Trying to calculate the material required to 3Dprint each "donut" layer of a hemisphere. I need to do this by using the spherical cap formula and not volumes of washer shapes.

Question

Adam Takeshita 2023-5-9

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1960514-trying-to-calculate-the-material-required-to-3dprint-each-donut-layer-of-a-hemisphere-i-need-to-d

回答： Nivedita 2023-8-23

ro = 63.95; %outer radius

ri = 63.05; %inner radius

ho = 0.9:0.9:63.9; %vary outer height by a step of 0.9

hi = 0:0.9:63; %vary inner height by a step of 0.9

v = zeros(1,length(ho));

for i = 1:length(ho)

out(i) = 1/3*pi*(ho(i)^2*(3*ro-ho(i))); %calculate outer cap volume

in(i) = 1/3*pi*(hi(i)^2*(3*ri-hi(i))); %calculate inner cap volume

v(i) = out(i) - in(i); %calculate remaining shell volume

end

Lay = diff(v) %calculate washer-shaped layer volume as h0 and hi vary. They are all coming out as constant, though it take less volume to print a layer at the top of a hemisphere vs the bottom

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Nivedita 2023-8-23

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1960514-trying-to-calculate-the-material-required-to-3dprint-each-donut-layer-of-a-hemisphere-i-need-to-d#answer_1292287

Hi Adam!

I understand that you are experiencing issues while trying to calculate the volume of a section of a hollow hemisphere.

From my understanding of your implemented code, since "ri" and "ro" are constants and the "hi" and "ho" values are incrementing by the same step value of 0.9, the difference between the consecutive values of "v" will be the same.

Also, since you want to calculate the volume in donut shaped layers, I would suggest varying the radii as well. Moving towards the top of the hemisphere, the radius should be smaller and towards the centre of the hemisphere, the radius should increase.

The code implemented computes the volume in bowl shaped hemispherical layers with varying heights rather than donut or disc shaped layers.

I hope I was able to answer your question.