3D-Animation plot with the hgtransform command

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Julián Francisco
Julián Francisco 2011 年 6 月 15 日
I continue to think of the possibilities for the 3d-animation (moving point along a curve). I have written the following code to try it using the hgtransform command but I do not understand why does not work.
t = 0:pi/50:10*pi;
x = sin(t);
y = cos(t);
z = t;
ah = axes;
set(ah,'XLim',[min(x) max(x)],'YLim',[min(y) max(y)],...
'ZLim',[min(z) max(z)]);
plot3(x,y,z,'Color','red');
hold on;
view(3);
hpoint = line('XData',x(1),'YData',y(1),'ZData',z(1),'Color','black','Marker',...
'o','MarkerSize',10,'MarkerFaceColor','black');
ht = hgtransform('parent',ah);
set(hpoint,'Parent',ht);
for i=2:length(x)
tx = x(i)-x(i-1);
ty = y(i)-y(i-1);
tz = z(i)-z(i-1);
trans = makehgtform('translate',[tx ty tz]),
set(ht,'Matrix',trans);
pause(0.01);
end
  4 件のコメント
Walter Roberson
Walter Roberson 2011 年 6 月 16 日
It rotates around for me. The only difficulty I observe is that the axis limits keep moving, shifting the plot back and forth.
Julián Francisco
Julián Francisco 2011 年 6 月 18 日
@Walter Roberson: Then I should freeze the axis limits.

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Patrick Kalita
Patrick Kalita 2011 年 6 月 17 日
I think the main reason why this doesn't do what you expect is this part right here:
tx = x(i)-x(i-1);
ty = y(i)-y(i-1);
tz = z(i)-z(i-1);
trans = makehgtform('translate',[tx ty tz]);
I think what you had indented was that at each iteration of the for-loop, you would translate the point from where it was at the last iteration to the next point. However, hgtransform doesn't work like that. The transformation don't accumulate. Once you set the transform's Matrix, that's it -- that's the transform. It doesn't matter what the Matrix used to be. (I think the documentation talks about that somewhere, but I can't seem to find it at the moment.)
So I think what you should do instead is first place the point at (0,0,0). On the first iteration, translate it to the first datapoint, on the next iteration translate to the second, etc. Like so:
t = 0:pi/50:10*pi;
x = sin(t);
y = cos(t);
z = t;
ah = axes;
set(ah,'XLim',[min(x) max(x)],'YLim',[min(y) max(y)],...
'ZLim',[min(z) max(z)]);
plot3(x,y,z,'Color','red');
hold on;
view(3);
hpoint = line('XData', 0,'YData', 0,'ZData', 0,'Color','black','Marker',...
'o','MarkerSize',10,'MarkerFaceColor','black');
ht = hgtransform('parent',ah);
set(hpoint,'Parent',ht);
for i=2:length(x)
tx = x(i);
ty = y(i);
tz = z(i);
trans = makehgtform('translate',[tx ty tz]);
set(ht,'Matrix',trans);
pause(0.01);
end
  3 件のコメント
Patrick Kalita
Patrick Kalita 2011 年 6 月 20 日
With the code as it is above, the point will never actually be seen at (0,0,0). The first time anything will actually be drawn is at the first call to "pause(0.01)". By that time the point will have already been translated to the first point of the trajectory. "Starting" the point at (0,0,0) is just to make the transformations easier -- it has absolutely no effect on what you would see on screen.
Julián Francisco
Julián Francisco 2011 年 11 月 5 日
@Patrick Kalita: Ok, you are right. Thank you back.

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