Math Stuffs

[Helz]

So basically this is about calculating the center of an arch with 3 known points of a circle. Its piratical application is drawing out the curve to create the frame for an arched bridge to lay down concrete. There is an 'easy enough' piratical way of doing it but I am trying to figure out a way to actually calculate the circles center point in order to run a string and draw the exact arch.

Like..

If you have 3 known points for your bridge. How could you calculate X

Circle.jpg

Its something like this. My first thought was to measure between points A and B and then pivot the B point at that controlled distance to be centered between points A and C but it was not perfect. It was either because the 2 points were not in line with the diameter of the circle or because it was not a perfect circle. Regardless I feel like a center point can be calculated to allow an arch to be sketched that connects the 3 points with the available information.

Kinda trying to turn the 'guesswork bullshit' they are teaching right now into actual calculations to turn blueprints into exact measurements in my degree program.. I am just too stupid to see how to do it at the moment.

[Firebringer]

So wait, are you looking for the radius of this? I mean, is X the center and you want to find distance between each point to the radius?

I mean, if you have X and Y coordinates for this, it should be simple enough.

Let Circle be X^2 + Y^2 + Dx + Ey +F = 0
You plug each of the points into this forumula.
Then you simplify, then eliminate one equation by making it equal to the other.
You will get a solution for either D or E, subsittute that in for the other. You will get F, then you will get the equation of the circle.

Once you get the equation of the circle, you have to adjust it to the make it a in a easier form to find radius. This involves moving the solo number to other side. Then grouping all Xs and Ys together, and square rooting other side (which will be the radius of the circle).

[Funce]

Orthogonal Vectors at each point and find the point of intersection of said vectors? EDIT: Derping from sleep, this may not work.

Orpz knows maths.

[Firebringer]

I need sleep as well!

[ika]

@Orpz

Do math

[Frozen Angel]

draw all three perpendicular bisectors of all three AB , BC and AC lines (two are needed - the third for you too see they intersect in one dot ) that dot is X

[MattZed]

The trick here is to realize that the center is the ONLY point that is an equal distance from all three points. (that is, assuming the points are not all part of a line)

So all we have to do is find the one point that's equi-distant from the others. Here's the technique for doing that that seems most intuitive to me:


To make life easier, let's say point A has coordinates (d,e), point B has (f,g), and point C has (i,j)

First, we find the line of points that are equally between points A and B.

Line1.png

There are a few ways to do this. One way to do this is find the midpoint of A and B, . Then we just have to find the slope of this line, which is going to be perpendicular to the line from A to B. We can compute, by the usual "rise over run" formula, that the slope of the line from A to B is . Thus, the slope of its perpendicular line is just the negative reciprocal, . Now, we can write an equation for the line of points equally between A and B using the high school "point-slope" formula. That is, we get:

Which we can "simplify" to:

But to make our lives easier, we'll say:

So our equation just becomes:

Now, we could have done this exact same procedure with points C and B instead of A and B. Then, we take the intersection of these two lines, which will be a single point. This point has an equal distance from A, B, and C, and thus must be the center!
Line 2.png

So all we have to do is compute the line of points equally between C and B, and then find the intersection of our two lines.


To find an equation for our secont, we just replacee d with i and e with j from our first equation. (since this is all that would change if we computed it from scratch) Thus, the equation for the line of points between C and B is just:

Which, making our lives easier, we'll write as:

To finish, we just need to find the intersection of and . Since the right-hand side of each equation equals y, they are equal to each other, so , or solving for x we get:

Plugging this value of x into the first equation (and simplifying a bit) gives us:

The point (x,y) will be the center of the circle.

[Frozen Angel]

your formulas are broken. but thats exactly why they use perpendicular bisectors. these are the points which has exactly equal distance from both ends