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Forums > C64 Coding > SID to calculate line slope
2010-08-16 08:38
TWW

Registered: Jul 2009
Posts: 545
SID to calculate line slope

To calculate a slope of a line you can use something like:


lda DeltaX
NoChange:
tax
(Plotpixel)
sec (Might not be neccessary if done right...)
txa
sbc DeltaY (Immediate)
bcs NoChange
(Change shit)
adc DeltaX (Immediate)
jmp NoChange

Which gives 11/15 cycles for each calculation/pixel(Oswald do you agree?^^). If unrolling one would have to add a cycle on the sbc/adc by use of ZP instead of Immediate values which means 12/14 cycles (No jmp if unrolled).

This is pretty good. In fact the two things I see as a possible optimalisation here is getting rid of the SEC by use of clever programming and the ADC DeltaX if you scale the relation between DeltaX and DeltaY so that the largest value = 256. You would require a division and mayhaps you get reduced accuracy (not confirmed though). Looking aside from the div. (tables perhaps?!) you'll end up with 9/11 (10/9 if unrolled(need to watch the branching here and build the routine so that you'd get a 9/10 instead) cycles fully optimized. (Am I missing something?)

However...


I read somewhere (Can't remeber where) that there is a technique where you can use the SID to calculate slopes. The SID is the chip i know the least about so I figgured I'd post here the folowing questions:

#1: Can it be done?
#2: If yes, How?
#3: How much time would the calc. use?

-TWW
 
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2010-08-18 10:13
TWW

Registered: Jul 2009
Posts: 545
Your talking about lines where DX>DY.

I can however promiss you that in vertical to diagonal lines (DY>DX) you only have 1(!) pixel to plot each bytes :)

A horizontal line however is a differen matter^^
2013-07-01 10:28
Rastah Bar
Account closed

Registered: Oct 2012
Posts: 336
I tabulated the length of the horizontal segments in lines for which DX>DY. Making use of that table also for lines where DY>DX, I can draw a cube (12 lines) on a 128x128 screen in slightly more than one frame. With tricks like letting the 1541 calculate the rotations, I think it is possible to show a rotating cube faster than 25 fps.
2013-07-01 19:11
Bitbreaker

Registered: Oct 2002
Posts: 508
Setup dy as - dy and dx as dx, then get err saved automatically to x and never bother about the state of carry as well. Code is best placed in zeropage.

-
    ;plot some pixel
    ;...

    txa
dx=*+1
    sbx #$00
    bcs -
    txa
dy=*+1
    sbx #$00
    ;advance x direction
    ;...

    jmp -


also note that you can combine the upoming subtraction of dx and addition of dy into a single addition/subtraction when you do an change in x direction.
2013-07-01 21:57
SIDWAVE
Account closed

Registered: Apr 2002
Posts: 2238
use a pulsewidth program, to stream into memory from sid code, then make whatever you want. structures.
2013-07-02 08:27
Rastah Bar
Account closed

Registered: Oct 2012
Posts: 336
So how many cycles per pixel are needed on average if you include EVERYTHING (initialization, slope calculation, changing bitmask, changing gfx pointer for plotting)?

I guess I need about 30-35 cycles per pixel in TOTAL when drawing 12 lines of a cube on a single color 128x128 pixel character screen. I do not unroll the loops and my code does not run in ZP.(For slope calculations I do not use the log/exp method, but just a plain divide. I should change that :-) ).
2013-07-02 09:18
Bitbreaker

Registered: Oct 2002
Posts: 508
Hard to give exact values, but for a cube in said size it is possible to render it > 50fps while even moving the 16x16 grid across the screen.
With all overhead i need around 9900 cycles per frame (without clear). If you know about the average linelength you can find out how many cycles i need for a pixel :-)
2013-07-02 10:57
Rastah Bar
Account closed

Registered: Oct 2012
Posts: 336
Sorry, but I don't understand how 9900 cycles is possible.
ora (ZP),y
sta (ZP),y
for the plotting is already 11 cycles. With the 9/10 cycles mentioned above we have at least 20 cycles per pixel. And this does not include any overhead, changing of bitmask etc.

12 lines times 50 pixels per line times 20 cycles is already 12000 cycles.
2013-07-02 11:02
chatGPZ

Registered: Dec 2001
Posts: 11386
so you just found out the trick is not doing that per pixel =P
2013-07-02 11:03
Bitbreaker

Registered: Oct 2002
Posts: 508
There's also horizontal lines and hidden surface removal done, the latter is something i include, but reading again that you draw 12 lines, it seems you draw all lines of the cube.
Also one can forgo on the ora if it is sure that the line is out of reach of other lines (what would cause clashes).
2013-08-01 16:18
Rastah Bar
Account closed

Registered: Oct 2012
Posts: 336
I made some improvements to my code and can draw and rotate 12 lines ("a cube") at about 30-35 fps.

Rotations about x,y, or z axes are over a fixed small angle though (of about 7 degrees). This I do to make the rotations fast. The angle is chosen such that sin(phi)=1/8, then cos(phi) is very close to 1-1/128-1/32768. Further, suppose you rotate first around the x-axis and then around the z-axis, then you can write:

x(i+2)=z(i)+(x(i+1)-z(i+1))*cos(phi)
y(i+2)=y(i)+(x(i+1)-z(i+1))*sin(phi)

where e.g. y(i) is the y-coord after the i-th rotation.
Now you need to calculate only 2 cos/sin terms, while with the rotation matrix there are 4 of such terms.

Similar equations can be derived for 2 consecutive rotations around any 2 axes.
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