This is a robot arm, meh, whatever, who cares. I know how to get R, as well as theta. Let's say that...

x = 6.0
y = 12.0
L1 = 12"
L2 = 10"

R = sqrtof(10^2+12^2) = 15.62
theta = tan^-1(y/x) = tan^-1(12/6) = 63.43º
How do I find theta1, and theta2?

Sorry for the yellow:

hellhawk123

United States84 Posts

January 31 2008 02:28 GMT

wait do you mean R is sqrt(6^2+12^2) = 13.42? then you know l1 l2 and R so use law of cosines

Raithed

China7078 Posts

January 31 2008 02:31 GMT

No, R = sqrtof(10^2+12^2) = 15.62. Not x, y, but a, b. I don't remember those laws. :[

Lemonwalrus

United States5465 Posts

January 31 2008 02:33 GMT

a^2 + b^2 = c^2 only works for right triangles.

Edit: oh, nvmd, I think.

Edit2: oh, nvmd the nvmd.

fight_or_flight

United States3988 Posts

January 31 2008 02:35 GMT

On January 31 2008 11:31 Raithed wrote:
No, R = sqrtof(10^2+12^2) = 15.62. Not x, y, but a, b. I don't remember those laws. :[

how is R equal to that?

you can only use pythagorean theorem for a right-triangle. I'm kind of confused on the question, you should word it better.

edit: lol at the theta

Raithed

China7078 Posts

January 31 2008 02:35 GMT

Oh, and thanks hellhawk123, but even though I know the Law of Cosines, how would I do it? I get confused at that point.

hellhawk123

United States84 Posts

January 31 2008 02:36 GMT

sorry but how are you getting R? I assumed x=x component of R and y=y component of R?

also law of cosines is given triangle w/ lengths a b c and respective angles alpha beta gamma, c^2=a^2+b^2-2abcos(gamma) (pythag theorem is when gamma=pi/2)

fight_or_flight

United States3988 Posts

January 31 2008 02:38 GMT

On January 31 2008 11:36 hellhawk123 wrote:
sorry but how are you getting R? I assumed x=x component of R and y=y component of R?

also law of cosines is given triangle w/ lengths a b c and respective angles alpha beta gamma, c^2=a^2+b^2-2abcos(gamma)

Correct. If R is given by the (x,y) components, then it should be easy to plug in a, b, c and solve for gamma.

Raithed

China7078 Posts

January 31 2008 02:41 GMT

OOPS, sorry sorry. I mistook a^2+b^2 but you're right it's x^2+y^2.

So basically it is:

13.42^2 = 10^2 + 12^2 - 2(10)(12)cos(gamma)
180 = 100+144 - 240cos(gamma)
180 = 244 - 240cos(gamma)

Oh fuck I fail at Algebra, I know the gamma has to be alone, so move the (244 - 240cos) to the other side?

hellhawk123

United States84 Posts

January 31 2008 02:48 GMT

#10

solving for gamma we have gamma=arccos((c^2-a^2-b^2)/(-2ab)) so just plug in. In your case, 180 = 244 - 240cos(gamma) -> cos(gamma)=(180-244)/(-240) -> gamma=arccos(0.26)=1.3 radians or 74.5 degrees

Raithed

China7078 Posts

January 31 2008 02:53 GMT

#11

Is arccos = cos^-1? And why does it have to be arccos?

Thanks hellhawk, so that would be theta1?

hellhawk123

United States84 Posts

January 31 2008 02:58 GMT

#12

yes. Arccos is the inverse of cos so if cos(a)=b, you take the inverse of both sides so arccos(cos(a))=arccos(b)), inverses cancel, a=arccos(b). The way you did it, it yields the angle opposite of R in the triangle consisting of the lines R, L1 and L2 which doesnt do us much good. You want to get the angle opposite of L2 (which is equal to theta - theta1) so just in the law of cosines equation, set c=L2 and the other 2 sides as a and b.

Raithed

China7078 Posts

January 31 2008 03:02 GMT

#13

Thanks a lot, I keep having an issue remembering/understanding this once the angles goes crazy different ways, hopefully in the future you can help further. I have another example but I'll post it tomorrow I guess, sick atm.

hellhawk123

United States84 Posts

January 31 2008 03:07 GMT

#14

no prob, and FYI, if you want more reliable help you can try http://www.artofproblemsolving.com/Forum/index.php Basically you post a problem and billions of math kids swarm, piranha-like, to answer it.

Kwidowmaker

Canada978 Posts

January 31 2008 03:50 GMT

#15

I'm not sure if it has been said, but you do want to opposite to R so that you can subtract that from 180 degrees to get theta 2.

Raithed

China7078 Posts

January 31 2008 03:51 GMT

#16

Not found?

Kwidowmaker

Canada978 Posts

January 31 2008 04:00 GMT

#17

Hmm? You can get it with cosine law, I'm sure.

I'm of the understanding that L1 is only the black and not the yellow line.

Raithed

China7078 Posts

January 31 2008 05:30 GMT

#18

No, I get it now, hellhawk showed me, what I mean is, the website gave me didn't work, which works now.

BluzMan

Russian Federation4235 Posts

January 31 2008 09:16 GMT

#19

You fucking know the lengths of all 3 sides of a triangle and you can't find the angles?

Slayer91

Ireland23379 Posts

January 31 2008 18:37 GMT

#20

Well obviously theta2 is the 180 minus angel opposite R.
theta1 is theta minus the angel opposite l2.
Finding those angles is a matter of plugging in the cosine rule, right?

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