Hard Light Productions Forums
Off-Topic Discussion => General Discussion => Topic started by: MicroPsycho on September 27, 2005, 03:09:34 pm
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Does anyone know how to solve for displacement between 2 positions algebraically?
the reason I ask is because in physics class I need to find displacement using a diagram and physically measure the dsiplacement, and I hate drawing diagrams, its both annoying and innacurate.
so instead of drawing a scale drwaing of, for example, 30km[E21'N] and 40km[W35'N] and then connect those points and measuring the displacement, what's the equation to solve it algebraically?
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So, there IS a difference between physical science and physics. As an eighth grader, i'm three years behind you. If I knew what you were talking about I would be able to help you, as math is my strong suit.
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What exactly are you given here? I'm not sure what you want to do.
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oh, I think I see now. Just split the vectors into rectangular components and take the distance. This should work:
sqrt( L1² - 2 L1 L2 cos[ (A1 - A2)*pi/180 ] + L2² )
where L1 and L2 are the lengths of the vectors and A1 and A2 are the respective angles in degrees. It doesn't matter how the angles are oriented as long as you use the same zero direction for all of them.
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Can't you use the horizontal distance and the vertical distance to form a triangle, and then calculate the displacement using pythagora's theorem?
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Ok, so I drew this drawing in photoshop to help. When D1 and D2 are give I need to find Delta D(displacement). The method I use in class is to draw D1 and D2 as shown and then connect them to get Delta D. This way requires me to measure the lenght of Delta D and its angle.
(http://img9.imageshack.us/img9/6104/displace5nm.jpg)
I hope this diagram makes sense. CP, that looks like it might work, I'll try it. And for the angles, do I write them as part of a whole 360 degrees (ie 69 degrees and 305 degrees, [E21'N] and [W35'N] respectively?)
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You can divide each up into two vectors and add them.
x = cos(55) * 40 + cos(291) * 30
y = sin(55) * 40 + sin(291) * 30
(Taking N as 0 degrees and going counter-clockwise)
Then the final vector can be expressed (x, y) or (sqrt(x^2 + y^2), tan^-1(y/x))
For the angles it doesn't matter how you express the angles as long as they're consistent. It also depends on your teacher's preference.