Math question for the trajectory of beamed propulsion.

I'm concerned with means to minimize the distance from the launch point
to a beamed propulsion craft so that the beamed energy reaching the
craft is not diminished too greatly.
Here's the scenario. The craft has to reach a certain speed in order
to reach orbital velocity. It's on the order of 7800 m/s. I'll just
call it
10,000 m/s for simplicity. You also would like the acceleration not to
be too
large The reason is that for beamed propulsion, for a given amount of
beamed energy
transmitted, a fixed thrust is going to be provided. From Newton's
equation F=ma, so
if you make the acceleration large, the mass, or payload will be
reduced. So
let's say the acceleration is kept below 40 m/s^2, or about 4 g's.
But then it takes time to build up the large velocity required. And
traveling at the limited acceleration allowed would force a long
distance to
be used to build up the necessary speed. Indeed for the space shuttle
for example the distance from the launch pad is on the order of 2000 km
for
powered thrust. This large distance is what I'm trying to avoid. A few
hundred km is
what I'm looking for.
So here's the suggestion. What's really needed for the beamed
propulsion is keeping the *straight-line* distance to the craft low.
Then what you
could do would have the craft travel in say a helical path. Then it
could be
traveling a long distance, the length of the helical path, while the
straight-line distance from start to finish might be low.
Yet this introduces more problems. For when the craft is turning in
that helical path, it is undergoing acceleration. And if the turning
radius
is small, the acceleration could be high, and your low acceleration
condition might be contradicted anyway.
So as a math problem what I'm looking for is a trajectory in 3-dim'l
space s(t) such that the magnitude of the acceleration is low, |s"(t)|
= 40,
the length *along* the trajectory curve is some given number say 2,000
km,
but the *straight-line* distance from beginning to end is minimized.
This is a calculus of variations problem, like the brachistochrone
problem. Note that without the condition that the acceleration is
limited we
could make the straight-line distance arbitrarily small by making the
curve make
many twists and turns while enclosed in a small box.
With the acceleration condition, I'm inclined to believe you can't do
any better than a straight-line but I don't see how to prove that.


- Bob Clark

"Robert Clark" wrote in message
ups.com...
So here's the suggestion. What's really needed for the beamed
propulsion is keeping the *straight-line* distance to the craft low.

....

This is a calculus of variations problem, like the brachistochrone
problem. Note that without the condition that the acceleration is
limited we
could make the straight-line distance arbitrarily small by making the
curve make
many twists and turns while enclosed in a small box.

Get a grip Robert, why do you always make everything
so hard? You just need to fly in a circle.

With the acceleration condition, I'm inclined to believe you can't do
any better than a straight-line but I don't see how to prove that.

Put your craft on the end of a hawser. Release when
you have enough speed. Set the radius so the inward
acceleration is as much as you can stand just as you
release. Ever seen someone throwing the hammer?

George

George Dishman wrote:
"Robert Clark" wrote in message
ups.com...
So here's the suggestion. What's really needed for the beamed
propulsion is keeping the *straight-line* distance to the craft low.

...

This is a calculus of variations problem, like the brachistochrone
problem. Note that without the condition that the acceleration is
limited we
could make the straight-line distance arbitrarily small by making the
curve make
many twists and turns while enclosed in a small box.

Get a grip Robert, why do you always make everything
so hard? You just need to fly in a circle.

With the acceleration condition, I'm inclined to believe you can't do
any better than a straight-line but I don't see how to prove that.

Put your craft on the end of a hawser. Release when
you have enough speed. Set the radius so the inward
acceleration is as much as you can stand just as you
release. Ever seen someone throwing the hammer?

George

Acceleration a= v^2/r, v the speed , r the radius for a body traveling
in the circle. So r = v^2/a. For v = 10,000 m/s and a = 40 m/s^2, r =
2,500,000 m, or 2500 km.
This is worse than going in a straight line at constant a = 40 m/s^2 :
s= v^2/2a = 1,250,000 m = 1250 km.


- Bob

Let us accelerate craft in a straight line. Let rate of aceleration be
A.

Let us now put a kink in the course. Let velocity be v craft having
accelerated for v/A seconds. Let us have a small acceleration a to the
side.

Total gain in velocity will be A-a^2/2 assuming aA

After a further t seconds velocity will be v+(A-a^2/2)t,at or
(v+A)t-t(a^2-a^2v/(At+V)

This is always less than v+At the velocity it would have if it had been
accelerated in a straight line. The distance travelled will be the
integral of this.

In fact small kinks reduce the distance to the accelerator but only by
reducing total velocity. Hence the optimal is a straight line.
Straightening out the king is always an improvement. One never improves
the situation by accentuating a kink.

wrote in message
ups.com...
Let us accelerate craft in a straight line. Let rate of aceleration be
A.

Let us now put a kink in the course. Let velocity be v craft having
accelerated for v/A seconds. Let us have a small acceleration a to the
side.

Total gain in velocity will be A-a^2/2 assuming aA

If the acceleration is transverse to the direction
of motion, it doesn't change the speed at all.

George

Yes Indeed. But the change in velocity is reduced because not all
components are in the direction of motion. This consideration is the
basis of the variational approach.

Wrt. our previous discusssion you go straight to Alpha Centauri, you do
not introduce variational kinks.

wrote:
Let us accelerate craft in a straight line. Let rate of aceleration be
A.

Let us now put a kink in the course. Let velocity be v craft having
accelerated for v/A seconds. Let us have a small acceleration a to the
side.

Total gain in velocity will be A-a^2/2 assuming aA

After a further t seconds velocity will be v+(A-a^2/2)t,at or
(v+A)t-t(a^2-a^2v/(At+V)

...

How do you derive A-a^2/2 for the change in velocity?


Bob Clark

Robert Clark wrote:
George Dishman wrote:
"Robert Clark" wrote in message
ups.com...
So here's the suggestion. What's really needed for the beamed
propulsion is keeping the *straight-line* distance to the craft low.

...

This is a calculus of variations problem, like the brachistochrone
problem. Note that without the condition that the acceleration is
limited we
could make the straight-line distance arbitrarily small by making the
curve make
many twists and turns while enclosed in a small box.

Get a grip Robert, why do you always make everything
so hard? You just need to fly in a circle.

With the acceleration condition, I'm inclined to believe you can't do
any better than a straight-line but I don't see how to prove that.

Put your craft on the end of a hawser. Release when
you have enough speed. Set the radius so the inward
acceleration is as much as you can stand just as you
release. Ever seen someone throwing the hammer?

George

Acceleration a= v^2/r, v the speed , r the radius for a body traveling
in the circle. So r = v^2/a. For v = 10,000 m/s and a = 40 m/s^2, r =
2,500,000 m, or 2500 km.
This is worse than going in a straight line at constant a = 40 m/s^2 :
s= v^2/2a = 1,250,000 m = 1250 km.


- Bob

However, we might be able to reduce the distance by using a
combination of a circle and a straight-line. You would use a circle to
build up to a certain velocity then travel in a straight-line for the
rest of the distance:

You gradually build up the speed going in the circle. As you are
buiding up speed in the circle the acceleration will be both along the
circle as well as the usual radial acceleration for a constant speed
around the circle. Since the total accleration has to be less than 40
m/s^2, you would keep the tangential acceleration small but the radial
acceleration close to but less than 40 m/s^2. Then when the speed
finally reaches the highest speed you want in the circle, you make the
acceleration be purely radial at 40 m/s^2. Then r = v^2/40.
As for the part of the trajectory in a straight-line, you would direct
this part radially inward into the circle you were just using. Then no
additional distance need be used as far as the distance from the launch
point is concerned, as long as this straight-line stays inside the
circle.
Then what we have is that the speed is v when going into the
straight-line. We want to build up the speed again at 40 m/s^2 to
10,000 m/s. Let's calculate the distance required to build up to a
speed given some initial speed. The formulas for constant acceleration
travel in a straight-line a

v = v0 + at, and
x = x0 +v0t +(1/2)at^2

For our scenario we can set x0 = 0. However, the v0 has to be the
speed we get from the circle portion of the trajectory, so we keep that
in the equations. Then t = (v-v0)/a and plugging this into the equation
for x we get: x = (v^2 - v0^2)/2a .
So if v is the initial speed you get coming from the circle, a = 40
m/s^2, and 10,000 m/s is the speed you want to reach, then x =
(10,000^2 - v^2)/80 is the distance it takes to reach the speed of
10,000 m/s. Now you want this distance to remain inside the circle, so
you want (10,000^2 - v^2)/80 = 2r = 2(v^2/40) , 10,000^2 -v^2 = 4v^2
, 10,000^2 = 5v^2, and
v^2 = (10,000^2)/5 . Now r = v^2/40 , and you want r as small as
possible so you take the smallest allowable v, which means v^2 =
10,000^2/5, and r = 500,000m = 500 km.
Then this would provide a shorter distance then traveling in a
straight-line alone.
However, this is for a path in 2 dimensions. For the actual path we
would need there to some vertical component to the path as well, at
least up to the altitude for orbit for example.
Still this 2-dim'l example may provide a means for finding the answer
in 3-dimensions.




Bob Clark

Robert Clark wrote:
....
Acceleration a= v^2/r, v the speed , r the radius for a body traveling
in the circle. So r = v^2/a. For v = 10,000 m/s and a = 40 m/s^2, r =
2,500,000 m, or 2500 km.
This is worse than going in a straight line at constant a = 40 m/s^2 :
s= v^2/2a = 1,250,000 m = 1250 km.


- Bob

However, we might be able to reduce the distance by using a
combination of a circle and a straight-line. You would use a circle to
build up to a certain velocity then travel in a straight-line for the
rest of the distance:

You gradually build up the speed going in the circle. As you are
buiding up speed in the circle the acceleration will be both along the
circle as well as the usual radial acceleration for a constant speed
around the circle. Since the total accleration has to be less than 40
m/s^2, you would keep the tangential acceleration small but the radial
acceleration close to but less than 40 m/s^2. Then when the speed
finally reaches the highest speed you want in the circle, you make the
acceleration be purely radial at 40 m/s^2. Then r = v^2/40.
As for the part of the trajectory in a straight-line, you would direct
this part radially inward into the circle you were just using. Then no
additional distance need be used as far as the distance from the launch
point is concerned, as long as this straight-line stays inside the
circle.
Then what we have is that the speed is v when going into the
straight-line. We want to build up the speed again at 40 m/s^2 to
10,000 m/s. Let's calculate the distance required to build up to a
speed given some initial speed. The formulas for constant acceleration
travel in a straight-line a

v = v0 + at, and
x = x0 +v0t +(1/2)at^2

For our scenario we can set x0 = 0. However, the v0 has to be the
speed we get from the circle portion of the trajectory, so we keep that
in the equations. Then t = (v-v0)/a and plugging this into the equation
for x we get: x = (v^2 - v0^2)/2a .
So if v is the initial speed you get coming from the circle, a = 40
m/s^2, and 10,000 m/s is the speed you want to reach, then x =
(10,000^2 - v^2)/80 is the distance it takes to reach the speed of
10,000 m/s. Now you want this distance to remain inside the circle, so
you want (10,000^2 - v^2)/80 = 2r = 2(v^2/40) , 10,000^2 -v^2 = 4v^2
, 10,000^2 = 5v^2, and
v^2 = (10,000^2)/5 . Now r = v^2/40 , and you want r as small as
possible so you take the smallest allowable v, which means v^2 =
10,000^2/5, and r = 500,000m = 500 km.
Then this would provide a shorter distance then traveling in a
straight-line alone.
...


This won't work, at least not as written. I'm trying to direct the
craft into the circle while using the velocity it attained in
travelling around the circle. However, this velocity vector around the
circle would be directed *tangentially* to the circle. So I couldn't
use this to give an additional boost radially into the circle.
What might do it is if this high speed took place while still in
significant atmosphere. Then we might be able to use lifting surfaces
on the craft to provide a velocity component perpindicular to the
direction of motion. Indeed the possibility of using lifting surfaces
would significantly increase the range of accelerations permitted
beyond that allowed by the beamed propulsion itself.


Bob Clark