More of a maths Q than Excel, But I need to link it to excel

The Idea Dude

Well-known Member
Joined
Aug 15, 2002
Messages
589
OK, here goes.

Draw a straight horizontal line of a known length (say 50). At the right hand end of that line, draw a vertical line of the same length. Join the two ends with a curve so that you end up with a picture that looks like the top left quadrant of a circle.

I need a formula that will identify the vertical distance between the horizontal line and the arc for any location along the horizontal line.

SO, as an example, for the point on the horizontal line that is 10 away from the vert line, I need to be able to get the vertical distance between that point and the arc.

Hope this makes sense. :)

Thanks
 

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Jay Petrulis

MrExcel MVP
Joined
Mar 17, 2002
Messages
2,040
Hi,

If the "curve" drawn is truly circular, you can use the Pythagorean Theorem to find the height (y) from any point (x) on the horizontal line.

Consider the point of intersection between the vertical and horizontal lines as the origin. The hypotenuse will always be the radius of the "circle."

So, for your example, the radius (z) = 50, and the horizontal distance (x) = 10, then the vertical distance is Sqr(50^2 - 10^2)

The mathematics:
x^2 + y^2 = z^2, solve for y

y = (z^2 - x^2)^(1/2)

In Excel, with the 50 in A1 and 10 in A2,
=SQRT(A1^2-A2^2)

I hope that I read this post correctly.
 

BigC

Well-known Member
Joined
Aug 4, 2002
Messages
851
If I understand you correctly, the curve is convex (ie bows outward from the ends of each line rather than in towards the intersection of the two)
If so, then:
1.) Let "z" = the distance from the intersection of the two lines to the point on the arc (already known as the height of the vertical line [50] = radius of arc)
2) Let "y" = the distance of the point on the horizontal line vertically below the arc point from the intersection of the other two lines (already known as per your example (10),
3) A simple right angled triangle can now be formed by drawing straight lines from the intersection of the two original lines to the point on the arc you wish to measure (z), and from that point vertically to the horizontal line (call this new vertical line "x").
Now, with 2 out of 3 values required, you can do the "hippopotumus" calculation, viz;
Square Root of z = x^2 + y^2, or inversely for your calc:
x = SqRoot of (z^2 - y^2)

x = SqRt((50^2)-(10^2))
x = SqRt (2500 - 100)
x = SqRt (2400)
x = 48.99

HTH
BigC
 

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