Math/Trigonometry Question

[LaggyLou]

I am trying to figure out the length of an arc of a circle. I do not know the radius of the circle. I know the length of the chord that has the endpoints of the arc as its endpoints, and I also know the distance from the cord to the apex of the arc.

Intuitively I bvelieve I have enough information to get the length of the arc, but I am just at a loss /images/graemlins/confused.gif. Van anyone help?

[PokerFink]

I'm thinking multiply the length of the cord by pi and then divide by 2?

[Evan]

This is wrong.

[jason_t]

[ QUOTE ]
I'm thinking multiply the length of the cord by pi and then divide by 2?

[/ QUOTE ]

This can't be correct as it doesn't use information about the distance of the chord to the apex of the arc. That is obviously important information.

[Evan]

R= (((A/2)^2) + (B^2)) / 2B
(R = radius) (A = Chord Length) (B = Arc Height)

Angle in Radians = 2 * Arcsin((Chord Length) / (2 * Arc Radius))

Arc Length = Arc Radius * Angle in Radians

[Jamey Saunders]

Don't take my word for this, because it's been 11 years since I had college trig, but...

If you know the distance from the chord to the apex (which, if I remember correctly is the center of the circle), and you know the length of the chord, then you know the lengths of the two legs of right triangles. The distance from the chord to the apex (let's call that X) and half the length of the chord (let's call the chord length Y) make up the sides of the triangle.

Knowing that, you could find the hypotenuse by solving C^2=X^2+(Y/2)^2. The hypotenuse would be the radius of the circle.

Where you go from here, I can't remember. But I do think I remember that there's a way to figure the opposing angle of a triangle if you know one angle (90*) and the lengths of a couple of sides. Once you've figured that angle, double it, and you've got the angle that includes the arc. Take that angle, and figure what percentage of the circumference it is. That percentage of the circumference of the circle should be the length of the arc.

That said, there's probably an easier way to do it. I'm waiting to see what others come up with.

[EDIT] Damn, Evan beat me to it, and he's much better than I am. Oh, well... /images/graemlins/smirk.gif

[TBag]

Edit - btw x is chord height / 2.
http://img110.echo.cx/img110/4869/math4td.jpg

I thought this worked, but Evan's is different.

When solving for radius, why do you divide by 2B?

[LaggyLou]

Thanks!!!!!!! OOT Rules, and I have now achieved the feat of billing my time for posting on 2+2!

Evan, if we ever meet, the drinks are on me!

[Evan]

[ QUOTE ]

Evan, if we ever meet, the drinks are on me!

[/ QUOTE ]
/images/graemlins/grin.gif

[TBag]

Yargh, friggin trig. Hated that class. Anyways, could you explain to me why you divided by 2B when solving for the radius?

Isn't it just pythagorean's?

[LaggyLou]

Hmmmm......maybe I'm not explaining it right. Here is the problem, and what I understand Evan's solution to be:
http://img175.echo.cx/my.php?image=arcproblem5pq.png

Hmmm. I Can't figure out why the image tag isn't working. Try it as a
link. (http://img175.echo.cx/my.php?image=arcproblem5pq.png)

[Evan]

If you read my post where I said that divding by 2B was an error, forget that. It is correct. I just suck at checking simple math (I also suck at mixing up radius and diameter /images/graemlins/shocked.gif). The correct formula is the one in my first post.

[purnell]

"Arc Length = Arc Radius * Angle in Radians "

I re-learned something today. /images/graemlins/cool.gif

[LaggyLou]

Ok, I'm doing something wrong. Let's call X half the chord length and y the arc height. As I understand it:

R=(x^2+y^2)/y -- is that right?

Then:

Angle = 2*Arcsin(x/arc Radius) {Chord length = 2x, so the 2's in your formula cancel out, right?}

Arc Length = R * Angle


.... Ok, so here are some real numbers:

x=.419, y=.251

Radius = .9504
Angle = .9131, yielding
Arc Length = .8678

But I don't see how that could be right, because the length of the hypotnuse of a rt triangle with X and y as the legs is .488428. The arc length has to be more than double that hypontuse, doesn't it?

[Evan]

Yea, I edited my "correction" post. Sorry for the confusing. Your first interpretation, the on in the image you posted, is correct. I blame whoever questioned why you dived by 2B. /images/graemlins/tongue.gif

As for an answer to that question, I don't know. I'm sure it's not incredibly hard to drive, but I am half studying for a final, so I can't do it right now.

[RacersEdge]

I think that drawing is right.

x = 1/2 cord lenght = known
c = cord height = known

In drawing, r (radius) = y + c

x^2 + y^2 = r^2

x^2 + (r-c)^2 = r^2

Now you know r since x,c are known.

sin(angle) = x/r

arc length = (angle/360)*2*pi*r

[tallstack]

I am responding to your calcs with reference to Tbag's diagram. Evan's equations are correct and you do need to divide by 2b (or 2y using your variables)

From TBag's diagram, we do not know the distance from the center of the circle to the chord, y. We know the distance from the chord to the outside of the circle. So in his diagram, y should be replaced with R-y.

If you use TBag's diagram then the correct equation is
R^2 = (r-y)^2 + x^2
or R^2 = (R^2 - 2Ry + y^2) + x^2
or 2Ry = y^2 + x^2
or R = (y^2 + x^2) / 2y

From your numbers,
x=.419, y=.251

Radius = .4752
Angle = 2.1589 rad
Arc Length = 1.0259

You are right that the arc length has to be more than twice the hypotenuse.

I hope that this helps.

Dave

[mostsmooth]

[ QUOTE ]
This is wrong.

[/ QUOTE ]
its not wrong if the chord goes through the center /images/graemlins/cool.gif