[Physltest] [Phys-L] Re: projectiles

[Jack Uretsky <jlu@HEP.ANL.GOV>]

Hi David-
I contend that the "solutions" cannot be equivalent because we each have
an expression for Cos(2a).  These expressions must  be consistent with
each other.  They are not.
        The difficulty with your derivation is that it has introduced
extraneous solutions.  You can see this by noting that your definition of
"s" gives it a range 0<s<2.  But s>1 in your first equation leads to
equating a positive and negative number.
                                        Regards,
                                                Jack


On Sat, 1 Jan 2005, David Bowman wrote:

> Regarding Jack's comments:
>
> > David-
> >       I don't recognizem my solution as identical to yours.  Either
> > one of us has made an error, or I gave up to soon on the algebra.
> >                                               Jack
>
> and
>
> > David, one of us has made an error.  The solutions are identical in
> > the special case of q=0, which explains why we got nearly identical
> > solutions for the case of small q.  We can communicate privately
> > until we agree on a solution.
> >                               Regards,
> >                                       Jack
>
> Jack, we *already* agree on a solution.  Your implicit equation for
> the angle a is clearly not *formally* identical to my explicit
> formula for it, but your equation *is* equivalent to mine in the
> sense that they both have the same value of a as the solution for a
> given value of q.  In fact, it is fairly straightforward to derive
> my formula from your equation.  Neither of us made an error.  You
> just gave up on the algebra a bit to soon.
>
> Below are a few more details of the steps in the derivation that I
> outlined in my previous response to Ken Fox (except I am now using
> Jack's notation for the variables rather than my/Anthony's notation).
>
> First let's define s == 1 - cos(2*a).  Substitution of 1 - s for
> cos(2*a) and s/2 for sin^2(a) into Jack's equation 2.) gives (after
> dividing both the numerator and denominator on the RHS by sin(a):
>
> 1 - s = q/(1 + sqrt(1 + 2*q/s))
>
> Multiplication on both sides by the denominator gives:
>
> (1 - s)*(1 + sqrt(1 + 2*q/s)) = q
>
> Isolating the sqrt() term gives:
>
> (1 - s)*sqrt(1 + 2*q/s) = q - 1 + s
>
> Squaring out both sides and canceling the common terms on both sides
> gives:
>
> 1 + 2*q/s - 4*q = (q - 1)^2
>
> Solving for the only appearance of s in the above equation gives:
>
> s = 2*q/((q - 1)^2 + 4*q - 1)
>
> Simplifying the denominator and canceling the common factor of q in
> the fraction on the right gives:
>
> s = 2/(q + 2)
>
> This means that 1 - s = 1 - 2/(q + 2) = q/(q + 2)
>
> Now since we had defined s as 1 - cos(2*a) this means that we can
> write a as:
>
> a = (1/2)*arccos(1 - s) = (1/2)*arccos(q/(q + 2))               QED
>
> Also, if we are willing to write this in terms of 1/q (to Ken Fox's
> chagrin about what happens when q --> 0) we can also write:
>
> a = (1/2)*arccos(1/(1 + 2/q))
>
> David Bowman

--
"Trust me.  I have a lot of experience at this."
                General Custer's unremembered message to his men,
                just before leading them into the Little Big Horn Valley
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