or gh = H+Sy.
sorry, i dont wanna sound mean but, those words are to high for me understand... but that formula is for the shadows just going to the "floor"? if yes, im trying to do a top down view version... (i coded a "collision engine" that works with every object..(rectangle))

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Tuesday, September 1, 2009 7:07 PM
• The formula Dx= Sx+(Bx-Sx)*(gh-Sy)/(By-Sy) calculates the x Coordinate(Dx) on the ground level y = gh, when the sun is at (Sx,Sy) and the top box corner is at (Bx,By).  Therefore a ray of light passes from the sun (Sx,Sy), just passing the box top (Bx,By) and hits the ground at (Dx,Dy), where the y for ground is Dy=gh.

This is a straight line passing through the sun, box top and the ground.  A straight line is any point (x,y) that satisfys the equation y = A*x+B, where A and B are constant numbers.

In this case A = (Sy-By)/(Sx-Bx) and B = Sy-A*Sx

So if the sun is at (100,50), and the box top is at (80,80), then A = (50-80)/(100-80) = -1.5 and B = 50-(-1.5*100) = 200

All points (x,y) on the ray satisfy y = -1.5x + 200

If the ground is at y=140, then the ray hits the ground at Dx = 100+(80-100)*(140-50)/(80-50) = 40

Hits the ground at (40,140)

y = -1.5x + 200 (the equation of the ray) is true for the sun (50= -1.5*100+200), the box top (80 =-1.5*80+200) and the ground (140 = -1.5*40+200)

Try drawing a diagram if unclear.

Tuesday, September 1, 2009 7:42 PM
• cool! thx! i'll check it out as soon i have time!
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Tuesday, September 1, 2009 7:48 PM
• Another thought - all of this is for shadows seen from the side with the sun and object in view.  More commonly this applies to light sources that are relatively close the the object.   Everything we said so far is based on the original code you posted (side on view).

If you have a plan view (seen from above), where the sun is above the view (above and behind the viewer), then some of this may be easier, you mentioned top down view a couple of posts ago.  For a plan view the sun may be assumed to be very far away and all objects will have the same direction shadow, proportional to their height.  The shadows only change (for all objects in the same way) if the sun moves and not depending on where the objects are.

Perhaps if you draw picture of what you want it may be possible to suggest a different approach.
Tuesday, September 1, 2009 8:03 PM

• You don't need any trigonometric formula here. You simply need to use similarity.

B = b * (H/h)
C = a * (H/h)

You also must note there's two shadows if the sun is just over the house.

If the sun is below the top of the house, the shadow should be so :

In this case, you'll need a triangle + a rectangle.
Fremy - Developer in VB.NET, C# and JScript ... - Feel free to try my extension
thx again for the formula!

i didnt work on the dynamic shadows project for some time. but i need it again now.

it would be awesome if you could tell me what the formula would be like when the light source (tip of h) is under the box/house.. cuz im really stuck at doing upside down shadows...

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Sunday, October 11, 2009 10:51 PM
• You need to fill the two grey shadow shapes (one triangle and one box).  Therefore you need the coordinates of their corners.

You know:

1] The top left corner of the box (BTLx,BTLy)

2] The width of the box (Bw)

3] The height of the box (Bh)

4] The window width (Wx)

5] The position of the sun (Sx,Sy)

Therefore the square shadow has corners (BTLx+Bw,BTLy) (Wx,BTLy) (Wx,BTLy+Bh) (BTLx+Bw,BTLy+Bh)

The bottom 2 corners of the triangle are also obvious (BTLx,BTLy) (Wx,BTLy)

Now the only non-trivial corner of the triangle:

It lies on a line from the sun, through the box corner, ending on the side of the window (x = Wx).

This line has an equation y = A.x+B, where A and B are constants.  As before we can find A and B since they are satisfied by the sun and box corner coordinates:

Sy = A.Sx + B and BTLy = A.BTLx + B

Hence if we subtract the second equation from the first (the Bs cancel) we get (Sy-BTLy) = A.(Sx-BTLx) and therefore A = (Sy-BTLy)/(Sx-BTLx), as long as the sun is to the left of the box (Sx < BTLx) we are OK.

We can find B by substituting our value of A back into the first equation above and rearranging : B = Sy - A.Sx = Sy - (Sy-BTLy)/(Sx-BTLx).Sx

Now the top point of the triangle will have x = Wx.

Hence its y coordinate will be y = A.Wx + B = (Sy-BTLy)/(Sx-BTLx).Wx + Sy - (Sy-BTLy)/(Sx-BTLx).Sx = Sy + (Sy-BTLy)/(Sx-BTLx)(Wx-Sx)

The triangle coordinates are: (BTLx,BTLy) (Wx,BTLy) (Wx,Sy + (Sy-BTLy)/(Sx-BTLx)(Wx-Sx))
Monday, October 12, 2009 2:17 PM
• thx!

are you sure i need to do it with a box? because i also did it wich 2 triangles.. like this: FCT935 (i know, the shadows arent right, but this should just show off the idea..)

or is the version with the box easyer to do?
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Monday, October 12, 2009 3:20 PM
• You can do it with triangles or with rectangles - I was just following the drawing you showed - the idea is always the same - calculate the corners of any shape you want to shade - their geometry just needs to be worked out - unfortunately there is no single formula for all cases, but the principal remains constant.

Monday, October 12, 2009 3:29 PM
• Try the following import CXM437 as an idea method, but the screen clearing may be an issue for any game.  Its nothing other than the geometry shown above, just treating the various cases carefully, sun above, below, etc.
• Marked as answer by Monday, October 19, 2009 9:29 PM
Monday, October 12, 2009 6:21 PM
• wow! works wonderful! just what i wanted to do! thx really much!

about the shadows. did you do them with 3 triangles? or 2 triangles and 1 box? i mean, it would also work with just 2 triangles, like i did...
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Tuesday, October 13, 2009 12:20 AM
• As you can see from the code I used 3 triangles.

GraphicsWindow.FillTriangle(C1x,C1y,C2x,C2y,(D1x+D2x)/2,(D1y+D2y)/2)
GraphicsWindow.FillTriangle(C1x,C1y,D1x,D1y,D2x,D2y)
GraphicsWindow.FillTriangle(C2x,C2y,D1x,D1y,D2x,D2y)

From the drawing (after finding the two box corners to use (C1 and C2) and the window face to project onto) the triangles are:

(C1,D1,D2) (C2,D1,D2) and (C1,C2,(D1+D2)/2), the last triangle using the midpoint of D1 and D2 is needed for some cases where the first 2 triangles leave a 'hole' in the shadow.  Try commenting out this triangle to see what I mean.

Tuesday, October 13, 2009 9:49 AM
• hmm.. good idea! but you can also do it with just 2 triangles, like i did.. there will be no hole in the shadow but you will see a fine line between the 2 triangles...
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Tuesday, October 13, 2009 5:19 PM
• Good persistence - looking at it more carefully I think you are right and just using (C1,C2,D1) and (C2,D1,D2) as you tried works, we could also use (C1,C2,D2 ) and (C1 ,D1,D2), but not (C1,C2,D1 ) and (C1 ,D1,D2) or (C1,C2,D2 ) and (C2 ,D1,D2) - Import NPW879.  There should be no fine joining line using a GraphicsWindow.PenWidth=0.
Tuesday, October 13, 2009 6:18 PM
• haha^^

but how did you manage to not get that fine line?
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Tuesday, October 13, 2009 7:41 PM
• It looks like the fine line is still there, but you can only see it clearly if the shadow colour is fairly dark compared to the lit background e.g. black on lightblue.  In this case you could also draw a shadow coloured line over the fine line e.g.

GraphicsWindow.BrushColor = GraphicsWindow.BackgroundColor
GraphicsWindow.FillRectangle(0,0,gw,gh)
GraphicsWindow.BrushColor = "Black"
GraphicsWindow.FillTriangle(C1x,C1y,C2x,C2y,D1x,D1y)
GraphicsWindow.FillTriangle(C2x,C2y,D1x,D1y,D2x,D2y)
GraphicsWindow.PenColor = "Black"
GraphicsWindow.PenWidth = 2
GraphicsWindow.DrawLine(C2x,C2y,D1x,D1y)

Tuesday, October 13, 2009 8:06 PM
• i knew it!

but, cant you see the end points of the line sticking out a little bit of the shadow? (i also tried this with the lines some times, but it never worked very well...)

well, i want those shadows to suck the smallest amount of performance as possible.. cuz this would be very useful for some little games..

btw, do shapes made with shapes.addrectangle suck more performance than graphicswindow.fillrectangle even when its not being moved around? because if it would suck the same amount of performance, we could, to remove the shadow flicker when the shadows are redrawn, add the shadow and then add the new one and then delete the old one, like you did for the animations in my game..
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Tuesday, October 13, 2009 8:12 PM
• With regard to performance - try it and see with a test program - my guess is that there won't be too much in it if you don't move the shape, but this is just a guess - its often not good to guess at how something is internally implemented so I would certainly test this performance first.

The shapes idea could solve the flicker and perhaps the small line bit sticking out since the shapes can have Pen borders as well as Brush filling - again prototype to see.

The main issue I see using shapes rather than painting to the background is that the shadows will be on top of all other shapes, whereas the GraphicsWindow.FillRectangle paints to the background and all shapes remain visible on top (e.g. the blue box in the shadows demo is always present on top of the shadows we draw).

The following (replacing all the current clearing and drawing code) shows the possible advantages and disadvantages of shapes (no indication of performance issues - need more dedicated tests for this).  Also, some playing with PenWidth or a Shapes.AddLine may improve the sticking out bit.

GraphicsWindow.BrushColor = "Black"
GraphicsWindow.PenColor = "Black"
GraphicsWindow.PenWidth = 2

Tuesday, October 13, 2009 8:34 PM
• damn!

i totally forgot that shapes are always over drawn things... it would be really useful (not just for this) if you could set the layer for each object in the next version....

i'll analyze your code as soon as i have time!
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Tuesday, October 13, 2009 8:37 PM
• Yes, this has been mentioned before I think as a possible enhancement - I think it would need some thought as to how it was implemented with the SmallBasic purpose in mind (simplicity as an educational teaching language).

A first time programmer isn't going to want to decide on Z (layer value) attributes to simply draw a circle.  So the best bet to me may be to introduce another Shapes property, say Shapes.Depth defaulted to 0.

If you want to see this feature then post in the Bugs and Suggestions (v0.6 to 0.7) thread with some description and any ideas you have on what it should do and some thinking on how it would work with the SmallBasic objectives in mind, then see what other people think.
Tuesday, October 13, 2009 9:15 PM
• i wouldnt worry about the simplicity because scratch also had that...

ok! i will!
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Wednesday, October 14, 2009 1:42 PM

but theres one problem.. i dont know how to make the background being repainted just once, instead of 2 times for each box... i hope you know what i mean..

but you will understand as soon as you test the code..
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Wednesday, October 14, 2009 7:26 PM
• If we clear the screen once every time before both sets of shadows are created then both shadows work.  We need to only do this when there has been some movement of the sun or box.

The flicker really need to use shapes and therefore the Z layer index option to be added.

Import ZSD244-1
Saturday, October 17, 2009 10:25 PM
• wow! nice solution!

its so awesome! now we can add as many objects as we want!

is this finished now? i guess..

can i add this to my example project compilation? (will be released soon) for sure i would credit you.
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Saturday, October 17, 2009 11:15 PM
• Of course - I look forward to the final project.
Sunday, October 18, 2009 2:51 PM
• cool! thx!

me too^^
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Sunday, October 18, 2009 5:44 PM