geometry problem

Hi Netmaker

This should work for you if you know the radius of the arc.

Let us label the center of the circle that the arc is part of "X".

Then AX the radius of the arc is the hypotenuse of a right triangle.

AD (1/2 of line ADC) is one of the legs of the right triangle.

Pythagoras said that AD² + DX² = AX² (radius)

Solving for DX: AX² - AD² = DX²

BDX is also a radius. Subtract DX from BDX = BD

TM

Tom,

Thank you for this.

I am just a bit confused with the formula.

Could you work out one for me given real dimensions ?

I have an arc that is 30'.

I have a line touching the bottom of the arc at the end points, that is 25'.

How far is the center of the arc from the center of the line?

thank you in advance for this. I know it is simple to you guys but I am just a dumb cajun with limited math skills.

merci beaucoup mon amis

Greg

net-maker@cox.net

Tom,

That's all it took my friend. I see where it's going from your diagram.

We have been designing nets for about 30 years and have many " so called" formulas for cutting tapers in webbing and such.

we are now getting into some very detailed deepwater net designs and have to know how much netting is interacting with the water so the scientists can estimate a drag quotient.

I design the bigger nets with a lot of estimating, as a 20 or 30 meter net can be +/- a few feet in this 'radius'. But now , we are doing some really small scale nets and scaling down nets from 90 M to 1 M is totally different.

i have been to Oregon along the coast and got to enjoy your sideways rain.

I still rate Astoria as one of the prettiest places in the USA. I hope the real estate guys don't frick it up like they do every other nice town.

Thank you again for the time.

some of these folks were very well meaning but they lost me in an instant with the high tech talk.

take care.

Greg

Hi Greg

Thank you. You had better hurry if you want a piece of historic Astoria. The Real- Aters are doing their best. We only tell people about the rain to hold down the population. I am surprised you actually found some. Well the horizontal kind is our very best. Net em long, deep and often. Tom

I am confused. If you define the points for the arc as A, B, and C; the chord as A and C; and D the midpoint of the chord; then you already have your answer! Why? Because you must know the coordinates of A, B, C, and D, right? So the length of the line that is D to B is simply:

Length = √ ∆X² + ∆Y²

Otherwise known as the Pythagorean Theorem. Where ∆X is the difference of the X coordinate of point D and the X coordinate of point B. ∆Y is the difference in the Y coordinate of point D and the Y coordinate of point B. Square the ∆X, square the ∆Y, add them together, and then take the square root of the sum. The result is the length of the line that bisects the arc and the chord. It's just that simple.

I should add one more thing. Suppose that point B lies along the arc, but is not equidistant from points A and C. Now you have a math challenge! You want to know the height of the line that is perpendicular to line A-C, bisects line A-C, and intersects the arc. That point along line A-C you designated as point D. It is easy to do if you know the coordinates of the origin of the arc. How do you do that?

You can determine the origin of an arc or circle given two chords. In this case there are 3 chords; A-B, A-C, and B-C.

If you calculate the center point for one chord you can calculate the line that bisects that chord at a right angle. That line will pass through the radius of the circle or arc! However, you don't know where that point is along the line. Using a second chord you can calculate the line that bisects the second chord's line at a right angle.

Given the equations for the two new lines it is a simple matter to calculate the intersection of those two new lines. Remember Y = mX + b? That intersection is the radius of the circle or arc!

Now that you know where the radius is you can calculate the length from the radius to any point (A, B, or C). Now, let's calculate the length of the line that bisects line A-C at a right angle and intersects the arc. If you were clever you would have chosen chord A-C as one of the lines to bisect and already know the coordinate of the point midway between points A and C. You called that point D. It is simple to calculate the length of the line from point D to the coordinates of the origin of the arc using the Pythagorean Theorem. Subtract that length from the length of the radius of the arc and you have your answer!

It is also easy to calculate where the line that emanates from point D intersects the arc given the above answers and data. Now that you know the theory for all of this you should be able to derive the formulas for yourself.

This problem occurs in the measuring the radius of a curve on a railway line using a tape measure and a piece of string. Line BD is known in this environment as the 'versine'. The platelayer's formula is:

V=C²/8R

where R is the radius of the curve, C the length of the line and V the versine.

The formula is derived from Pythagoras' Theorem and is approximate (having dismissed a term in V² as being tiny in comparison with R² along the way for this particular application), though sufficiently accurate enough for someone playing with string and railway lines.

If something more accurate is required, then trigonometry leads to a precise answer.

Thank you so very much.

netmaker