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What's Up 2007 Sky Watchers Catalog
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A stargazer's guide to the night sky. You can download the freely available PDF file (23.4 MB) by clicking the image to the left. Or, alternatively, you can buy the book over at Lulu.com
For more information, direct your favorite web browser to astrowhatsup.com
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Resultants For 3+ Vectors
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A few nights ago, it occurred to me that with a slight addition of complexity to my previous algorithm, it
could be expanded to calculate the resultant of a theoretically infinite number of vectors. In the image to
the right, R1 is the resultant for U and V.
Next, we combine R1 and Z to get R2. If you think about this, it
really does make sense. If you have to have two people of different strengths each pulling a rope tied to a
large rock and pulling from different directions; the rock only moves one direction. That resultant force is
obtained by combining the two forces exerted upon it by the two people. The same thing would happen if one
person (probably a very strong person) were pulling by himself. So logically, a resultant is truly equal
to the first plus the second vector. So, if you want to add three vectors together, you just add the first two
and use the resultant from that to add to the third. And the process can be repeated ad infinitum.
To test this, I've added a link to the navigation bar labeled Vector Example. The
program is setup for four vectors, but you can do fewer by putting in zeros for one or more vectors.
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More on Resultants
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Here I will articulate the process I have refined for calculating a resultant vector from any two vectors.
I have also written a program in Java to demonstrate that the algorithm works. The program can be found
here.
As you can see from the picture on the bottom right, there's a lot going on. But, the logic that drives
it is clean.
In the previous article, I described how to find the resultants magnitude if the degree was known, but only
if the two original vectors were at an acute angle. This new method works for any two vectors.
First off, we need to determine the magnitude of R. Surprisingly, we don't need it's degree, for
this method, to discover it. This involves copying U from the beginning of V to the end,
which is allowed under the parallelogram law.
And because the degree of U is 350°, we know that U's clone, we'll call it
U1, is also 350°. Because U1 has a magnitude of 2 we use the
equation: U1a = 2 cos(350) = 1.97 and
U1b = 2 sin(350) = -0.35. We now know that the end of R is
1.97 north and 0.35 west of the end of V.
But, this information is totally useless if V's endpoint is unknown. So we use the exact same
process on V as we did on U1: Va = 3 cos(110) = -1 and
Vb = 3 sin(110) = 2.82. To get the end point of R we simply do a little
addition U1a + Va = 0.97 and
U1b + Vb = 2.47.
So we now know that R ends at point (2.47, 0.97).
Now, we apply the Pythagorean theorem: 2.472 + 0.972 = 7 and
the square root of 7 being about 2.646. So now
we know that the magnitude of R is 2.646. Notice that we did all that
without the benefit of a degree for R. However, a vector isn't a vector until it has both
magnitude and direction. We can determine that direction with the equation:
degree = tan-1(2.47 / 0.97) = 68.5. And with that, all necessary
information regarding the resultant vector R.
R = U + V
U: M = 2 D = 350
V: M = 3 D = 110
R: M = 2.646 D = 68.5
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Extracting The Resultant From Two Vectors Using Parallelograms
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I've been reading the book,
"About Vectors". And, while
reading, the author make reference to vector addition and even goes so far as to say that, using
parallelograms, one can find the result of addition, or resultant (represented by vector OR to the right.)
What he does not do, is explain how this is
done (well, at least not in as much as I have read.) So, I set about the task of discovering it myself. And
after MUCH trial and error I came up with a solution. What I came up with may not be "The Way It's Done" in
formal math, but it works.
First and foremost, one must recognize that because angle OTR is not 90°, we can't just use the
pythagorean theorem -- not yet, at least. What we have to do is find an triangle that makes exactly the
same hypotenuse with a vertex of 90°. This is no small task, you can tell from the line-segments QT
and QR, that guessing won't get very good answers. What you can do, is work the pythagorean theorem
backwards -- you already have the hypotenuse of angle TQR. But, there's one problem, in the
theorem you have a2 + b2 = h2; we have h,
but a and b are both unique numbers, and we know neither. Normally, this is where things
would end, because first year algebra tells us that we can't solve an equation with more than one variable
present.
But, we're not through yet. First, most should know that all angles within any triangle add up to
180°. Therfore, since we already know that our new triangle (in grey) is a right triangle, we only
need to know one more angle and then we can use subtraction to determine the other. We can tell that angle
RTQ is 60°, because line TR is parallel with OS. Our simple subtraction leaves us with 30° for
angle QRT.
Now that we know all the angles in our triangle and the hypotenuse, we can use a little trigonometry and
find the length of sides a and b. Because line b is on the opposite side of angle RTQ,
we use sin to determine the length of side b.
b = 4 sin(60) = 3.4641
a = 4 sin(30) = 2
(6 + 2)2 + 3.46412 = 8.7182
And the resultant is found. I'm still not sure why this is called "Adding Vectors," if in fact that is
what this is (my understanding of this is still rather foggy.) Much thanks to
David Teeple for remembering his high-school math and
Wikipedia for remembering everything that ever was.
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