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Alexander Camargo

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- #1

Alexander Camargo

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Engineering news on Phys.org

- #2

berkeman

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This looks like it is for your schoolwork assignment, so please show us your work so far. Thank you.

- #3

Alexander Camargo

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It's not a school project, I have knowledge of mcnp, I just don't have experience with complex geometriesberkeman said:This looks like it is for your schoolwork assignment, so please show us your work so far. Thank you.

- #4

Alex A

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The tip could be two spherical shells connected to two cylinders. Can you see how you would do that?

- #5

Alexander Camargo

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no, sorry. Do I put everything in just one cell?Alex A said:

The tip could be two spherical shells connected to two cylinders. Can you see how you would do that?

my biggest problem is this curve.

When I add two cylinders, it looks like this.

- #6

Alex A

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Your dimensions seem to be off, are you putting diameter numbers into the radius? Show us your cell and surface definitions.

- #7

Alexander Camargo

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Alex A said:Depende para que serve, a camada externa de 3 cm de espessura parece precisar ser de um material, então pode ser uma célula.

Suas dimensões parecem estar erradas, você está colocando números de diâmetro no raio? Mostre-nos suas definições de célula e superfície.

I tried this, maybe I'm going the wrong way, but I'm trying to join these two surfaces to make the curve on the inside.

- #8

Alex A

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I'm not using Vised so our approaches may be very different.

The diagram has rotational symmetry, so I would try to construct it out of spheres, cylinders and cones. The diagram is missing some dimensions so some of the following is calculated and some has been guessed.

The purple section is made from spheres, blue and yellow sections are cylinders, green section is cones.

A cylinder is easy to join to a sphere at the full width, the cone is a bit more complicated. Every join uses a py plane surface. I've used a 1/1 angle, because it looks about right. It also makes the geometry maths easier. Every 1cm along the axis the cone is 1cm closer or further away. The inner cone needs to come in by 1.85cm so the point of the cone is 1.85cm higher up than the join point. -6cm + 1.85cm = -4.15

The inner cone needs to widen to meet the inner cylinder lower down. That has a radius of 3.6, from the other cylinder of radius 1.85. 3.6-1.85=1.75 and with an angle of 1/1 the height of the cone section is then also 1.75

The other cone section maths produces the same result. This shoulder looks a little thin, and could be improved by adding a band which would make it match the diagram better, but I'm not going to do that.

If the shaded area is needed as a single cell that can be done by a union of all the pieces. Because it is so long it needs to span more than one line. For example

This is just the shaded outer part. Other structures are needed, most can be defined with the existing surfaces, but you may need to add another plane and a cylinder or two. You may also need to correct my guesses as to the cylinder section lengths if you have actual measurements.

If you use this or you don't use this good luck either way!

Edit, I will just add that given this is a pinpoint ion chamber it probably makes more sense that the dimensions are in millimeters not centimeters. In which case all distances should be divided by 10.

The diagram has rotational symmetry, so I would try to construct it out of spheres, cylinders and cones. The diagram is missing some dimensions so some of the following is calculated and some has been guessed.

The purple section is made from spheres, blue and yellow sections are cylinders, green section is cones.

A cylinder is easy to join to a sphere at the full width, the cone is a bit more complicated. Every join uses a py plane surface. I've used a 1/1 angle, because it looks about right. It also makes the geometry maths easier. Every 1cm along the axis the cone is 1cm closer or further away. The inner cone needs to come in by 1.85cm so the point of the cone is 1.85cm higher up than the join point. -6cm + 1.85cm = -4.15

The inner cone needs to widen to meet the inner cylinder lower down. That has a radius of 3.6, from the other cylinder of radius 1.85. 3.6-1.85=1.75 and with an angle of 1/1 the height of the cone section is then also 1.75

The other cone section maths produces the same result. This shoulder looks a little thin, and could be improved by adding a band which would make it match the diagram better, but I'm not going to do that.

If the shaded area is needed as a single cell that can be done by a union of all the pieces. Because it is so long it needs to span more than one line. For example

Code:

```
2 2 0.001 (200 -101 100):(-200 201 -103 102):
(-201 202 301 -302):(-202 203 104 -105)
```

This is just the shaded outer part. Other structures are needed, most can be defined with the existing surfaces, but you may need to add another plane and a cylinder or two. You may also need to correct my guesses as to the cylinder section lengths if you have actual measurements.

If you use this or you don't use this good luck either way!

Edit, I will just add that given this is a pinpoint ion chamber it probably makes more sense that the dimensions are in millimeters not centimeters. In which case all distances should be divided by 10.

Last edited:

- #9

Alexander Camargo

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You opened my mind, thank you very much for the ideas. I will definitely use it. When I complete it, I'll post it here for anyone who wants to use the input with a pinpoint chambers.Alex A said:Não estou usando o Vised, então nossas abordagens podem ser muito diferentes.

O diagrama tem simetria rotacional, então eu tentei construí-lo a partir de esferas, cilindros e cones. O diagrama está faltando algumas dimensões, então a parte do seguinte é calculada e a parte foi adivinhada.

[ANEXO=completo]348493[/ANEXO]

A seção roxa é feita de esferas, as garrafas azul e amarela são cilindros e a seção verde é feita de cones.

Um cilindro é fácil de unir a uma esfera na largura total, o cone é um pouco mais complicado. Cada união usa uma superfície plana py. Eu uso um ângulo de 1/1, porque parece correto. Também torna a matemática geométrica mais fácil. A cada 1 cm ao longo do eixo, o cone fica 1 cm mais perto ou mais longe. O cone interno precisa entrar em 1,85 cm para que o ponto do cone fique 1,85 cm mais alto do que o ponto de união. -6cm + 1,85cm = -4,15

O cone interno precisa ser alargado para encontrar o cilindro interno mais abaixo. Isso tem um raio de 3,6, do outro cilindro de raio 1,85. 3,6-1,85=1,75 e com um ângulo de 1/1 a altura da seção do cone também é 1,75

A outra matemática da seção cônica produz o mesmo resultado. Este ombro parece um pouco fino, e poderia ser melhorado adicionando uma faixa que o combinaria melhor com o diagrama, mas não vou fazer isso.

Se uma área sombreada por necessidade como uma única célula, isso pode ser feito por uma união de todas as peças. Por ser tão longo, precisa abranger mais de uma linha. Por exemplo [código]2 2 0,001 (200 -101 100):(-200 201 -103 102):

(-201 202 301 -302):(-202 203 104 -105)[/código]

Esta é apenas a parte externa sombreada. Outras estruturas são fáceis, a maioria pode ser definida com as superfícies existentes, mas você pode precisar adicionar outro plano e um cilindro ou dois. Você também pode precisar concordar com meus palpites quanto aos comprimentos das garrafas do cilindro se tiver medidas reais.

Se você usar isso ou não, boa sorte de qualquer maneira!

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