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I'm trying to calculate the relative velocity ($V_R$) between an exact velocity ($V_0$) and a velocity range ($V_1$).

The exact velocity ($V_0$) is represented simply by ($course$, $speed$).

The velocity range ($V_1$) is represented by a range of courses and a range of speeds, like so: $([course_{min}, course_{max}], [speed_{min}, speed_{max}])$

I'd like to obtain the relative velocity ($V_R = V_1 - V_0$), which would also be represented by $([course_{min}, course_{max}], [speed_{min}, speed_{max}])$.

Note that $(course, speed)$ is very similar to polar coordinates, with the only difference being that $course$ is zero when facing up (north) and increases clockwise.

I built a spreadsheet to see what patterns would emerge for different ranges of courses and speeds, and I came to the conclusion that the minimum and maximum relative courses & speeds often occur at the "corners" of course/speed space ($(course_{min}, speed_{min})$, $(course_{min}, speed_{max})$, $(course_{max}, speed_{min})$, $(course_{max}, speed_{min})$), but not always.

I'm wondering if there's a relatively simple equation to find $V_R$, or if I'll just need to perform a "brute force" calculation, where I loop through the courses and speeds (at some level of precision) and pick out the min/max values.

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It's not really clear to me what you are trying to model. Do these velocities change with time? Is your "exact" velocity the constant, average velocity? Is your velocity "range" $V_1$ the instantaneous velocity at time $t$? Is there a constraint on these aside from that the object they model starts at some time $t_0$ and position $x_0$ (where $x$ is distance along a line or arc length along a curve) and ends at some time $t_1$ and position $x_1$? If this is your model, then $x=x(t)$ is a function of $t$ and $$v_\text{avg}=\frac{x_1-x_0}{t_1-t_0},\qquad v_\text{inst}=\frac{dx}{dt}$$. –  bgins Apr 1 '12 at 18:55
    
Could you show the rest of your work as I'm not really sure what you are asking. Presumably, you know how to compute $V_R$ given one vector $V_1$, yes? So why is it a problem to determine the range of possible values? –  Raskolnikov Apr 1 '12 at 18:56
    
Usually, we assume that $\frac{dx}{dt}$ is continuous, but it can even approach infinity at a point. For example, $x(t)=\sqrt{t}$ for $t\in[0,1]$ has average velocity $1$ but $x'(0)=\infty$. –  bgins Apr 1 '12 at 19:00
    
@bgins: I think that is irrelevant to OP's question. His is purely a trigonometry question. But I don't see why he is stuck on it, as he seems to be able to compute things. –  Raskolnikov Apr 1 '12 at 19:06
    
Actually, I've just noticed your other question about the annular sector, and I realize now what you expect. But that is not possible. As you can see, if $V_1$ is part of such a sector, subtracting $V_0$ will just translate that sector. But since polar coordinates are always described w.r.t. the same fixed origin, the expression for the translated domain in terms of polar coordinates will become unwieldy. You will not be able to put it in the form you desire. –  Raskolnikov Apr 1 '12 at 19:12

1 Answer 1

To help advance the question a bit, I have made a graphic of what DanM has in mind. Say that $V_1$ should be in the following course/speed range: $([\pi/6,\pi/2],[1,2])$. This would give rise to the orange annular sector in the picture. But now, he wants to shift the values of this range by a vector $V_0$ with course/speed given by $(3\pi/4,\sqrt{2})$ (in Cartesian coordinates, this is just $(-1,1)$). Then, in what annular sector does the shifted orange sector fit? I have colored it blue in the picture.

Annular sector 1

In this case, the extreme values of the range in term of course/speed correspond to the corners of the orange range. But it is easy to construct a case where this is not so. Take the same orange sector, but now with $V_0$ having course/speed $(\pi/4,3)$. The picture becomes

Annular sector 2

The lower bound for the range of speeds is now determined by the circle boundary and not a corner. And to end, a picture of what happens when $V_0$ is just on the inner boundary:

Annular sector 3

To summarize, DanM's question is: when I change the origin of my coordinate system, in what minimal annular sector defined w.r.t. that new origin does the old sector, defined w.r.t. the old origin, belong?

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