Perspective Projection in Android in an augmented reality application(Android 中的透视投影在增强现实应用程序中)

问题描述

目前我正在编写一个增强现实应用程序，但在将对象显示在屏幕上时遇到了一些问题.令我非常沮丧的是，我无法将 gps 点转换为我的 android 设备上的相应屏幕点.我已经阅读了很多关于 stackoverflow 的文章和许多其他帖子(我已经问过类似的问题)，但我仍然需要你的帮助.

我做了维基百科中解释的透视投影.

我要如何处理透视投影的结果才能获得最终的屏幕点?

前段时间看维基百科的文章也让我很困惑.这是我尝试以不同的方式解释它:

情况

让我们简化一下情况.我们有:

• 我们的投影点 D(x,y,z) - 你所说的 relativePositionX|Y|Z
• 大小为 w * h
的图像平面
• ，我们得到:
  • MB/CD = EM/EC <=> X/x = f/z (2)

  同时使用 (1) 和 (2)，我们现在有:

  • X = (x/z) * ( (w/2)/tan(α) )

  如果我们回到维基百科文章中使用的符号，我们的等式相当于:

  • b_x = (d_x/d_z) * r_z

  您会注意到我们缺少 s_x/r_x 的乘法.这是因为在我们的例子中，显示尺寸"和记录面"是相同的，所以s_x/r_x = 1.

  <块引用>

  注意:Y 的推理相同.

  <小时>

  实际使用

  一些备注:

  • 通常使用α = 45deg，即tan(α) = 1.这就是为什么这个术语没有出现在许多实现中的原因.

  • 如果你想保持你显示的元素的比例，保持f对于X和Y都保持不变，即而不是计算:

    • X = (x/z) * ( (w/2)/tan(α) ) 和 Y = (y/z) * ( (h/2)/tan(α))

    ...做:

    • X = (x/z) * ( (min(w,h)/2)/tan(α) ) 和 Y = (y/z) * ( (min(w,h)/2)/tan(α) )
    <块引用>

    注意:当我说显示尺寸"和记录表面"是相同的"，这不太正确，min操作是为了补偿这个近似值，适应方形表面 r 到潜在矩形表面 s.

    注意 2:Appunta 不使用 min(w,h)/2，而是使用 screenRatio=(getWidth()+getHeight())/2 如您所见.两种解决方案都保留了元素比率.焦点，因此视角，只会有点不同，取决于屏幕本身的比例.您实际上可以使用任何您想要的功能定义 f.

  • 您可能已经在上图中注意到，屏幕坐标在这里定义在 [-w/2 ;w/2] 用于 X 和 [-h/2 ;h/2] 表示 Y，但您可能想要 [0 ;w] 和 [0 ;h] 代替.X += w/2 和 Y += h/2 - 问题已解决.

  <小时>

  结论

  我希望这将回答您的问题.如果需要版本，我会留在附近.

  再见！

  <块引用>

  <自我推销警报 > 其实我前段时间做了一个 文章关于 3D 投影和渲染.实施在Javascript，但应该很容易翻译.

  Currently I'm writing an augmented reality app and I have some problems to get the objects on my screen. It's very frustrating for me that I'm not able to transform gps-points to the correspending screen-points on my android device. I've read many articles and many other posts on stackoverflow (I've already asked similar questions) but I still need your help.

  I did the perspective projection which is explained in wikipedia.

  What do I have to do with the result of the perspective projection to get the resulting screenpoint?

  解决方案

  The Wikipedia article also confused me when I read it some time ago. Here is my attempt to explain it differently:

  ---

  The Situation

  Let's simplify the situation. We have:

  • Our projected point D(x,y,z) - what you call relativePositionX|Y|Z
  • An image plane of size w * h
  • A half-angle of view α

  ... and we want:

  • The coordinates of B in the image plane (let's call them X and Y)

  A schema for the X-screen-coordinates:

  E is the position of our "eye" in this configuration, which I chose as origin to simplify.

  The focal length f can be estimated knowing that:

  • tan(α) = (w/2) / f (1)

  ---

  A bit of Geometry

  You can see on the picture that the triangles ECD and EBM are similar, so using the Side-Splitter Theorem, we get:

  • MB / CD = EM / EC <=> X / x = f / z (2)

  With both (1) and (2), we now have:

  • X = (x / z) * ( (w / 2) / tan(α) )

  If we go back to the notation used in the Wikipedia article, our equation is equivalent to:

  • b_x = (d_x / d_z) * r_z

  You can notice we are missing the multiplication by s_x / r_x. This is because in our case, the "display size" and the "recording surface" are the same, so s_x / r_x = 1.

  Note: Same reasoning for Y.

  ---

  Practical Use

  Some remarks:

  • Usually, α = 45deg is used, which means tan(α) = 1. That's why this term doesn't appear in many implementations.

  • If you want to preserve the ratio of the elements you display, keep f constant for both X and Y, ie instead of calculating:

    • X = (x / z) * ( (w / 2) / tan(α) ) and Y = (y / z) * ( (h / 2) / tan(α) )

    ... do:

    • X = (x / z) * ( (min(w,h) / 2) / tan(α) ) and Y = (y / z) * ( (min(w,h) / 2) / tan(α) )

    Note: when I said that "the "display size" and the "recording surface" are the same", that wasn't quite true, and the min operation is here to compensate this approximation, adapting the square surface r to the potentially-rectangular surface s.

    Note 2: Instead of using min(w,h) / 2, Appunta uses screenRatio= (getWidth()+getHeight())/2 as you noticed. Both solutions preserve the elements ratio. The focal, and thus the angle of view, will simply be a bit different, depending on the screen's own ratio. You can actually use any function you want to define f.

  • As you may have noticed on the picture above, the screen coordinates are here defined between [-w/2 ; w/2] for X and [-h/2 ; h/2] for Y, but you probably want [0 ; w] and [0 ; h] instead. X += w/2 and Y += h/2 - Problem solved.

  ---

  Conclusion

  I hope this will answer your questions. I'll stay near if it needs editions.

  Bye!

  < Self-promotion Alert > I actually made some time ago an article about 3D projection and rendering. The implementation is in Javascript, but it should be quite easy to translate.

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