Using lift with beamed propulsion.

I was interested to read the discussion on this forum on beamed propulsion. I would be interested in getting some feedback on the feasibility of the idea proposed below.
It's based on the fact that lift for aircraft can be higher than the thrust for the craft, so why not use this to increase the acceleration produced by beamed propulsion?
There are a few points I'm uncertain about. First, I was assuming that the energy delivered to the craft would be much less than the amount emitted at transmission because of the distances of 100's to 1000's of kilometers. However, I take it after reading the Kare/Parkin paper most of the beam power will arrive at the vehicle because of focusing.
Still if you did reduce the distance by a factor of 10 and using the same transmitter, would the power arriving at the vehicle be increased by a factor of 100?
Secondly, how much acceleration are the laser and microwave systems expected to produce?
Thirdly, my analysis was only a preliminary one. The scenario is made more complicated by the fact the vehicle has to stay oriented to keep the velocity vector along the centerline of the vehicle. Then we would have to determine if the control surfaces are sufficient to keep this orientation or would a portion of the beamed propulsion thrust be needed. Additionally the path would not be in a straight-line which would complicate the calculation of the thrust and lift directions.



- Bob Clark

==============================================
From: Robert Clark
Date: Fri, Jun 30 2006 6:17 pm
Email: "Robert Clark" <rgregorycl...@yahoo.com>
Groups: sci.astro, sci.space.policy, sci.physics
Subject: Re: Using lift to increase speeds

William.M...@gmail.com wrote:
> Orbital speed is where centripetal force equals gravity force and is
> given by;

> v = sqrt(GMe/r)

> Which can be derived from the following three equations;

> F = G*m*Me/r^2 - gravitational force
> a = v^2/r - centripetal acceleration
> F = ma - relating mass and acceleration

> a = F/m = GMe/r^2 - gravitational acceleration
> a = v^2/r - centripetal acceleration

> Setting the two accelerations equal

> v^2/r = GMe/r^2
> v^2 = GMe/r
> v = sqrt(GMe/r)

> If we increase velocity by 41.4% we double the centripetal
> acceleration, which means that if we were to fly an aircraft at Mach 33
> we'd need wings to hold it in the atmosphere! Since wings lift
> aircraft all the time against gravity, it seems reasonable to believe
> that wings could hold an aircraft down. Everything would seem quite
> normal to the occupants, except down would be up to them, and the lift
> would be directed toward the Earth's center.

> The vehicle if possible would be capable of circumnavigating the Earth
> in 60 minutes - and delivering payloads to targets anywhere in 30
> minutes or less.

> Would such a craft be possible?

Yes. I speculated about this possibility for the use with beamed
propulsion:

From: Robert Clark
Date: Sat, Nov 19 2005 2:23 pm
Email: "Robert Clark" <rgregorycl...@yahoo.com>
Groups: sci.astro, sci.physics, sci.math
Subject: Math question for the trajectory of beamed propulsion.
http://groups.google.com/group/sci.astr ... 32000ef7f7

This would also be applicable to the scenario where electrical power
for propulsion is transmitted though long cables:

From: Robert Clark
Date: Fri, May 27 2005 12:10 pm
Email: "Robert Clark" <rgregorycl...@yahoo.com>
Groups: sci.astro, sci.space.policy, sci.physics,
sci.electronics.design, sci.electronics.misc
Subject: Re: Long cables to power "ioncraft" to orbit?
http://groups.google.com/group/sci.astr ... 463e87dde6

The problem is that though the height to orbit might be 100 km, the
horizontal distance travelled might be 2000 km in order to build up
sufficient speed for orbital velocity.
The proposals for beamed propulsion I've seen though do not use
lifting surfaces for the craft:

Riding Laser Beams to Space.
http://www.space.com/businesstechnology ... 00705.html

However, the lift to drag ratios at hypersonic speeds suggest we might
be able to increase the thrust and therefore the acceleration by
several times if the craft was designed for aerodynamic lift. See the
graph showing lift to drag ratio versus Mach number here:

Waverider Design.
http://www.aerospaceweb.org/design/wave ... ider.shtml

With airplanes you have the thrust directed horizontally to overcome
the drag force against forward motion and the lift provides the force
to keep the airplane aloft. Since subsonic L/D ratios can be 15 to 1
and higher the thrust required from the engines is much less than the
actual weight of the plane.
However, with beamed propulsion a key problem is the dimunition of the
power with distance, which decreases with the square of the distance so
you want to keep the distance short. The idea then in this case using
aerodynamic lift would be to use the thrust produced by the beamed
propulsion to overcome gravity and drag and use the lift force to
provide the higher acceleration to reach orbital velocity in a shorter
distance. Essentially the craft would be pointed upwards so that the
wings/lifting surfaces provide the "lift" in the horizontal direction.
The graph on the "Waverider Design" page shows the L/D ratio can be
about 7 to 8 at hypersonic speeds. For instance if the beamed
propulsion provided a thrust of 1 g to counter gravity plus 4 g's
against drag for a total of 5 g's in the vertical direction, then the
horizontal acceleration could be as much as 8*4 = 32 g's.
Note though it would be important to keep the craft oriented so that
so that the velocity vector is always pointed through the forward
centerline of the craft. When lift and drag calculations are made it's
always in regard to the craft moving so the airstream is flowing more
or less parallel over the wings/lifting surfaces, according to angle of
attack. If instead the airstream was flowing perpindicular to the plane
of the wings the lift would be much less and drag would be much greater
so the L/D ratio would be severely reduced. The aerodynamic control
surfaces would be used to keep the craft properly oriented.
Estimates for beamed propulsion are about 1 megawatt of power to send
1 kilogram to orbit. If say such beamed propulsion provided thrust for
5 g's of acceleration then the lifting force could provide 32 g's, or a
factor of 6 more. So the distance required would be smaller by a factor
6. This means the power required would be smaller by a factor 6^2 = 36.
Then 36 times greater mass could be lifted for the same power. This is
dependent though on how much acceleration beamed propulsion could
provide. If it were 7 g's then the lifting acceleration would be 8*6 =
48 g's, about a factor of 7 more. Then the power required would be less
by 7^2 = 49, and 49 times greater mass could be lifted.
There are apparently megawatt class lasers already in operation:

Mid-Infrared Advanced Chemical Laser (MIRACL).
http://www.fas.org/spp/military/program/asat/miracl.htm

Let's say they are at the 10 megawatt stage now. Then this could
accelerate 10 kilos to orbit. Then with aerodynamic lift it could lift
perhaps 360 kilos to orbit, which is the size of a small sized
satellite.

Bob Clark

Here's one scenario in which this could be useful. I was trying to find a trajectory that would minimize the *straight-line* distance to a laser/microwave power trasmitter that nevertheless could provide a long distance of travel for a slow build up of speed. For this purpose I wanted the acceleration in the velocity direction to be low since I wanted the beamed power to provide this acceleration.
The obvious thing to try would be for the craft to travel in a circle and let
the beamed power just slowly build up the speed by the craft's going around
and around in a circle, while the distance to the transmitter stayed constant.
Let's say you wanted the radius to be no more than 100 km say, much shorter
than the 2000 km or so horizontal distance required for the space shuttle. But
if you wanted the final speed to be say 7000 m/s then the radial acceleration
would be v^2/r = 7000^2/100,000 = 49,000,000/100,000 = 490 m/s^2, about 49
g's. But then you're in a worse situation than before because of the high
power required to maintain this acceleration IF this high acceleration were
provided by the beamed power system.
So the idea is not to use the beamed power for this radial acceleration but
instead to provide this by the lifting force. Note that this lifting force
would be radial since it is perpindicular to the velocity vector.
So the idea would be for the beamed power to provide a little more
acceleration than the drag so there would be a slow, gradual build up of
speed around the circle and even at a maximum speed of 7000 m/s, the beamed
power would only have to provide 1/8 the acceleration provided by the lift or
49/8, about 6 g's.



Bob Clark

No, if the beam is focused then it is not an inverse square relationship. The inverse square assumes energy is radiated equally in all directions, like light from a light bulb with no reflector or shade. If the beam were perfectly focused, there would be no loss with distance at all, power a million miles away would be the same as one mile away. Of course perfect focus is not possible, but pretty good focus is. So you might see a 1.1x increase in power with a 10x decrease in distance, but not 10x.

Yes. But Kare and Parkin in their paper "A Comparison of Laser and Microwave Approaches to CW Beamed Energy Launch" speak of trading "power for aperture" on page 7. This leads me to think if you kept the same size lens, but reduced the distance by 10 then the power of the transmitter could indeed be reduced by 100. Or you could reduce the lens diameter by 10, thus reducing the area by 100, and keep the transmitter power the same.
In the Kare/Parkin paper they discuss using an aperture of 462 m so reducing this diameter by a factor of 10 would result in a major reduction in cost and complexity.


Bob Clark

If the beam is focused well, then changing the distance by a factor of 10 should have almost no effect on the received power. It is just like shining a flashlight on a solar cell. If the flashlight beam makes a spot of light as small as or smaller than the solar cell from 10 feet away, then all the power is received, limited only by how efficiently the solar cell converts light to electricity. Flashlight beams are not that well focused so the spot of light it makes gets bigger 100 feet away, and if it becomes larger than the solar cell, part of the light will not land on the solar cell and be lost. That is the reason you would lose power with distance. But if the beam is so well focused that the size of the spot does not change much with distance, like a laser pointer, then the power received 10 feet away is the same as that received a 100 feet away because then entire beam still fits on the solar cell.

Yes but the beam becomes less focused the further you are away. Think of a magnifying lens focusing sunlight on a piece of paper. For a fixed magnifying lens, there is only one distance to the paper that will give you tight focus. Move it closer or farther to the lens and the focus becomes diffuse.
Actually this means you can be too close too. But I suppose making it closer means you can make the lens smaller.


Bob Clark

OK, focus is the wrong word, collimated is better. A collimated beam is parallel, not changing diameter at all over long distances. Laser pointers are well collimated; the spot they make is small at 1 foot and nearly as small at 100 feet. Flashlights are less well collimated; they make small spots at 1 foot and slightly larger spots at 100 feet. This is nothing at all like your magnifying glass example where the size of the spot changes dramatically with distance. In fact, it is the opposite of the magnifying glass. In that case you are taking sunlight which is arriving in nearly parallel beams of light and focusing them down to one small spot. To make a collimated beam you put a point source of light at the focal point of the glass and the light from it passes through the glass to project a collimated beam in the other direction.