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New shape of the target #745
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Hey @mimboch, I've adapted the basic target demo using expressions from the maths parser to obtain the following target: which was created with this input deck:
Hope this helps, |
Thank you so much for your assistance . |
Can you be more specific about the curve-shape? For an exponential, I would need to know where you want it to start from, which point on the circle you want it to end, and the exponential scale-length If the older ball-rod target above doesn't work for you, then your results must be sensitive to the nature of this curve. If I just give you a random curve, it will probably have the same problem as the simple ball-rod. |
Hello, |
Hello Stuart, leftCirclePosition = [0, -2]; % Bottom-left corner of the left circular end % Arcs % Transformations % Visualization figure; hold on; % Right circular end % Top and bottom arcs % Combine arcs into a shape % Transform the left shape to create the right shape % Plot the shapes % Visualization settings % Functions function [x, y] = topArc(startPoint, endPoint, radius) |
Hello,
I'm trying to simulate the following shape of the jet gas target, with circular ends and a flat section in the middle .
I would be very grateful if anyone could give me a hand.
Thank you in advance.
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