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Geometry Problem
Howard Rush:
I was trying to figure out the headwind component on an airplane at the outside upper corner of a square eight. I am insufficiently smart to do this. Anybody else want to do it? Here are some simplifying assumptions:
Wind is uniform, and parallel to the ground.
Lines are straight, and the wing is aligned with them.
The airplane's path is parallel to the ground, and angle of attack is zero.
The airplane is at 45 degrees elevation and at 45 degrees around the circle from downwind, as it would be just before starting a zero-radius corner.
I think the pitch angle will be nonzero. A wind component something like .707 Vw or (.707)^2 Vw or (.707)^3 Vw gets added vectorially to ground speed.
Tim Wescott:
Why Howard, this is easy, if only you use quaternions*
Define the windward direction to be y and down to be z (use your favorite version of the right hand rule to find x : http://xkcd.com/199/). So the half-angle rotation vector of your personal pointing vector to the vector referenced by this projection is
r_g = cos(22.5) + k * sin(22.5)
These will be raised up by 45 degrees from the vertical, so in a reference frame that uses r_va as it's x direction, the half-angle rotation vector from you to the plane will be
r_u = cos(22.5) + j * cos(22.5)
To find the vector from one to one's plane, one need merely multiply r_g * r_u * i * r_u' * r_g' (using the well-known** rules of quaternion multiplication), to get
i / 2 - j / 2 - k / sqrt(2)
That's a nice check, and gives us the expected results (note that the first time I did this I inexplicable forgot the order of multiplication for the rotations -- and of course, since quaternions are not commutative, the answer to r_u * r_g * i * r_g' * r_u' is different).
Of course, the plane's direction of flight is in a frame of reference that's rotated 90 degrees from the lines:
r_p = cos(45) - k * sin(45)
(as a check, in level flight the plane's 'nose' vector is r_p * i * r_p' = -j -- that's exactly as expected).
So the airplane's pointing vector with respect to the earth is simply
r_g * r_u * r_p * i * r_p' * r_u' * r_g' = (-i - j)/sqrt(2)
Dang. Well, that's a really good check, but you wanted the wind's pointing vector with respect to the plane, didn't you. Oh well -- just do the rotations in the reverse direction and the reverse order, then, keeping in mind that the wind's vector is i:
r_p' * r_u' * r_g' * i * r_g * r_u * r_p = -i / sqrt(2) + j / 2 + k / 2.
And we know this is right, because we have used math, and math is never wrong!!
It's also intuitively about right: there's a good amount of wind against the plane's nose (-i), a good amount of wind blowing in the starboard direction across the plane (+j), and a good amount of wind down (+z), which is your "downward component".
There. Another problem in 3D geometry solved easily and plainly by use of simple arithmetic.
* Note: sane people*** need not read this post. However, if you are sane and you take the trouble to wrap your brain around it, you can read it again later in comfort, secure in the knowledge that you won't have any remaining sanity to lose!
** heh heh
*** "Quaternions came from Hamilton after his really good work had been done; and, though beautifully ingenious, have been an unmixed evil to those who have touched them in any way, including Clerk Maxwell." — Lord Kelvin, 1892. (from http://en.wikipedia.org/wiki/Quaternions#Quotes)
Howard Rush:
Thanks, Tim. I'll pore over this later, but it reminds me of some stories, as many things do to old men. I undertook a study of quaternions a couple of years ago, but got sidetracked by having to go back to fill in other stuff I'd forgotten, and never got that filled in. I do remember something about Hamilton figuring it out and writing it on a bridge on his way to the pub, because he knew he would be cognitively impaired later and would forget. As for the right-hand rule, I remember stopping to talk to a teacher about something involving vectors, meanwhile mentally using my free hand to perform the right-hand rule to support the conversation. Trouble is, I was holding some books in my right hand, so I was attempting to use my left hand for the right-hand rule.
Tim Wescott:
I wouldn't recommend a study of quaternions to anyone who isn't doing a lot of 3D rotations, and even then they're not to be embraced lightly. I just know them because I had a two-year long project for a customer doing a bunch of inertial navigation stuff and while they certainly make the math hard, the arithmetic ends up being easy (and fast for the computer). Your loved ones just have to accept that you're going to be wandering around the house mumbling to yourself and making strange gestures with the fingers of your right hand.
In fact, the reason I could do those calculations so fast is because I have what amounts to a four-function calculator for quaternions written for Scilab, as a consequence of the contract.
I guess I just need to thank my lucky stars every day that there are no practical applications for octonions.
Chris Wilson:
What ............. the heck ................. is an 'octonion?'
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