Math question

[Toocool100]

Lets say i have an arc as shown in the picture. I know the length and the radius of the arc. But, how do i figure out what the shortest distance between its two points (ends) are, as shown by the question mark?

thx

[Miigga]

Draw a straight line from the start to the end...?

[DanAbnormal]

Make two right-angled triangles subtended by 3 radiuses, with one of them bisecting the chord and the other two joining the chords ends and use the sin or cosine rule for one of these triangles and find half of the chord length. THen times by 2 as there are two triangles.

[Toocool100]

At 5/8/06 02:22 PM, Miigga wrote: Draw a straight line from the start to the end...?

i want to know the distance of that line

[Popbrain1]

At 5/8/06 02:20 PM, Toocool100 wrote: Lets say i have an arc as shown in the picture. I know the length and the radius of the arc. But, how do i figure out what the shortest distance between its two points (ends) are, as shown by the question mark?

thx

Do you have the points's coordonate ?

If so , draw a triangle like this , and use the Pytagorus theory.

Like ...

[N-Antichrist]

I'm better with Pythagoras' theorum

[Innocence-Lost]

At 5/8/06 02:22 PM, Miigga wrote: Draw a straight line from the start to the end...?

its a trick question the shortest path on a arc is always a curve

[DanAbnormal]

At 5/8/06 02:24 PM, DanAbnormal wrote: Make two right-angled triangles subtended by 3 radiuses, with one of them bisecting the chord and the other two joining the chords ends and use the sin or cosine rule for one of these triangles and find half of the chord length. THen times by 2 as there are two triangles.

Like so.

[Toocool100]

At 5/8/06 02:24 PM, -Popbrain- wrote:

At 5/8/06 02:20 PM, Toocool100 wrote: Lets say i have an arc as shown in the picture. I know the length and the radius of the arc. But, how do i figure out what the shortest distance between its two points (ends) are, as shown by the question mark?

thx

Do you have the points's coordonate ?

i dont have the points coordinates

[Popbrain1]

At 5/8/06 02:25 PM, Noob-Antichrist wrote: I'm better with Pythagoras' theorum

Sorry if I wrote it badly ....

in French we say '' Relation de Pythagore '' So , its hard to translate.. :S

[Casualty]

actually it's 2.r² - 2.r.cos(S) where S = the "span" of the arc, I dunno how it's called.

[Toocool100]

At 5/8/06 02:26 PM, DanAbnormal wrote:
Like so.

problems...

[HeartbreakHoldout]

If you know the radius, you can work out the angle of the segment, if two radii meet at the centre (see diagram, me too lazy to describe right). When you work out the angle, you can use cosine rule to work out the line that you need.
If that doesn't clarify, then i'll have another go doing it more clearly.

[Casualty]

This way, I THINK:

[josh483]

Not 100% sure but if you know the Length of the arc .. i think you will need to possibly use Radians as in When Pi=180. So with the length of arc figure it out in Radians. Following that Sqaure the length of arc and length of radius. Then Take them away from each other. Sqaure root that awnser and it should be good.

[DanAbnormal]

At 5/8/06 02:31 PM, Toocool100 wrote:

At 5/8/06 02:26 PM, DanAbnormal wrote:
Like so.

problems...

Lets imagine the radius is 6.

Using the rules of circles, any angle subtended by the diameter (shown in blue) creates a right angle. The radius bisects this angle, so it makes the one you want 45 degrees.
Now you can find the adjacent, using this angle and the known radius. as it is adjacent and hypotenuse, we use the Cos.

Cos45 = adjacent/6

So the adjacent is 4,34

times this by 2

=8.84

[1]

At 5/8/06 02:40 PM, DanAbnormal wrote: LOL I AM GOOD AT MATHS

Why must you taunt me...

[Lopas1]

i don't like math...i doesn't understand a crap of it...

[TheDoctor]

Don't you just use the good ol' SOHCATOA?

[1]

At 5/8/06 02:44 PM, -TheDoctor- wrote: Don't you just use the good ol' SOHCATOA?

Please tell me you're only good at writing.

[StrikerF2]

Use a ruler?

[Casualty]

My technique worked, but it seems TooCool is not posting anymore. Oh well..

[Toocool100]

At 5/8/06 03:06 PM, -Casualty- wrote: My technique worked, but it seems TooCool is not posting anymore. Oh well..

thx for the help...

not posting because im still hard at work at my project... -.-

[DanAbnormal]

At 5/8/06 03:06 PM, -Casualty- wrote: My technique worked, but it seems TooCool is not posting anymore. Oh well..

How did you find the angle of A?

[JoeyNukes]

Sqaure(x2-x1)to the second power+(y2-y1)to the second power

[Casualty]

At 5/8/06 03:12 PM, DanAbnormal wrote:

At 5/8/06 03:06 PM, -Casualty- wrote: My technique worked, but it seems TooCool is not posting anymore. Oh well..

How did you find the angle of A?

TooCool knew it, or knew how to compute it...