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Parametrics


Date: 12/18/96 at 22:58:51
From: Neil
Subject: Parametrics

Hi, 

I have a question involving parametrics (at least I think it's 
parametrics!).  I have no clue how to solve it so I'll get to the 
point:

You're in 3-D space and there are 3 objects. You're at point A. 
You want to get to point B. And you know the coordinates to point B 
from point C. Point B is 34+ degrees above point C and is also 
bearing 45 degrees from point C. It is 8354 units from point C. It is
moving at a speed of 6.56 (in which case 6.56*6.56*6.56 would be the 
number of units you'd go in 2 seconds) and is heading 123 degrees from 
point C at a elevation of -46 degrees.  

What heading (degrees, elevation, and speed) would I have to set in 
order to meet point B somewhere along the line?

This is extremely difficult to explain so I'll include a diagram of 
the degrees from point C.

                  0 degrees
                  |
                  |      B (I assume point B would be somewhere
                  |                           around this area)
                  |
   270 degrees----C------90 degrees
                  |
                  |
                  |
                  |
                180 degrees 

Thank you for your time.


Date: 12/20/96 at 09:39:12
From: Doctor Jerry
Subject: Re: Parametrics

Hi Neil,

I'm not sure you've given me enough information to solve the problem, 
but I can make some comments and invite you to respond.

I see no harm in assuming that C is the origin of a coordinate system. 
Initially, the moving object is at the point r0. I'll think of r0 as 
a position vector. To figure out r0, I'll first find a vector in the 
direction of the vector from the origin to r0. This would be the sum 
of the "bearing vector" {cos(45d),cos(45d),0} and the "elevation 
vector" {0,0,cos(56d)}. The sum of these vectors is 
{cos(45d),cos(45d),cos(56d)}. Next, I'll calculate the length L of 
this vector.  Then:

r0 = (8354/L){cos(45d),cos(45d),cos(56d)} = {5155.8,5155.8,4077.3}

To get the velocity vector of the object, we do something similar.  
The direction of the velocity vector is (if I've interpreted your 
statements correctly) the same as that of the vector 
{cos(123d),cos(33d),cos(-46d)}. Letting W be the length of this 
vector, the velocity vector will be:

v = (6.56/W){cos(123d),cos(33d),cos(136d)} = {-2.9004,4.4662,-3.8307}

So, the position r of the moving object at any time t (assume distance 
is in meters, speed in meters per second, and time in seconds) is:

r= r0 + t*v = {5155.8 - 2.9004t,5155.8 + 4.4662t,4077.3 - 3.8307t}

You said "in which case 6.56*6.56*6.56 would be the number of units 
you'd go in 2 seconds."  I don't understand what you mean. If your 
speed is 6.56 meters/second, then in 2 seconds you will have gone 
2*6.56 meters. From the vector r, after 2 seconds the object would be 
at the point: 

r = r0 + 2*v = {5150.0,5164.7,4069.7}

To meet the object at some point along its path, we need to know the 
starting point A of person P. Then, as long as the speed of P is 
larger than 6.56 meters/s, it will be possible for P to catch up to 
the moving object. If P starts from A and always heads toward the 
moving object, his or her path will be curved. An equation for the 
path can be found but it requires some calculus and differential 
equations. I'm guessing that you mean that P leaves A at the same 
time the moving object was at r0 and moves on a straight line. If 
this is true, then you need to determine the velocity vector V of P so 
that solutions exist for the three equations implicit in:

A + t*V = r0 + t*v

Let's suppose we know the length v of V = {v1,v2,v3}, which is the 
speed of P. So, we have four equations in four unknowns. The four 
equations are:

a1 + t*v1 = 5155.8 - 2.9004t
a2 + t*v2 = 5155.8 + 4.4662t
a3 + t*v3 = 4077.3 - 3.8307t
m^2 = v1^2 + v2^2 + v3^2

So, did you think it was this hard?  Maybe there's an easier way? I 
realize that I've assumed that you know something about vectors.  If 
this isn't true, I'm sorry.  However, everything I've said can be 
interpreted in terms of easy trigonometry.  

-Doctor Jerry,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
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