Parachute/Aerobraking theory question. Please answer!

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BBRAZEAU

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I you made a parachute with a rigid frame and hollow center tube like an umbrella and had it fire a projectile out of the end of the tube facing skyward at say 150,000 ft. in altitude would the reaction force of the aerobraking effect be enough to give a counter force for the projectile to accelerate at say 1/2 the rate of a normal ballistic reaction? Or more importantly what scale would surface area of parachute have to be in regard to projectile mass/shape, barrel length/fit, charge size/burn rate? The answer does not have to be mathematically verified. I'm just looking for an idea if this is feasible.
 
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billslugg

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<p>Welcome to SDC! I hope&nbsp;we can give the answer to your question.&nbsp;</p><p>It makes little difference how big the parachute is. There is no way that you could possibly transfer the spike of force to the atmosphere. You could not make the assembly rigid enough. The solution is trivial though. You simply provide a&nbsp;reaction mass at the gun barrel. The heavier you make it, the faster the projectile will exit the tube and the slower the tube will recoil. Conservation of momentum.</p> <div class="Discussion_UserSignature"> <p> </p><p> </p> </div>
 
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BBRAZEAU

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<p>"There is no way that you could possibly transfer the spike of force to the atmosphere"</p><p>Wouldn't the spike force be a function of the burn time and the encapsulation (fit) between projectile and barrel? Decrease burn force increase burn time. manipulate the gas/pressure equation of the semi ballistic reaction to 1/4 or 1/8 the rate of a normal ballistic reaction. Could any gain be made over a projectile launch thru an open ended tube?</p><p>I think that the forces a parachute encounter are pretty strong and sudden are they not? The assembly I envision would not transfer forces to the members of the arm frame work. It would have the normal attachment lines going from tips of parachute down to the bottom of the tube. The frame work would only hold the parachute open so the reaction force/mass was there immediately.</p><p>&nbsp;</p><p>&nbsp;</p>
 
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DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>"There is no way that you could possibly transfer the spike of force to the atmosphere"Wouldn't the spike force be a function of the burn time and the encapsulation (fit) between projectile and barrel? Decrease burn force increase burn time. manipulate the gas/pressure equation of the semi ballistic reaction to 1/4 or 1/8 the rate of a normal ballistic reaction. Could any gain be made over a projectile launch thru an open ended tube?I think that the forces a parachute encounter are pretty strong and sudden are they not? The assembly I envision would not transfer forces to the members of the arm frame work. It would have the normal attachment lines going from tips of parachute down to the bottom of the tube. The frame work would only hold the parachute open so the reaction force/mass was there immediately.&nbsp;&nbsp; <br />Posted by BBRAZEAU</DIV></p><p>The problem that you have is that the recoil forces are the result of conservation of momentum and the precise nature of the reactive forces.&nbsp; With no opposing force the gun/umbrella with simply react with the combined momentum of the projectile and the propelling gasses that exist the muzzle of the gun.&nbsp; The umbrella will provide a reacting force that is governed by the speed with respect to the atmosphere and the density of the atmosphere.&nbsp; However, if you start from rest, then there is no speed to create the reacting force, and so the initial recoil forces will not be affected by the aerodynamics resistance of the umbrella.&nbsp; In short, the aerodynamic forces will not be appreciable at low velocities and to control recoil you need to control the dynamics at low velocities.</p><p>The reason that parachust forces are strong and sudden is that parachutes open at high velocities relative to the air.&nbsp; That is not the case for your problem.<br /></p> <div class="Discussion_UserSignature"> </div>
 
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BBRAZEAU

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<p>Forgive my obstinacy but. wouldn't the gas expand in both directions until the projectile exits the muzzle end of the barrel? And if this is the case then, depending on how long the barrel were wouldn't the accelerations which are high, try to move the umbrella down at the same&nbsp;rate as the projectile moves up and thus create its own reaction mass/force with a portion of it's acceleration over a time transient? Also I'm not positive what the shock wave shape&nbsp;of a parachute moving thru air&nbsp;is but isn't its reaction force more a function of the volume of air that the canopy displaces with the velocity of the of the parachute being more a limited factor in the equation&nbsp;past an optimal acceleration?</p><p>&nbsp;In short&nbsp;if the equation for the distribution of the forces with respect to time were plotted wouldn't the up (parachute reaction) side&nbsp;be an exponentially increasing force while the down (projectile mass) side be&nbsp;more constant and thus create a net gain?</p>
 
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DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Forgive my obstinacy but. wouldn't the gas expand in both directions until the projectile exits the muzzle end of the barrel?Replying to:<BR/><DIV CLASS='Discussion_PostQuote'></p><p>No.&nbsp; There is high pressure, hot rapidly expanding gas behiing the projectile (while in the bore) and much cooler, lower pressure gas being compressed and expelled from the barrel ahead of the projectile.&nbsp; The pressure behind the projectile, due to combustion of the propellant, is MUCH higher and the gas is MUCH hotter than the gas in front of the projectile.&nbsp; Flame temperatures of the propellants involved are on the order of 3500K.</p><p>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>&nbsp;And if this is the case then, depending on how long the barrel were wouldn't the accelerations which are high, try to move the umbrella down at the same&nbsp;rate as the projectile moves up and thus create its own reaction mass/force with a portion of it's acceleration over a time transient? </DIV></p><p>One way to look at this problem is to consider the effect of the gas pressure on the projectile and on the breech of the gun.&nbsp; The pressures are the same.&nbsp; The&nbsp;effective areas are the same.&nbsp; There is a frictional component between the projectile and the bore which is reacted as a retarding force on the gun,but that can be safely neglected in a first-order analysis.&nbsp; So basically the force on the projectile equals the force on the gun, although they are in opposite directions.&nbsp; But the gun is usually MUCH more massive than the projectile, and so the acceleration of the gun is correspondingly less.&nbsp; When the bullet exits the muzzle, the gas from the combustion of the propellant expands very rapidly and exits the muzzle at the local speed of sound (in the hot gas) which adds substantially to the momentum of the material expelled from the barrel (projectile plus gas) and contributed to recoil.&nbsp; But what is conserved is momentum, not acceleration or speed.&nbsp; The rearward momenteum of the gun is equal to the forward momentum of the projectile and expelled gasses.&nbsp;&nbsp;Because the gun is MUCH heavier than the projectile and gasses, the speed of the gun is MUCH lower. (Momentum is mass times velocity).</p><p>&nbsp;Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>&nbsp;Also I'm not positive what the shock wave shape&nbsp;of a parachute moving thru air&nbsp;is but isn't its reaction force more a function of the volume of air that the canopy displaces with the velocity of the of the parachute being more a limited factor in the equation&nbsp;past an optimal acceleration?&nbsp;In short&nbsp;if the equation for the distribution of the forces with respect to time were plotted wouldn't the up (parachute reaction) side&nbsp;be an exponentially increasing force while the down (projectile mass) side be&nbsp;more constant and thus create a net gain? <br />Posted by BBRAZEAU</DIV></p><p>The reaction force of a parachute is a funtion of two major things.&nbsp; One is velocity, which determines dynamics pressure (dynamic pressure is 1/2*air density*velocity^2).&nbsp; The other is projected area normala to the velocity.&nbsp; The reation force from the parachute, since the area is basically constant, will be primarily dependent on the velocity.&nbsp;</p><p>I don't understand your last sentence.&nbsp; But at the moment that the projectile leaves the barrel, if one assumes that the parachute is stationary with respect to the atomosphere, the reaction will be determined completely by conservation of momentum, the momentum of the projectile and expelled gasses, and there will be no aerodynamic effect.&nbsp; Aerodynamics will only be non-zero once the parachute has accelerated and has a velocity relative to the air, which creates drag. </p><p><br /><br />&nbsp;</p> <div class="Discussion_UserSignature"> </div>
 
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BBRAZEAU

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<p>Correction: I said "wouldn't the gas expand in both directions until the projectile exits the muzzle end of the barrel?" What I should have said was: gas would try to expand in all directions but since its constrain in a barrel it will have only 2 appreciable vectors down and up. I still believe this to be true. Having said that I agree with every thing you said in your first paragraph but think it does more to substantiate my&nbsp;idea than dispel it.</p><p>In your second paragraph "When the bullet exits the muzzle, the gas from the combustion of the propellant expands very rapidly and exits the muzzle at the local speed of sound (in the hot gas) which adds substantially to the momentum of the material expelled from the barrel (projectile plus gas) and contributed to recoil.&nbsp; But what is conserved is momentum, not acceleration or speed."</p><p>Question: What does substantially represent here 30-50-60%? I find this hard to comprehend since the millisecond it leaves the muzzle the compressed gases are free to expand in any direction not just 2. </p><p>Question: How does the projectile get from one end of the barrel to the other so quickly without having built up acceleration, momentum ,and speed? Where your explanation of conservation of momentum uses the mass/ weight of the gun, mine uses the opposing force of areobraking. the fact that the distance of recoil on the breech side before sufficient force is developed to create momentum on the muzzle side will be larger than that of heavy gun does not mean there is no momentum transferred just that it will be less. If you had an extremely light weight&nbsp;gun clamped solid and fired it the bullet would go further than if you had the gun suspended in air by rubber bands, but the bullet would still exit the barrel in the latter with a lot of momentum. Would it not?</p><p>&nbsp;</p>
 
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DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>...Question: What does substantially represent here 30-50-60%? I find this hard to comprehend since the millisecond it leaves the muzzle the compressed gases are free to expand in any direction not just 2.</DIV></p><p>To calclulate the effect of the gasses precisely you need to know the weight of the propellant and quite a bit about the composition of the propellant and the thermodynamic properties of the combustion gasses.&nbsp;&nbsp; As a rule of thumb about half of the recoil is due to the gasses, or, more precisely&nbsp;you can look at the momentum&nbsp;of the gasses which is about the weight of the initial propellant with a velocity of about 4700 ft/s (this velocity is about right for rifles and is taken from Hatcher's Notebook, a useful reference in such matters.)</p><p>The gasses don't really start to expand radially until they have left the muzzle.&nbsp; The initial expansion is is pretty much along the axis of the barrel, and that is what contributes to momentum.&nbsp; If you draw a free body diagram, it is only what happens precisely at the muzzle exit that determines the momentum of the gun.&nbsp; And in any case the radial expansion of the gasses is symmetric and hence the contribution to momentum of the component of velocity normal to the axis is zero.</p><p>&nbsp;Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>&nbsp;Question: How does the projectile get from one end of the barrel to the other so quickly without having built up acceleration, momentum ,and speed? Where your explanation of conservation of momentum uses the mass/ weight of the gun, mine uses the opposing force of areobraking. the fact that the distance of recoil on the breech side before sufficient force is developed to create momentum on the muzzle side will be larger than that of heavy gun does not mean there is no momentum transferred just that it will be less. If you had an extremely light weight&nbsp;gun clamped solid and fired it the bullet would go further than if you had the gun suspended in air by rubber bands, but the bullet would still exit the barrel in the latter with a lot of momentum. Would it not?&nbsp; <br />Posted by BBRAZEAU</DIV></p><p>The projectile clearly undergoes extreme acceleration and has high speed with corresponding momentum by the time it leaves the barrel.&nbsp; Conservation of momentum is a cornerstone of classical mechanicsm, and is the central principle behind recoil.</p><p>I have no idea what you mean in the second sentence above.&nbsp; The momentum of the propellant/gas combination in the forward direction will be precisely equal to the momentum of the gun the rearward direction.&nbsp; Because the gun is much heavier than the projectile and gasses combined the velocity of the gun will be correspondingly lower, and because energy is 1/2 m*v^2 the energy of the gun will also be less than the kinetic energy of the projectile/gas combination -- which is why heavy guns seem to&nbsp;"kick" less that light guns.&nbsp; But ther is no aerobraking until the gun has some velocity and that velocity will be very small unless the gun is very very&nbsp;light or the projectile and propellant are quite heavy.&nbsp; </p><p>The difference in velocity between a light gun clamped solidly and a heavy gun suspended by rubber bands would be miniscule.&nbsp; It is common practice to measure bullet velocity using a chronograph and a rifle that is rather free to recoil backwards (simply sitting on bags at the forward and aft end).&nbsp; This loose suspension of the rifle has virtually no measureable effect on bullet velocity.</p><p>The bullet momentum will be virtually identical in either of the cases that you propose.&nbsp; Whether that is a lot of momentum or not depends on your standards.&nbsp; A 150 grain 30-06 bullet leaving the muzzle at 2800 fps has a momentum of 1.86 lb-sec.&nbsp; A 20-lb cannon projectile leaving the muzzle at 5000 fps has a momentum of 3108 lb-sec.<br /></p> <div class="Discussion_UserSignature"> </div>
 
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DrRocket

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<p>Here is an alternate way to look at the problem.&nbsp; It is very conservative, and overestimates the effect of the "umbrella" but illustrates the basic physics and ought to provide an upper bound on the effedtiveness of the umbrella concept.</p><p>If you put an umbrella on the gun near and surrounding the muzzle, then you still have conservation of momentum going for you. The momentum of the propellant and gasses in the forward direction will still match the momentum of the gun plus whatever moves with it in the rearward direction.&nbsp; If the umbrella is fixed rigidly to the muzzle and moves rearward with muzzle, without deformation, then it will carry with it some mass of air.&nbsp; So the momentum of the projetile/gas combinatin will be matched by the momentum of the gun/air combination.&nbsp; The effect is to add the mass of some amount of air to the mass of the gun.&nbsp; Now the precise amount of air involved will be rather difficult to calculate, since the air is compressible, and some will flow around the&nbsp;umbrella.&nbsp; But if you simply assume that a column of air covered by the umbrella of some reasonable length moves with the barrel as a rigid "slug" that will give you an upper bound for the effectiveness of the umbrella idea.&nbsp; </p> <div class="Discussion_UserSignature"> </div>
 
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BBRAZEAU

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<p>"The difference in velocity between a light gun clamped solidly and a heavy gun suspended by rubber bands would be miniscule.&nbsp;" How about the difference between light gun clamped solidly and light gun suspended by rubber bands.Which is what I meant.<br /><img src="http://sitelife.space.com/ver1.0/Content/images/store/8/11/b890ef8b-2963-4a07-bb49-9acb34df9185.Medium.jpg" alt="" />"The effect is to add the mass of some amount of air to the mass of the gun." Precisely, And the amount of work this mass can do&nbsp;is governed by volume of air and delta t that it has before projectile reaches the muzzle end. I'm thinking very long barrel on the order of 100-150 L/D</p>
 
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DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>"The difference in velocity between a light gun clamped solidly and a heavy gun suspended by rubber bands would be miniscule.&nbsp;" How about the difference between light gun clamped solidly and light gun suspended by rubber bands.Which is what I meant."The effect is to add the mass of some amount of air to the mass of the gun." Precisely, And the amount of work this mass can do&nbsp;is governed by volume of air and delta t that it has before projectile reaches the muzzle end. I'm thinking very long barrel on the order of 100-150 L/D <br />Posted by BBRAZEAU</DIV></p><p>An L/D of 100 or so is long but not unheard of.&nbsp; My 22-25- has a 26" barrel and hence and L/D of about 118.&nbsp; When barrels start to have very long L/D you either need to make them very stiff (i.e. large in outside diameter) or give up quite a bit of accuracy.&nbsp; This is because the chief cause of variability in point of impact is the vibrational modes of the barrel which combine with small differences in residence time of the projectile to create significantly different states at the time that the projectile exits from the muzzle.</p><p>A light gun clamped solidly and a light gun suspended by rubber bands will still not be that different unless the gun is very light indeed.&nbsp; But the variability in the dynamics will make a huge difference in accuracy -- and the consistency will determine which is the more accurate.&nbsp; I would bet on the solidly clamped gun in most cases, but I can see possible situations where the freely suspended gun might be more accurate.</p><p>If you want to reduce recoil a very effective means is a muzzle brake.&nbsp; Muzzle brakes vent a portion of the exhaust gasses rearward and provide thereby a force in the forward direction and also reduce the substantial contribution of the gasses with respect to the non-vented case (they take some of what would be forward momentum and convert it to rearward momentum).&nbsp; The down side is that they are really loud -- but that is not a problem with a cannon since cannon muzzle blast already requires that crews be protected.</p><p>Mass does not do work.&nbsp; Only force applied over a distance does work.<br /></p> <div class="Discussion_UserSignature"> </div>
 
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DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I you made a parachute with a rigid frame and hollow center&nbsp;tube&nbsp;like an umbrella and had it fire a projectile out of the end of the tube facing skyward at say 150,000 ft. in altitude would the reaction force of the aerobraking effect be enough to give a counter force for the projectile to accelerate at say 1/2 the rate of a normal ballistic reaction? Or more importantly what scale would surface area of parachute have to be in regard to projectile mass/shape, barrel length/fit, charge size/burn rate?&nbsp;The answer does not have to be mathematically verified. I'm just looking for an idea if this is feasible. <br />Posted by BBRAZEAU</DIV></p><p>One thing you need to consider is that at 150,000 feet the air is VERY thin so you are actually working in pretty fair vacuum.&nbsp; A parachute works best in high density air.</p><p>This table shows how thin the air is at high altitudes --&nbsp;only a bit more than&nbsp;0.1% as dense near 150,000 ft as at sea level.</p><p><table border="0" class="wikitable"><tbody><tr><th rowspan="2">Subscript <em>b</em></th><th colspan="2">Height Above Sea Level (<em>h</em>)</th><th colspan="2">Mass Density (<span class="texhtml">&rho;</span>)</th><th rowspan="2">Standard Temperature (<em>T'</em>)<br />(K)</th><th colspan="2">Temperature Lapse Rate (<em>L</em>)</th></tr><tr><th>(m)</th><th>(ft)</th><th>(kg/m&sup3;)</th><th>(slugs/ft&sup3;</th><th>(K/m)</th><th>(K/ft)</th></tr><tr><td align="center">0
 
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BBRAZEAU

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<p>Thanks for the link to nasa. Good info.</p><p>"A parachute works best in high density air." Yes, so the chosen height must give max air density acheivable while still meeting the other parameters of the desired result... Put some mass "x kg." into oribit as cheaply and reliably as possible.</p><p>Mass="x kg" (<u><strong>cargo only</strong></u> so g forces experienced are not limiting factor)( nothing too heavy so scale of concept is managable)</p><p>Use&nbsp;helium balloons to lift assembly to 150 k.</p><p>Use aero-braking (parachute/umbrella) + the mass of a light weight launch tube to supply a "semi" ballistic reaction to&nbsp;boost payload and rocket. ( may need sabot behind rocket). low air density has +'s and -'s on these reactions.</p><p>Use 1 or 2 small stages from rocket to achieve orbit.</p>
 
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DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Thanks for the link to nasa. Good info."A parachute works best in high density air." Yes, so the chosen height must give max air density acheivable while still meeting the other parameters of the desired result... Put some mass "x kg." into oribit as cheaply and reliably as possible.Mass="x kg" (cargo only so g forces experienced are not limiting factor)( nothing too heavy so scale of concept is managable)Use&nbsp;helium balloons to lift assembly to 150 k.Use aero-braking (parachute/umbrella) + the mass of a light weight launch tube to supply a "semi" ballistic reaction to&nbsp;boost payload and rocket. ( may need sabot behind rocket). low air density has +'s and -'s on these reactions.Use 1 or 2 small stages from rocket to achieve orbit. <br />Posted by BBRAZEAU</DIV></p><p>OK.&nbsp; Why didn't say in the beginning that your objective was to put a payload into orbit?</p><p>That means a BIG gun, It means a complex system.&nbsp; It also means hardening the payload against the extremely high g forces of a gun launch.&nbsp; It means not only providing high speed, but also controlling the vector to achieve the proper orbit.&nbsp; It means launching a package with at least divert propulsion and control systems.&nbsp; It means getting that gun to a high altitude by some outside methods.</p><p>In short this is idea is very complex. and IT AIN'T GONNA WORK.</p> <div class="Discussion_UserSignature"> </div>
 
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BBRAZEAU

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<p>If payload is kept small/light. If the SEMI BALLISTIC "g" forces (remember this is <strong><u>NOT</u></strong> meant to give velocities of a gun) are not so extreem as require payload hardening. If, as I stated in my last thread, the rocket had a stage to give vector and final velocity.</p><p>This is not such a crazy idea. As there are people doing this sort of thing now and in the past 1)."JP aeorspace"(balloon launched rocket) 2).HARP, High Altitude Reasearch Project (ground based gun launch of projectiles only)&nbsp; 3).The McGill group Marlet3 which was a disgrading sabot solid-propellent rocket (but, again ground based)</p><p>I know you can't achieve a cannon firing at 150k ft. in altitude, but thats not what I'm suggesting. In an excerpt from the article with some of the above info in it "<span style="font-size:12pt;font-family:'TimesNewRoman'">The fact that a projectile leaving the muzzle of the space cannon loses energy from that instant on means that it has its highest velocity during the part of its flight path that moves through the densest parts of the atmosphere. " <font face="Verdana" size="1">my idea remedies this. "<span style="font-size:12pt;font-family:'TimesNewRoman'">One way of minimizing these losses is to launch the projectiles from the top of a mountain. Calculations show that launch energy requirements are cut by almost a third if the cannon's muzzle is placed on a mountaintop at an altitude of 4.6 kilometers (15,000 feet). <font face="Verdana" size="1">my idea remedies this. What I am suggesting is that by lofting the assembly to altitudes where friction is decreased and supplimenting initial rocket launch with a toned down ballistic reaction gains can be made!</font></span></font></span></p>
 
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DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>If payload is kept small/light. If the SEMI BALLISTIC "g" forces (remember this is NOT meant to give velocities of a gun) are not so extreem as require payload hardening. If, as I stated in my last thread, the rocket had a stage to give vector and final velocity.This is not such a crazy idea. As there are people doing this sort of thing now and in the past 1)."JP aeorspace"(balloon launched rocket) 2).HARP, High Altitude Reasearch Project (ground based gun launch of projectiles only)&nbsp; 3).The McGill group Marlet3 which was a disgrading sabot solid-propellent rocket (but, again ground based)I know you can't achieve a cannon firing at 150k ft. in altitude, but thats not what I'm suggesting. In an excerpt from the article with some of the above info in it "The fact that a projectile leaving the muzzle of the space cannon loses energy from that instant on means that it has its highest velocity during the part of its flight path that moves through the densest parts of the atmosphere. " my idea remedies this. "One way of minimizing these losses is to launch the projectiles from the top of a mountain. Calculations show that launch energy requirements are cut by almost a third if the cannon's muzzle is placed on a mountaintop at an altitude of 4.6 kilometers (15,000 feet). my idea remedies this. What I am suggesting is that by lofting the assembly to altitudes where friction is decreased and supplimenting initial rocket launch with a toned down ballistic reaction gains can be made! <br />Posted by BBRAZEAU</DIV></p><p>The effect of aerodynamic drag and the relationship between drag, speed, and altitude are very well know.&nbsp; All rockets are analyzed for thiese effects early in the development (Post 6D is the most common computer code along with some drag codes).&nbsp; Aerodynamic drag is the primary reason that when trajectories are optimized the best trajectory is commonly quite steep so as to gain altitude and reduce drag quickly.</p><p>You idea addresses a very well-known concern and one that is a design driver for modern launch systems.&nbsp; A gun launch is not a particularly good solution.&nbsp; It has been considered.&nbsp; There are too many other system-level problems with the idea.&nbsp; Other ideas including electromagnetic propulsion in place of the chemical propulsion of a conventional gun have also been considered.&nbsp; There really isn't anything new in this arena.&nbsp; <br /></p> <div class="Discussion_UserSignature"> </div>
 
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BBRAZEAU

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<p>Replying to:</p><p>"There are too many other system-level problems with the idea.&nbsp; Other ideas including electromagnetic propulsion in place of the chemical propulsion of a conventional gun have also been considered.&nbsp; There really isn't anything new in this arena."</p><p>What are some of the system-level problems?... Having to harden payloads or rocket fuel? Addressed by not accelerating as fast. At what "g" force do you need to harden payloads to the point where weight becomes a factor 100-500-1000? Stay below this acceleration. Which plays into the idea because&nbsp;potential reaction mass is smaller than ground based systems.</p><p>The method of aceleration in the barrel is open and there is much new in this arena. electromagnetic,coil,ram accelerators,mass drivers,light gas guns (my favorite, because <u>I think</u> it would have lightest weight and adjustable accereration), but these are all meant to accelerate faster and all are ground based <strong>requiring</strong> the need to accelerate faster. Move the whole thing to 150 k ft., Jump start the acceleration with a comprimised, "tube based" launch system that takes advantage of Aero-braking and the need for extreemly high inital velocities is aleveated. In fact any gains made would be icing on the cake. I mean&nbsp;Nasa's "pegasaus rocket" basically launches from a much lower altitude with only rocket power!</p>
 
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DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Replying to:"There are too many other system-level problems with the idea.&nbsp; Other ideas including electromagnetic propulsion in place of the chemical propulsion of a conventional gun have also been considered.&nbsp; There really isn't anything new in this arena."What are some of the system-level problems?... Having to harden payloads or rocket fuel? Addressed by not accelerating as fast. At what "g" force do you need to harden payloads to the point where weight becomes a factor 100-500-1000? Stay below this acceleration. Which plays into the idea because&nbsp;potential reaction mass is smaller than ground based systems.The method of aceleration in the barrel is open and there is much new in this arena. electromagnetic,coil,ram accelerators,mass drivers,light gas guns (my favorite, because I think it would have lightest weight and adjustable accereration), but these are all meant to accelerate faster and all are ground based requiring the need to accelerate faster. Move the whole thing to 150 k ft., Jump start the acceleration with a comprimised, "tube based" launch system that takes advantage of Aero-braking and the need for extreemly high inital velocities is aleveated. In fact any gains made would be icing on the cake. I mean&nbsp;Nasa's "pegasaus rocket" basically launches from a much lower altitude with only rocket power! <br />Posted by BBRAZEAU</DIV></p><p>NASA had absolutely nothing whatever to do with the Pegasus Rocket.&nbsp; That was a joint venture (totally private and commercial)&nbsp;of Orbital Sciences and the Hercules Aerospace Company.&nbsp; <br /></p> <div class="Discussion_UserSignature"> </div>
 
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