使用未知固定点的给定距离进行 3D 三边测量

发布于 2024-10-17 04:26:24 字数 954 浏览 12 评论 0原文

我是这个论坛的新手，不是一个以英语为母语的人，所以请友善！ :)

这是我目前面临的挑战：我想根据 2 点之间的一组给定距离计算 3D 欧几里德空间中未知点的（近似）相对坐标。在我的第一种方法中，我想忽略可能的多个解决方案，只是随机采用第一个解决方案。

例如：给定的一组距离：（我认为它创建了一个以直角三角形为基础的金字塔）

P1-P2-距离

1-2-30
2-3-40
1-3-50
1-4-60
2-4 -60
3-4-60

第 1 步： 现在，如何计算这些点的相对坐标？
我认为第一个点为 0,0,0，所以第二个点为 30,0,0。
之后，可以通过找到点 1 和 2 的 2 个圆的交点以及它们到点 3 的距离（分别为 50 和 40）来计算第三个点。我如何从数学上做到这一点？（尽管我用这些简单的数字来简单地表示我脑海中的情况）。此外，我不知道如何以正确的数学方式得到答案，第三点位于 30,40,0（或 30,0,40，但我会忽略它）。
但获得第四点却没有那么容易。我认为我必须使用 3 个球体来计算交叉点才能得到该点，但我该怎么做呢？

第2步： 在我弄清楚如何计算这个“简单”示例之后，我想使用更多未知点...对于每个点，到另一个点的最小给定距离为 1，以将其“链接”到其他点。如果坐标由于其自由度而无法计算，我想忽略所有可能性，除了我随机选择的一种，但相对于已知距离。

第三步： 现在最后阶段应该是这样的：由于现实生活中的情况，每个测量的距离都有点不正确。因此，如果给定点对的距离超过 1 个，则对距离进行平均。但由于距离不精确，确定点的确切（相对）位置可能会很困难。所以我想将不同的可能位置平均到“最佳”位置。

你能帮助我一步步完成我的挑战吗？

原文

I am new to this forum and not a native english speaker, so please be nice! :)

Here is the challenge I face at the moment:
I want to calculate the (approximate) relative coordinates of yet unknown points in a 3D euclidean space based on a set of given distances between 2 points.
In my first approach I want to ignore possible multiple solutions, just taking the first one by random.

e.g.:
given set of distances: (I think its creating a pyramid with a right-angled triangle as a base)

P1-P2-Distance

1-2-30
2-3-40
1-3-50
1-4-60
2-4-60
3-4-60

Step1:
Now, how do I calculate the relative coordinates for those points?
I figured that the first point goes to 0,0,0 so the second one is 30,0,0.
After that the third points can be calculated by finding the crossing of the 2 circles from points 1 and 2 with their distances to point 3 (50 and 40 respectively). How do I do that mathematically? (though I took these simple numbers for an easy representation of the situation in my mind). Besides I do not know how to get to the answer in a correct mathematical way the third point is at 30,40,0 (or 30,0,40 but i will ignore that).
But getting the fourth point is not as easy as that. I thought I have to use 3 spheres in calculate the crossing to get the point, but how do I do that?

Step2:
After I figured out how to calculate this "simple" example I want to use more unknown points... For each point there is minimum 1 given distance to another point to "link" it to the others. If the coords can not be calculated because of its degrees of freedom I want to ignore all possibilities except one I choose randomly, but with respect to the known distances.

Step3:
Now the final stage should be this: Each measured distance is a bit incorrect due to real life situation. So if there are more then 1 distances for a given pair of points the distances are averaged. But due to the imprecise distances there can be a difficulty when determining the exact (relative) location of a point. So I want to average the different possible locations to the "optimal" one.

Can you help me going through my challenge step by step?

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没有心的人 2024-10-24 04:26:24

您需要使用三角学 - 具体来说，是“余弦规则”。这将为您提供三角形的角度，从而让您解决第三点和第四点。

规则规定，

c^2 = a^2 + b^2 - 2abCosC

其中 a、b 和 c 是边的长度，C 是与边 c 相对的角度。

在您的例子中，我们需要 1-2 和 1-3 之间的角度 - 两条线在 (0,0,0) 处交叉的角度。它将是 90 度，因为您有 3-4-5 三角形，但让我们证明：

50^2 = 30^2 + 40^2 - 2*30*40*CosC
CosC = 0
C = 90 degrees

这是线 (0,0,0)-(30,0,0) 和 (0,0,0 )- 第 3 点；沿该线延伸边 1-3 的长度（即 50），您将得到第二个点 (0,50,0)。

找到第四点有点棘手。我能想到的最简单的算法是首先找到点的 (x,y) 分量，然后使用毕达哥拉斯的 z 分量很简单。

考虑 (x,y,0) 平面上有一个点位于点 4 的正下方 - 将此点称为 5。您现在可以创建 3 个直角三角形 1-5-4、2-5- 4、和3-5-4。

你知道 1-4、2-4 和 3-4 的长度。因为它们是直角三角形，所以比率 1-4 : 2-4 : 3-4 等于 1-5 : 2-5 : 3-5。使用三角方法找到点 5 - “正弦规则”将为您提供 1-2 和 1-2 之间的角度。 1-4、2-1 和 2-4 等。

“正弦规则”规定（在直角三角形中）

a / SinA = b / SinB = c / SinC

因此，对于三角形 1-2-4，尽管您不知道长度 1-4 和 2-4，你确实知道比例1-4 : 2-4。同样，您也知道其他三角形中的比率 2-4 : 3-4 和 1-4 : 3-4。

我会让你去解决第 4 点。一旦你有了这一点，你就可以使用毕达哥拉斯轻松地解决 4 的 z 分量 - 你将得到边 1-4、1-5，长度 4-5 将是z 分量。

You need to use trigonometry - specifically, the 'cosine rule'. This will give you the angles of the triangle, which lets you solve the 3rd and 4th points.

The rules states that

c^2 = a^2 + b^2 - 2abCosC

where a, b and c are the lengths of the sides, and C is the angle opposite side c.

In your case, we want the angle between 1-2 and 1-3 - the angle between the two lines crossing at (0,0,0). It's going to be 90 degrees because you have the 3-4-5 triangle, but let's prove:

50^2 = 30^2 + 40^2 - 2*30*40*CosC
CosC = 0
C = 90 degrees

This is the angle between the lines (0,0,0)-(30,0,0) and (0,0,0)- point 3; extend along that line the length of side 1-3 (which is 50) and you'll get your second point (0,50,0).

Finding your 4th point is slightly trickier. The most straightforward algorithm that I can think of is to firstly find the (x,y) component of the point, and from there the z component is straightforward using Pythagoras'.

Consider that there is a point on the (x,y,0) plane which sits directly 'below' your point 4 - call this point 5. You can now create 3 right-angled triangles 1-5-4, 2-5-4, and 3-5-4.

You know the lengths of 1-4, 2-4 and 3-4. Because these are right triangles, the ratio 1-4 : 2-4 : 3-4 is equal to 1-5 : 2-5 : 3-5. Find the point 5 using trigonometric methods - the 'sine rule' will give you the angles between 1-2 & 1-4, 2-1 and 2-4 etc.

The 'sine rule' states that (in a right triangle)

a / SinA = b / SinB = c / SinC

So for triangle 1-2-4, although you don't know lengths 1-4 and 2-4, you do know the ratio 1-4 : 2-4. Similarly you know the ratios 2-4 : 3-4 and 1-4 : 3-4 in the other triangles.

I'll leave you to solve point 4. Once you have this point, you can easily solve the z component of 4 using pythagoras' - you'll have the sides 1-4, 1-5 and the length 4-5 will be the z component.

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ぶ宁プ宁ぶ 2024-10-24 04:26:24

我最初假设您知道所有点对之间的距离。

正如您所说，您可以选择一个点 (A) 作为原点，沿 x 轴定向第二个点 (B)，然后放置第三个点 (< code>C) 沿 xy 平面。您可以按如下方式求解 C 的坐标：

given: distances ab, ac, bc
assume
A = (0,0)
B = (ab,0)
C = (x,y)  <- solve for x and y, where:
  ac^2 = (A-C)^2 = (0-x)^2 + (0-y)^2 = x^2 + y^2
  bc^2 = (B-C)^2 = (ab-x)^2 + (0-y)^2 = ab^2 - 2*ab*x + x^2 + y^2

-> bc^2 - ac^2 = ab^2 - 2*ab*x
-> x = (ab^2 + ac^2 - bc^2)/2*ab
-> y = +/- sqrt(ac^2 - x^2)

为了使其准确工作，您需要避免点 {A,B,C} 位于一条直线或接近该直线的情况。

求解 3 空间中的其他点是类似的 - 您可以扩展距离的毕达哥拉斯公式，取消二次元，并求解所得的线性系统。但是，这并不能直接帮助您完成步骤 2 和 3...

不幸的是，我也不知道步骤 2 和 3 的良好精确解决方案。您的总体问题通常是过度约束（由于冲突的噪声距离）和约束不足（由于缺少距离）。

您可以尝试迭代求解器：从所有点的随机放置开始，将当前距离与给定距离进行比较，然后使用它来调整您的点以改进匹配。这是一种优化技术，所以我会查找有关数值优化的书籍。

I'll initially assume you know the distances between all pairs of points.

As you say, you can choose one point (A) as the origin, orient a second point (B) along the x-axis, and place a third point (C) along the xy-plane. You can solve for the coordinates of C as follows:

given: distances ab, ac, bc
assume
A = (0,0)
B = (ab,0)
C = (x,y)  <- solve for x and y, where:
  ac^2 = (A-C)^2 = (0-x)^2 + (0-y)^2 = x^2 + y^2
  bc^2 = (B-C)^2 = (ab-x)^2 + (0-y)^2 = ab^2 - 2*ab*x + x^2 + y^2

-> bc^2 - ac^2 = ab^2 - 2*ab*x
-> x = (ab^2 + ac^2 - bc^2)/2*ab
-> y = +/- sqrt(ac^2 - x^2)

For this to work accurately, you will want to avoid cases where the points {A,B,C} are in a straight line, or close to it.

Solving for additional points in 3-space is similar -- you can expand the Pythagorean formula for the distance, cancel the quadratic elements, and solve the resulting linear system. However, this does not directly help you with your steps 2 and 3...

Unfortunately, I don't know a well-behaved exact solution for steps 2 and 3, either. Your overall problem will generally be both over-constrained (due to conflicting noisy distances) and under-constrained (due to missing distances).

You could try an iterative solver: start with a random placement of all your points, compare the current distances with the given ones, and use that to adjust your points in such a way as to improve the match. This is an optimization technique, so I would look up books on numerical optimization.

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