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Using lift with beamed propulsion.
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Space Walker
Joined: Sat Jul 01, 2006 12:18 am
Posts: 213 
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 straightline 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: MidInfrared 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 
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Space Walker
Joined: Sat Jul 01, 2006 12:18 am
Posts: 213 
Here's one scenario in which this could be useful. I was trying to find a trajectory that would minimize the *straightline* 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 
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Moon Mission Member
Joined: Tue Oct 05, 2004 5:38 pm
Posts: 1361 Location: Austin, Texas 
RGClark wrote: ...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? 
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Space Walker
Joined: Sat Jul 01, 2006 12:18 am
Posts: 213 
campbelp2002 wrote: RGClark wrote: ...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? 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 
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Moon Mission Member
Joined: Tue Oct 05, 2004 5:38 pm
Posts: 1361 Location: Austin, Texas 
RGClark wrote: ...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. 
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Space Walker
Joined: Sat Jul 01, 2006 12:18 am
Posts: 213 
campbelp2002 wrote: RGClark wrote: ...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. 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 
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Moon Mission Member
Joined: Tue Oct 05, 2004 5:38 pm
Posts: 1361 Location: Austin, Texas 
RGClark wrote: 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. 
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