Ballistics and hills

[John Jones]

I was challenged to solve the following problem, which was said not to
be easy. I found it quite stretching although only using O level maths
(s = u*t - 0.5*f*t^2 and calculus-derivatives).

The answer is surprisingly elegant. My technique for getting it was
extremely hairy. Is there an elegant method of deriving it?

Problem
=======
There's a cannon on top of a hemispherical hill. It fires a ball, which
just grazes the hill, before reaching the flat ground below.

What is the *minimum* velocity at which the cannon fires the ball to do
this?
Assume: no air resistance, cannon of zero height, ball of zero radius.

Hint: if the cannon fired horizontally the velocity would be sqrt(g*h)
where g is the gravitational acceleration (32 ft/s/s, 981 cm/s/s) and h
is the height of the hill (and its radius therefore).

HTH
JJ

[William Elliot]

> I was challenged to solve the following problem, which was said not to
> be easy. I found it quite stretching although only using O level maths
> (s = u*t - 0.5*f*t^2 and calculus-derivatives).

What's u and f?

> The answer is surprisingly elegant. My technique for getting it was
> extremely hairy. Is there an elegant method of deriving it?

Do you have a comprehensible problem statement?

> There's a cannon on top of a hemispherical hill. It fires a ball, which
> just grazes the hill, before reaching the flat ground below.

What does it mean to graze a hill when you're already on top of it?

> What is the *minimum* velocity at which the cannon fires the ball to do
> this?

Do what?

> Assume: no air resistance, cannon of zero height, ball of zero radius.

> Hint: if the cannon fired horizontally the velocity would be sqrt(g*h)
> where g is the gravitational acceleration (32 ft/s/s, 981 cm/s/s) and h
> is the height of the hill (and its radius therefore).

Huh? The original velocity of a projectile is not dependent upon gravity
nor height of hill.

[Virgil]

> > I was challenged to solve the following problem, which was said not to
> > be easy. I found it quite stretching although only using O level maths
> > (s = u*t - 0.5*f*t^2 and calculus-derivatives).

> > The answer is surprisingly elegant. My technique for getting it was
> > extremely hairy. Is there an elegant method of deriving it?

> > There's a cannon on top of a hemispherical hill. It fires a ball, which
> > just grazes the hill, before reaching the flat ground below.

> > What is the *minimum* velocity at which the cannon fires the ball to do
> > this?

Unless that hill is more parabolic than semi-circular in vertical cross
section, no amount of hairyness of technique is sufficient to solve the
problem.
--

[John Jones]

On Thu, 14 Feb 2013 19:38:35 -0800, "William Elliot" <ma...@panix.com>
wrote in article <Pine.NEB.4.64.13...@panix2.panix.com>...

>
> > I was challenged to solve the following problem, which was said not to
> > be easy. I found it quite stretching although only using O level maths
> > (s = u*t - 0.5*f*t^2 and calculus-derivatives).
>
> What's u and f?

this is/was a standard formula for movement under acceleration in a
straight line: s is the distance travelled, u is the starting speed,
v the closing speed, t the time, and f the signed rate of
acceleration/deceleration. Other formulae are v = u + f*t;
v^2 - u^2 = 2*f*s.

>
> > The answer is surprisingly elegant. My technique for getting it was
> > extremely hairy. Is there an elegant method of deriving it?
>
> Do you have a comprehensible problem statement?

See below.

>
> > There's a cannon on top of a hemispherical hill. It fires a ball, which
> > just grazes the hill, before reaching the flat ground below.
>
> What does it mean to graze a hill when you're already on top of it?

The ball is fired into the air, follows a parabolic trajectory in free
air for a while, then just hits/just misses hitting the hill (ie the
trajectory is tangent to the hill at a single point, so the ball can
continue in its downward path - another assumption) and then bashes onto
the ground.

>
> > What is the *minimum* velocity at which the cannon fires the ball to do
> > this?
>
> Do what?

There is a continuum of velocities, by which I mean speed, that allow
the above kind of trajectory - for each the cannon needs to point in
different upward angles. For each angle there is a speed that achieves
"grazing". There is an angle at which the required initial speed is
smallest.

>
> > Assume: no air resistance, cannon of zero height, ball of zero radius.
>
> > Hint: if the cannon fired horizontally the velocity would be sqrt(g*h)
> > where g is the gravitational acceleration (32 ft/s/s, 981 cm/s/s) and h
> > is the height of the hill (and its radius therefore).
>
> Huh? The original velocity of a projectile is not dependent upon gravity
> nor height of hill.

Maybe this isnt a good hint. What it says is that if the cannon angle is
zero, then the ball will *begin* by grazing the hill, but it wont ever
graze it again (so still a solution albeit trivial). If the original
speed is high, it will zoom outwards well beyond the hill. If low, it
will hit the hill pretty rapidly (the curvature of the corresponding
parabola becomes a lot tighter than a circle) and stop before it hits
the ground (and so that isnt a solution as stated). The speed at which
it just fails to hit the hill is as stated. However, that speed is not
the smallest - there are smaller initial speeds which do the trick if
the cannon is initially pointed upwards.

Why was it a hint anyway? Because (if I am right) the minimum speed
required by the problem is a dimensionless multiple (less than 1) of
that speed. In that case the angle of the cannon is not zero (but it is
a good round number of degrees).
HTH
JJ

[John Jones]

On Thu, 14 Feb 2013 22:47:44 -0700, "Virgil" <vir...@ligriv.com> wrote
in article <virgil-F4000C....@BIGNEWS.USENETMONSTER.COM>...

Hi Virgil,
Please see my response to William Elliot.
In particular:

The ball is fired into the air, follows a parabolic trajectory in free
air for a while, then just hits/just misses hitting the hill (ie the
trajectory is tangent to the hill at a single point, so the ball can
continue in its downward path - another assumption) and then bashes onto
the ground.

Regards
JJ

[William Elliot]

On Fri, 15 Feb 2013, John Jones wrote:
> On Thu, 14 Feb 2013 19:38:35 -0800, "William Elliot" <ma...@panix.com>
> wrote in article <Pine.NEB.4.64.13...@panix2.panix.com>...

> > > There's a cannon on top of a hemispherical hill. It fires a ball, which
> > > just grazes the hill, before reaching the flat ground below.
> >
> > What does it mean to graze a hill when you're already on top of it?

> The ball is fired into the air, follows a parabolic trajectory in free
> air for a while, then just hits/just misses hitting the hill (ie the
> trajectory is tangent to the hill at a single point, so the ball can
> continue in its downward path - another assumption) and then bashes onto
> the ground.

From where is the ball fired?
At what height does the ball graze the hill?

It seems to me that the ball would land the same
distance from the hill as the cannon is from the hill.

But that's assuming the ball grazes the hill on
a great circle passing through the top of the hill
that's normal to the trajectory.

Hm. In what direction, relative to the center of
the hemishere from the cannon, is the ball ejected?

[Mike Terry]

"William Elliot" <ma...@panix.com> wrote in message
news:Pine.NEB.4.64.13...@panix3.panix.com...

> On Fri, 15 Feb 2013, John Jones wrote:
> > On Thu, 14 Feb 2013 19:38:35 -0800, "William Elliot" <ma...@panix.com>
> > wrote in article <Pine.NEB.4.64.13...@panix2.panix.com>...
>
> > > > There's a cannon on top of a hemispherical hill. It fires a ball,
which
> > > > just grazes the hill, before reaching the flat ground below.
> > >
> > > What does it mean to graze a hill when you're already on top of it?
>
> > The ball is fired into the air, follows a parabolic trajectory in free
> > air for a while, then just hits/just misses hitting the hill (ie the
> > trajectory is tangent to the hill at a single point, so the ball can
> > continue in its downward path - another assumption) and then bashes onto
> > the ground.
>
> From where is the ball fired?

From the location specified in the problem, i.e. the top of the hill. (I
can see this stated a few lines above :) )

> At what height does the ball graze the hill?

You could answer this once you've worked out the answer to the problem,
because then you would know the trajectory details. The trajectory is to be
chosen so that the initial firing velocity can be minimised while still
meeting the problem constraints. (So the detail you are asking for is part
of solving the problem.)

>
> It seems to me that the ball would land the same
> distance from the hill as the cannon is from the hill.

No, the ball is fired from the top of the hill.

OK, I'll have my own go at describing the problem for you:

There is a hemispherical hill. A cannon is ON TOP of the hill, i.e. it is
at the highest point of the hill, and the hill spreads out below it
symmetrically until it meets the flat plain on which the hill stands.
Suppose it is desired that the cannon shoot the ball in such a way that the
ball "clears the hill", i.e. the ball travels and lands on the plain beyond
the hill, without having hit the hill. Obviously if the cannon is Very
Powerful this is easily achieved, and there will be many firing angles that
work. If the cannon is Very Weak clearly it can't achieve this, and
regardless of firing angle the ball trajectory will intersect the hill.

The problem is to find the minimum firing velocity which is the boundary
between being able to clear the hill and not being able to clear the hill.
[The firing angle for the canon is not given as part of the problem - we
would expect that at the critical firing velocity there would be a unique
trajectory where the ball will just graze the hill during its trajectory,
i.e. the parabolic trajectory of the ball will be tangental to the
hemispherical hill at some point.]

There are the usual assumptions stated in the OP: no air resistance,
zero-sized cannon ball etc. to make it a well defined mechanics problem,
such as you might see in an applied maths exam paper. I suppose such a
paper would have had a nice picture which would have avoided your
confusions...

Mike.

[Macavity]

Let the hill have height h and be denoted by
x^2 + y^2 = h^2. The relevant part to us is in the first quadrant (of the vertical plane in which the cannon ball is fired).

The trajectory of the ball can then be described as
y = b - (x-a)^2, where (a, b) is the topmost point - also with a > 0, b > h.

As this must pass through (0, h), we have h = b-a^2 and can simplify the trajectory to
y = h + a^2 - (x-a)^2 = h + x (2a - x)

By conservation of energy between the initial point and the topmost point, we have
gh + u^2/2 = gb, where u is the initial speed. I don't think equations of motion are additionally needed, lets see...

Thus, we must have u = Sqrt[2g(b-h)] = a Sqrt(2g)
So minimising u is the same as minimising a.

Now for the "grazing" condition, we must have the two curves intersect, and their tangents match.
For the tangents to match, we must have dy/ dx = 2(a-x) and dy/dx = - x/y to be the same
and for intersection, we have the two equations:
y = h + x(2a-x)
y^2 + x^2 = h^2

Eliminating y, we have a quartic in x,
[h+x(2a-x)]^2 + x^2 - h^2 = 0

Rather than equating the tangents, which looks messy, lets try another approach… This quartic could have up to four solutions, corresponding to the parabola intersecting the circle in four points - we need the case where one of the solution points (x,y) is (0, h) - satisfied by construction, another is in the third quadrant, and there is a solution in the first quadrant with a multiplicity of two.

Clearly x = 0 is a solution. Removing this factor from the quartic, we have the cubic
x(2a-x)^2 + 2h(2a-x) + x = 0, which should have two distinct roots, one in third quadrant, and another in the first quadrant with a multiplicity of two. Simplifying,
P(x) = x^3 - 4ax^2 + (4a^2 - 2h + 1)x + 4ah = 0

Now the root with multiplicity of two must be also a root of P'(x), so
P'(x) = 3x^2 - 8ax + (4a^2 - 2h+1) = 0 has a root x > 0.

=> 64a^2 - 12(4a^2 - 2h+1) > 0
or a^2 > (3/4) (1-2h).

So, if I have done all those calculations right (please check),
the minimum value of a is Sqrt[(3/4) (1-2h)]
and hence the minimum for u must be Sqrt[(3g/2) (1-2h)]

Do let me know if the hairy method gave the same answer...

HTH.

[Macavity]

On reading again, found too many things to fix, the above conservation itself is wrong, as at the highest point there is still a horizontal velocity, and the condition I found on P'(x) is merely required, not sufficient, so need to go back to the standard approach on this one. or the drawing board.

Please ignore the last post...

[Ray Koopman]

h = initial height (ft)
a = angle of elevation
v = muzzle velocity (ft/sec)
g = acceleration due to gravity (ft/sec^2)
T = time (sec)

cannonball trajectory is {x[T],y[T]), with

x = v T cos a
y = h + v T sin a - g/2 T^2

Write y as a function of x

y = h + t x - u x^2, where

t = tan a, u = g/2 (sec a)^2 x^2 / v^2.

Pick an angle A in the first quadrant,
and let (x,y) = h*(cos A, sin A).

Solve for (t,u) so that the parabola
is tangent to the circle at (x,y)

t = x/y + 2(y-h)/x, u = 1/y + (y-h)/x^2

In t and u, substitute for x and y in terms of A

t = 2(tan A - sec A) + cot A,
u = ((tan A - sec A) sec A + csc A)/h.

From the definitions of t and u we have

v = sqrt[g/2 (1 + t^2)/u].

To minimize v, minimize (1 + t^2)/u.

(1 + t^2)/u = h w, where w = 4 sin A + csc A - 3

A = pi/6 gives w = 1, which is the minimum in the first quadrant.

The minimum muzzle velocity is sqrt[g/2 h].
The corresponding angle of elevation is pi/6.

[John Jones]

On Mon, 18 Feb 2013 01:25:38 -0800 (PST), "Ray Koopman" <koo...@sfu.ca>
wrote in article <28145133-962a-4be4-8572-
ea4c88...@kk9g2000pbc.googlegroups.com>...

>
> On Feb 14, 3:30 am, John Jones <a1...@hotamil.com> wrote:
> > I was challenged to solve the following problem, which was said not to
> > be easy. I found it quite stretching although only using O level maths
> > (s = u*t - 0.5*f*t^2 and calculus-derivatives).
> >
> > The answer is surprisingly elegant. My technique for getting it was
> > extremely hairy. Is there an elegant method of deriving it?
> >
> > Problem
> > =======
> > There's a cannon on top of a hemispherical hill. It fires a ball, which
> > just grazes the hill, before reaching the flat ground below.
> >
> > What is the *minimum* velocity at which the cannon fires the ball to do
> > this?
> > Assume: no air resistance, cannon of zero height, ball of zero radius.
> >
> > Hint: if the cannon fired horizontally the velocity would be sqrt(g*h)
> > where g is the gravitational acceleration (32 ft/s/s, 981 cm/s/s) and h
> > is the height of the hill (and its radius therefore).
> >
> > HTH
> > JJ
>

excuse me for top-posting here, but just to explain that my annotations
are for those who like me need everything spelled out... :-)

> h = initial height (ft)
> a = angle of elevation
> v = muzzle velocity (ft/sec)
> g = acceleration due to gravity (ft/sec^2)
> T = time (sec)
>
> cannonball trajectory is {x[T],y[T]), with
>
> x = v T cos a
> y = h + v T sin a - g/2 T^2
>
> Write y as a function of x
>
> y = h + t x - u x^2, where
>
> t = tan a, u = g/2 (sec a)^2 x^2 / v^2.

u = g/2 (sec a)^2 / v^2

>
> Pick an angle A in the first quadrant,
> and let (x,y) = h*(cos A, sin A).

(obviously x,y now a general point on the hill as well as on the
parabola)

>
> Solve for (t,u) so that the parabola
> is tangent to the circle at (x,y)

to expand:
the tangent of the parabola is dy/dx = t - 2*u*x
the tangent of the hill at x,y is -x/y (from a diagram, general
knowledge,
or since (x^2 + y^2) = h^2, dy/dx = - pdx(x^2+y^2)/pdy(x^2+y^2))
so t - 2*u*x = -x/y
and from the parabolic equation
t*x - u*x^2 = (y-h)
so these are two linear equations for t,u in x and y solved as follows:

> t = x/y + 2(y-h)/x, u = 1/y + (y-h)/x^2

this is the crux of why your method is so much better than mine.

>
> In t and u, substitute for x and y in terms of A
>
> t = 2(tan A - sec A) + cot A,
> u = ((tan A - sec A) sec A + csc A)/h.

if it helps, just to make calculation of (1+t^2)/u easier (!),
these simplify to
t = (1 - s)^2/(sc)
h*u = (1 - s)/(sc^2)
where s=sinA c=cosA

>
> From the definitions of t and u we have
>
> v = sqrt[g/2 (1 + t^2)/u].
>
> To minimize v, minimize (1 + t^2)/u.
>
> (1 + t^2)/u = h w, where w = 4 sin A + csc A - 3

ie w = (2s - 1)^2/s + 1
which is least when (2s - 1)=0 given s = sinA>0

>
> A = pi/6 gives w = 1, which is the minimum in the first quadrant.
>
> The minimum muzzle velocity is sqrt[g/2 h].
> The corresponding angle of elevation is pi/6.

Aha, great!
HTH
JJ

[Mike Trainor]

On Thu, 14 Feb 2013 11:30:33 -0000, John Jones <a1...@hotamil.com>
wrote:

>Problem
>=======
>There's a cannon on top of a hemispherical hill. It fires a ball, which
>just grazes the hill, before reaching the flat ground below.
>
>What is the *minimum* velocity at which the cannon fires the ball to do
>this?
>Assume: no air resistance, cannon of zero height, ball of zero radius.
>
>Hint: if the cannon fired horizontally the velocity would be sqrt(g*h)
>where g is the gravitational acceleration (32 ft/s/s, 981 cm/s/s) and h
>is the height of the hill (and its radius therefore).
>
>HTH
>JJ

John, Rather intriguing problem, but I cannot find an 'elegant'
answer! This is what I did (may have goofed, but the method is what
matters). Further, my news server did not ship in Ray Koopman's
post.

Anway, what I did is to start with the trajectory of the projectile:

y = h + x* tan A - g*x**2/(2*v**2 * (cos A)**2)

where v is the iniital speed and A the angle of projection.

Now, we need this parabola to touch the circle of radius h.
Let the point on the circle have a polar angle B. Then, the
contact point is

(h cos B, h sin B).

Shove this into the trajectory to get one condition:

sin B = 1 + tan A cos B - g*h*(cos B)**2/(2*v**2 * (cos A)**2)

The slope of the trajectory is easy to get.

y' = tan A - g*x/(v * cos A)**2

Also, the
slope of the circle at the point of contact is - cot B. So, the
second condition is

- cot B = tan A - g*h*cos B/(v**2 * (cos A)**2))

Now, get rid of tan A between the two conditions and do
some algebra. We end up with

v**2 = (g*h/2) (cos B/cos A)**2 * (1 - sin B)

This does not seem quite right as I do not like the factor
of 2. But, the bigger issue is how can one find the
minimum of v, for all A and B?

Thanks for a great problem.

mt

[John Jones]

On Tue, 26 Feb 2013 16:17:56 -0500, "Mike Trainor"
<mtra...@hotmail.com> wrote in article
<s87qi8tmir7il81hg...@4ax.com>...

Mike - A and B are not independent.
Above there are two equations for these two unknowns.
Ray's solution relied on solving for these, although he used different
characterisations.

I havent checked, but I dont like the 2 either. The answer is in fact
when A = B = 30 degrees, when (cos B/cos A)**2 * (1 - sin B) = 1/2

If the result had been
v**2 = (g*h) (cos B/cos A)**2 * (1 - sin B)
then the minimum v (if proven A=B=30):
v**2 = (g*h)/2 would be correct.

Why A=B=30? dunno. But it's nice.
HTH
JJ

[Mike Trainor]

On Wed, 27 Feb 2013 17:09:18 -0000, John Jones <a1...@hotamil.com>

Thanks I will look for a solution by eliminating B in terms
of A. Clearly, the tangency condition links A and B, even
though v occurs in it. But, we can group the terms in v
and isolate is and express v as a function of A alone
(in principle!).

The factor of 2 is all right. I did the simple case of A = 0.
Then, B = pi/2 (the top of the hill is the tangential point.

The parabola is now simply

y = h - g*x**2/(2*v**2)

When the projectile hits the 'ground' y = 0 and
so the x coordinate of the point of impact is given by

x**2 = 2*h*v**2/g

To avoid hitting the hill on the way down, we require

x => h

and thus v => sqrt(g*h/2)

The minimum is just gh/2.

Well, at least in the limiting case, the factor is right!

Back to work now.

thanks
mt