問題描述

目前我正在編寫一個(gè)增強(qiáng)現(xiàn)實(shí)應(yīng)用程序，但在將對(duì)象顯示在屏幕上時(shí)遇到了一些問題.令我非常沮喪的是，我無法將 gps 點(diǎn)轉(zhuǎn)換為我的 android 設(shè)備上的相應(yīng)屏幕點(diǎn).我已經(jīng)閱讀了很多關(guān)于 stackoverflow 的文章和許多其他帖子(我已經(jīng)問過類似的問題)，但我仍然需要你的幫助.

我做了維基百科中解釋的透視投影.

我要如何處理透視投影的結(jié)果才能獲得最終的屏幕點(diǎn)?

前段時(shí)間看維基百科的文章也讓我很困惑.這是我嘗試以不同的方式解釋它:

情況

讓我們簡(jiǎn)化一下情況.我們有:

• 我們的投影點(diǎn) D(x,y,z) - 你所說的 relativePositionX|Y|Z
• 大小為 w * h
的圖像平面
• ，我們得到:
  • MB/CD = EM/EC <=> X/x = f/z (2)

  同時(shí)使用 (1) 和 (2)，我們現(xiàn)在有:

  • X = (x/z) * ( (w/2)/tan(α) )

  如果我們回到維基百科文章中使用的符號(hào)，我們的等式相當(dāng)于:

  • b_x = (d_x/d_z) * r_z

  您會(huì)注意到我們?nèi)鄙?s_x/r_x 的乘法.這是因?yàn)?strong>在我們的例子中，顯示尺寸"和記錄面"是相同的，所以s_x/r_x = 1.

  <塊引用>

  注意:Y 的推理相同.

  <小時(shí)>

  實(shí)際使用

  一些備注:

  • 通常使用α = 45deg，即tan(α) = 1.這就是為什么這個(gè)術(shù)語(yǔ)沒有出現(xiàn)在許多實(shí)現(xiàn)中的原因.

  • 如果你想保持你顯示的元素的比例，保持f對(duì)于X和Y都保持不變，即而不是計(jì)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) )
    <塊引用>

    注意:當(dāng)我說顯示尺寸"和記錄表面"是相同的"，這不太正確，min操作是為了補(bǔ)償這個(gè)近似值，適應(yīng)方形表面 r 到潛在矩形表面 s.

    注意 2:Appunta 不使用 min(w,h)/2，而是使用 screenRatio=(getWidth()+getHeight())/2 如您所見.兩種解決方案都保留了元素比率.焦點(diǎn)，因此視角，只會(huì)有點(diǎn)不同，取決于屏幕本身的比例.您實(shí)際上可以使用任何您想要的功能定義 f.

  • 您可能已經(jīng)在上圖中注意到，屏幕坐標(biāo)在這里定義在 [-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 - 問題已解決.

  <小時(shí)>

  結(jié)論

  我希望這將回答您的問題.如果需要版本，我會(huì)留在附近.

  再見！

  <塊引用>

  <自我推銷警報(bào) > 其實(shí)我前段時(shí)間做了一個(gè) 文章關(guān)于 3D 投影和渲染.實(shí)施在Javascript，但應(yīng)該很容易翻譯.

  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?

  解決方案

  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.

  這篇關(guān)于Android 中的透視投影在增強(qiáng)現(xiàn)實(shí)應(yīng)用程序中的文章就介紹到這了，希望我們推薦的答案對(duì)大家有所幫助，也希望大家多多支持html5模板網(wǎng)！

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