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VPython对象革命
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Stack Overflow用户
提问于 2018-05-30 17:06:52
回答 1查看 1.2K关注 0票数 0

现在必须使用VPython,我想建立一个太阳系的模型。

目前我有所有的行星和轨道环,然而,实际的轨道是我所发现的非常困难的。

代码语言:javascript
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GlowScript 2.7 VPython
from visual import *



# Declaring Celestial Body Objects
Sun     = sphere(pos =     vec(0, 0, 0), radius =   10, color = color.yellow)
Mercury = sphere(pos =    vec(25, 0, 0), radius =    2, color =  color.green)
Venus   = sphere(pos =    vec(40, 0, 0), radius =  2.5, color =    color.red)
Earth   = sphere(pos =    vec(50, 0, 0), radius = 2.65, color =   color.blue)
Mars    = sphere(pos =    vec(70, 0, 0), radius =  2.3, color =    color.red)
Jupiter = sphere(pos =    vec(90, 0, 0), radius =    3, color = color.orange)
Saturn  = sphere(pos =   vec(105, 0, 0), radius =  2.9, color = color.orange)
Uranus  = sphere(pos = vec(117.5, 0, 0), radius =  2.9, color = color.orange)
Neptune = sphere(pos =   vec(135, 0, 0), radius =  2.8, color =   color.blue)
Pluto   = sphere(pos =   vec(165, 0, 0), radius =  1.5, color =  color.white)



# Declaring Orbital Rings of Celestial Body Objects
Mercury.ring = ring(pos = vec(0, 0, 0), axis = vec(0, 1, 0), size = vec(0.1, Mercury.pos.x * 2, Mercury.pos.x * 2))
Venus.ring   = ring(pos = vec(0, 0, 0), axis = vec(0, 1, 0), size = vec(0.1, Venus.pos.x * 2, Venus.pos.x * 2))
Earth.ring   = ring(pos = vec(0, 0, 0), axis = vec(0, 1, 0), size = vec(0.1, Earth.pos.x * 2, Earth.pos.x * 2))
Mars.ring    = ring(pos = vec(0, 0, 0), axis = vec(0, 1, 0), size = vec(0.1, Mars.pos.x * 2, Mars.pos.x * 2))
Jupiter.ring = ring(pos = vec(0, 0, 0), axis = vec(0, 1, 0), size = vec(0.1, Jupiter.pos.x * 2, Jupiter.pos.x * 2))
Saturn.ring  = ring(pos = vec(0, 0, 0), axis = vec(0, 1, 0), size = vec(0.1, Saturn.pos.x * 2, Saturn.pos.x * 2))
Uranus.ring  = ring(pos = vec(0, 0, 0), axis = vec(0, 1, 0), size = vec(0.1, Uranus.pos.x * 2, Uranus.pos.x * 2))
Neptune.ring = ring(pos = vec(0, 0, 0), axis = vec(0, 1, 0), size = vec(0.1, Neptune.pos.x * 2, Neptune.pos.x * 2))
Pluto.ring   = ring(pos = vec(0, 0, 0), axis = vec(0, 1, 0), size = vec(0.1, Pluto.pos.x * 2, Pluto.pos.x * 2))



# Infinite Loop
while 1 == 1:

    Mercury.rotate(angle = radians(360), axis = vec(Mercury.pos.y, Mercury.pos.x, 0), origin = vec(0, 0, 0))
    rate(50)

print("Error! Escaped While Loop!")

当我用Mercury.rotate(angle = 0.0174533, axis = vec(0, Mercury.pos.x, 0), origin = vec(0, 0, 0))切换出旋转方法时,它会正确地旋转.但只有四分之一的轮值。我读过关于的一切,但是N/A。

在四分之革命之后,当角度较大时,行星有时会决定以暴力的方式“夺取”。这似乎是某种障碍。

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回答 1

Stack Overflow用户

发布于 2018-11-15 18:18:41

您应该编写axis=vec(0,1,0)。旋转轴需要总是指向向上。

票数 0
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/50610101

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