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Why not increase contact surface when reentering the atmosphere?


Why can't you just parachute down right from orbit?How does the surface area of a heat shield impact the ideal angle of attack for reentry?How does skipping off the atmosphere work?Why do spacecraft enter the atmosphere violently instead of a smooth spiral?Dissipating large amounts of energy fast, in spaceWhat's the atmospheric drag coefficient of a Falcon 9 at launch (sub-sonic, large fairing)Why not simulate certain regions of the atmosphere of Venus?Which crewed spacecraft provides the gentlest decent and/or landing?How can an *increase* in atmospheric temperature cause an *increase* in the atmospheric mass density?Why can't you just parachute down right from orbit?Orbital reentry glider with no heat shield













23












$begingroup$


If a craft were to increase the surface area where contact is made with air during reentry, I imagine the heat quantity per area unit would decrease, making the use of (heavy) heat shields less of a problem. It would probably come at the expense of greater g-forces, is it the limiting factor ? If not, where am I wrong ?










share|improve this question









$endgroup$







  • 1




    $begingroup$
    If you increase surface area and this increases the g-force, an increased heat load is the result. So heat load per area would not neccessarily decrease.
    $endgroup$
    – Uwe
    Mar 26 at 21:13










  • $begingroup$
    Related: space.stackexchange.com/questions/30370/…
    $endgroup$
    – leftaroundabout
    Mar 28 at 0:08















23












$begingroup$


If a craft were to increase the surface area where contact is made with air during reentry, I imagine the heat quantity per area unit would decrease, making the use of (heavy) heat shields less of a problem. It would probably come at the expense of greater g-forces, is it the limiting factor ? If not, where am I wrong ?










share|improve this question









$endgroup$







  • 1




    $begingroup$
    If you increase surface area and this increases the g-force, an increased heat load is the result. So heat load per area would not neccessarily decrease.
    $endgroup$
    – Uwe
    Mar 26 at 21:13










  • $begingroup$
    Related: space.stackexchange.com/questions/30370/…
    $endgroup$
    – leftaroundabout
    Mar 28 at 0:08













23












23








23


2



$begingroup$


If a craft were to increase the surface area where contact is made with air during reentry, I imagine the heat quantity per area unit would decrease, making the use of (heavy) heat shields less of a problem. It would probably come at the expense of greater g-forces, is it the limiting factor ? If not, where am I wrong ?










share|improve this question









$endgroup$




If a craft were to increase the surface area where contact is made with air during reentry, I imagine the heat quantity per area unit would decrease, making the use of (heavy) heat shields less of a problem. It would probably come at the expense of greater g-forces, is it the limiting factor ? If not, where am I wrong ?







reentry atmosphere atmospheric-drag heat-shield heat






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Mar 26 at 21:08









MagixMagix

316312




316312







  • 1




    $begingroup$
    If you increase surface area and this increases the g-force, an increased heat load is the result. So heat load per area would not neccessarily decrease.
    $endgroup$
    – Uwe
    Mar 26 at 21:13










  • $begingroup$
    Related: space.stackexchange.com/questions/30370/…
    $endgroup$
    – leftaroundabout
    Mar 28 at 0:08












  • 1




    $begingroup$
    If you increase surface area and this increases the g-force, an increased heat load is the result. So heat load per area would not neccessarily decrease.
    $endgroup$
    – Uwe
    Mar 26 at 21:13










  • $begingroup$
    Related: space.stackexchange.com/questions/30370/…
    $endgroup$
    – leftaroundabout
    Mar 28 at 0:08







1




1




$begingroup$
If you increase surface area and this increases the g-force, an increased heat load is the result. So heat load per area would not neccessarily decrease.
$endgroup$
– Uwe
Mar 26 at 21:13




$begingroup$
If you increase surface area and this increases the g-force, an increased heat load is the result. So heat load per area would not neccessarily decrease.
$endgroup$
– Uwe
Mar 26 at 21:13












$begingroup$
Related: space.stackexchange.com/questions/30370/…
$endgroup$
– leftaroundabout
Mar 28 at 0:08




$begingroup$
Related: space.stackexchange.com/questions/30370/…
$endgroup$
– leftaroundabout
Mar 28 at 0:08










2 Answers
2






active

oldest

votes


















61












$begingroup$

I've done a lot of work on this subject with researchers and engineers at JPL, NASA Langley, and NASA Ames. There are some interesting things that come out of high-fidelity CFM (Computational Fluid Mechanics) modeling of entries or re-entries, and also from flight experience. This FAA tutorial segment is a good general reference for the principles involved.



There is no single entry circumstance (atmosphere-relative entry velocity, entry flight path angle, atmospheric temperature profile, atmospheric composition, etc.) that tells unambiguously what the entry conditions (heat flux profile, maximum heat flux, integrated heat load, inertial force profile ["g-load" profile], stagnation pressure profile, etc.) will be. It's a multivariate problem.



One of the important entry circumstances is the vehicle's ballistic coefficient, the vehicle's mass divided by its effective area, i.e. the actual frontal area times the drag coefficient. This online paper by Dinesh Prabhu (one of my Ames colleagues) shows a type of chart they use a lot, this one specific to an Earth entry and heat shield geometry. The axes are two entry circumstances: entry flight path angle and ballistic coefficient $beta$. The three sets of curves you can read off that chart are entry conditions: peak heating rate, integrated heat load, and peak deceleration rate in g's.



One interesting result is that for an exponential atmosphere without huge deviations from the typical temperature profile, the deceleration profile is only mildly sensitive to the ballistic coefficient! "How can that be??" you say. As @Uwe pointed out, a larger surface area for roughly the same mass (and hence lower ballistic coefficient) will produce a larger drag force, thus a larger deceleration rate.



If the velocities and air densities are the same.



But they're not. The vehicle with the lower ballistic coefficient starts seeing non-trivial deceleration at lower atmospheric densities, and thus higher altitudes. By the time it gets down to the altitude where the higher-$beta$ vehicle is seeing non-trivial deceleration, the lower-$beta$ vehicle is now traveling slower.



The net result is that for the lower-$beta$ vehicle the time profile of deceleration looks very much like that of the higher-$beta$ vehicle, just earlier and at higher altitudes. This means that the lower-$beta$ vehicle's deceleration is, on average, in less dense atmosphere, so the heating rate and integrated heat load per unit area do indeed decrease.



This fact has led to the "ballute" concept for atmospheric entry and aerocapture (example papers here, here, and here). The fundamental concept is that you use a balloon-like structure to create a huge surface area, so the major part of deceleration occurs at very high altitudes and at very low atmospheric densities, so heating rates and heat loads are much lower. If they get low enough, flexible materials can handle them, so the deceleration system can be enclosed in a relatively small container until needed, and its total mass is less than that of a system that must handle more challenging heating environments.



NASA and collaborators are still looking for a material that is light enough, flexible enough, and sufficiently resistant to heating that you can build such a ballute that actually saves mass. For a while they were considering polybenzoxazole but found there were problems in manufacturing thin, flexible, uniform sheets. The search continues.






share|improve this answer









$endgroup$




















    10












    $begingroup$

    There is a fixed amount of energy which has to be dissipated. You can, to some extent, choose how fast this is done -- more air resistance (either by getting into thicker air or having a bigger surface) dissipates it faster, with higher g forces. Less air resistance dissipates it slower, but you do have to make sure to get rid of it all before you hit the ground. Choosing a slower (or at least not faster) dissipation with a bigger area (eg a pointy vehicle with wings that spreads the load out along the sides and wings) will mean you can use a lighter heat shield per unit area, but, you need more of it, so it might be heavier overall. Finally, aerodyanmic stability across a huge range of speeds may limit the shapes you can use.






    share|improve this answer









    $endgroup$








    • 4




      $begingroup$
      I suppose you don't HAVE to get rid of all the energy before you hit the ground, but the people inside would really appreciate it if you did.
      $endgroup$
      – corsiKa
      Mar 27 at 18:20










    • $begingroup$
      Afaik the possibility of hitting the ground with a significant speed (coming from the orbital kinetical energy and not from the low athmosphere freefall) is practically impossible. 0.001 bar athmosphere could kill the Columbia, I don't think that anything man-made could exist what could reach the ground with living humans therein. I think most of the deceleration should happen in $approx$ 40-60 km altitude.
      $endgroup$
      – peterh
      Mar 27 at 18:50










    • $begingroup$
      @corsiKa well technically you don't have to get rid of ALL of the energy, considering some energy is transferred to the earth or ocean on touchdown in anything other than a retropropulsive hoverslam landing. Still, the energy left is several orders of magnitude smaller than before re-entry.
      $endgroup$
      – Sdarb
      Mar 27 at 21:40










    • $begingroup$
      If you've got a wheeled gliding vehicle, you could tolerate quite a bit of residual horizontal speed (up to several hundred meters per second), the exact amount depending on the maximum certificated tyre speed and the heat tolerance of the tyres and brake assemblies.
      $endgroup$
      – Sean
      Mar 27 at 22:33






    • 1




      $begingroup$
      @Sean: So basically the Space Shuttle. Still, the problem is that shedding 99% of the speed means shedding 99.99% of the kinetic energy. And those are realistic figures - starting at Mach 25 and landing at Mach 0.25.
      $endgroup$
      – MSalters
      Mar 28 at 11:07











    Your Answer





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    2 Answers
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    active

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

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    61












    $begingroup$

    I've done a lot of work on this subject with researchers and engineers at JPL, NASA Langley, and NASA Ames. There are some interesting things that come out of high-fidelity CFM (Computational Fluid Mechanics) modeling of entries or re-entries, and also from flight experience. This FAA tutorial segment is a good general reference for the principles involved.



    There is no single entry circumstance (atmosphere-relative entry velocity, entry flight path angle, atmospheric temperature profile, atmospheric composition, etc.) that tells unambiguously what the entry conditions (heat flux profile, maximum heat flux, integrated heat load, inertial force profile ["g-load" profile], stagnation pressure profile, etc.) will be. It's a multivariate problem.



    One of the important entry circumstances is the vehicle's ballistic coefficient, the vehicle's mass divided by its effective area, i.e. the actual frontal area times the drag coefficient. This online paper by Dinesh Prabhu (one of my Ames colleagues) shows a type of chart they use a lot, this one specific to an Earth entry and heat shield geometry. The axes are two entry circumstances: entry flight path angle and ballistic coefficient $beta$. The three sets of curves you can read off that chart are entry conditions: peak heating rate, integrated heat load, and peak deceleration rate in g's.



    One interesting result is that for an exponential atmosphere without huge deviations from the typical temperature profile, the deceleration profile is only mildly sensitive to the ballistic coefficient! "How can that be??" you say. As @Uwe pointed out, a larger surface area for roughly the same mass (and hence lower ballistic coefficient) will produce a larger drag force, thus a larger deceleration rate.



    If the velocities and air densities are the same.



    But they're not. The vehicle with the lower ballistic coefficient starts seeing non-trivial deceleration at lower atmospheric densities, and thus higher altitudes. By the time it gets down to the altitude where the higher-$beta$ vehicle is seeing non-trivial deceleration, the lower-$beta$ vehicle is now traveling slower.



    The net result is that for the lower-$beta$ vehicle the time profile of deceleration looks very much like that of the higher-$beta$ vehicle, just earlier and at higher altitudes. This means that the lower-$beta$ vehicle's deceleration is, on average, in less dense atmosphere, so the heating rate and integrated heat load per unit area do indeed decrease.



    This fact has led to the "ballute" concept for atmospheric entry and aerocapture (example papers here, here, and here). The fundamental concept is that you use a balloon-like structure to create a huge surface area, so the major part of deceleration occurs at very high altitudes and at very low atmospheric densities, so heating rates and heat loads are much lower. If they get low enough, flexible materials can handle them, so the deceleration system can be enclosed in a relatively small container until needed, and its total mass is less than that of a system that must handle more challenging heating environments.



    NASA and collaborators are still looking for a material that is light enough, flexible enough, and sufficiently resistant to heating that you can build such a ballute that actually saves mass. For a while they were considering polybenzoxazole but found there were problems in manufacturing thin, flexible, uniform sheets. The search continues.






    share|improve this answer









    $endgroup$

















      61












      $begingroup$

      I've done a lot of work on this subject with researchers and engineers at JPL, NASA Langley, and NASA Ames. There are some interesting things that come out of high-fidelity CFM (Computational Fluid Mechanics) modeling of entries or re-entries, and also from flight experience. This FAA tutorial segment is a good general reference for the principles involved.



      There is no single entry circumstance (atmosphere-relative entry velocity, entry flight path angle, atmospheric temperature profile, atmospheric composition, etc.) that tells unambiguously what the entry conditions (heat flux profile, maximum heat flux, integrated heat load, inertial force profile ["g-load" profile], stagnation pressure profile, etc.) will be. It's a multivariate problem.



      One of the important entry circumstances is the vehicle's ballistic coefficient, the vehicle's mass divided by its effective area, i.e. the actual frontal area times the drag coefficient. This online paper by Dinesh Prabhu (one of my Ames colleagues) shows a type of chart they use a lot, this one specific to an Earth entry and heat shield geometry. The axes are two entry circumstances: entry flight path angle and ballistic coefficient $beta$. The three sets of curves you can read off that chart are entry conditions: peak heating rate, integrated heat load, and peak deceleration rate in g's.



      One interesting result is that for an exponential atmosphere without huge deviations from the typical temperature profile, the deceleration profile is only mildly sensitive to the ballistic coefficient! "How can that be??" you say. As @Uwe pointed out, a larger surface area for roughly the same mass (and hence lower ballistic coefficient) will produce a larger drag force, thus a larger deceleration rate.



      If the velocities and air densities are the same.



      But they're not. The vehicle with the lower ballistic coefficient starts seeing non-trivial deceleration at lower atmospheric densities, and thus higher altitudes. By the time it gets down to the altitude where the higher-$beta$ vehicle is seeing non-trivial deceleration, the lower-$beta$ vehicle is now traveling slower.



      The net result is that for the lower-$beta$ vehicle the time profile of deceleration looks very much like that of the higher-$beta$ vehicle, just earlier and at higher altitudes. This means that the lower-$beta$ vehicle's deceleration is, on average, in less dense atmosphere, so the heating rate and integrated heat load per unit area do indeed decrease.



      This fact has led to the "ballute" concept for atmospheric entry and aerocapture (example papers here, here, and here). The fundamental concept is that you use a balloon-like structure to create a huge surface area, so the major part of deceleration occurs at very high altitudes and at very low atmospheric densities, so heating rates and heat loads are much lower. If they get low enough, flexible materials can handle them, so the deceleration system can be enclosed in a relatively small container until needed, and its total mass is less than that of a system that must handle more challenging heating environments.



      NASA and collaborators are still looking for a material that is light enough, flexible enough, and sufficiently resistant to heating that you can build such a ballute that actually saves mass. For a while they were considering polybenzoxazole but found there were problems in manufacturing thin, flexible, uniform sheets. The search continues.






      share|improve this answer









      $endgroup$















        61












        61








        61





        $begingroup$

        I've done a lot of work on this subject with researchers and engineers at JPL, NASA Langley, and NASA Ames. There are some interesting things that come out of high-fidelity CFM (Computational Fluid Mechanics) modeling of entries or re-entries, and also from flight experience. This FAA tutorial segment is a good general reference for the principles involved.



        There is no single entry circumstance (atmosphere-relative entry velocity, entry flight path angle, atmospheric temperature profile, atmospheric composition, etc.) that tells unambiguously what the entry conditions (heat flux profile, maximum heat flux, integrated heat load, inertial force profile ["g-load" profile], stagnation pressure profile, etc.) will be. It's a multivariate problem.



        One of the important entry circumstances is the vehicle's ballistic coefficient, the vehicle's mass divided by its effective area, i.e. the actual frontal area times the drag coefficient. This online paper by Dinesh Prabhu (one of my Ames colleagues) shows a type of chart they use a lot, this one specific to an Earth entry and heat shield geometry. The axes are two entry circumstances: entry flight path angle and ballistic coefficient $beta$. The three sets of curves you can read off that chart are entry conditions: peak heating rate, integrated heat load, and peak deceleration rate in g's.



        One interesting result is that for an exponential atmosphere without huge deviations from the typical temperature profile, the deceleration profile is only mildly sensitive to the ballistic coefficient! "How can that be??" you say. As @Uwe pointed out, a larger surface area for roughly the same mass (and hence lower ballistic coefficient) will produce a larger drag force, thus a larger deceleration rate.



        If the velocities and air densities are the same.



        But they're not. The vehicle with the lower ballistic coefficient starts seeing non-trivial deceleration at lower atmospheric densities, and thus higher altitudes. By the time it gets down to the altitude where the higher-$beta$ vehicle is seeing non-trivial deceleration, the lower-$beta$ vehicle is now traveling slower.



        The net result is that for the lower-$beta$ vehicle the time profile of deceleration looks very much like that of the higher-$beta$ vehicle, just earlier and at higher altitudes. This means that the lower-$beta$ vehicle's deceleration is, on average, in less dense atmosphere, so the heating rate and integrated heat load per unit area do indeed decrease.



        This fact has led to the "ballute" concept for atmospheric entry and aerocapture (example papers here, here, and here). The fundamental concept is that you use a balloon-like structure to create a huge surface area, so the major part of deceleration occurs at very high altitudes and at very low atmospheric densities, so heating rates and heat loads are much lower. If they get low enough, flexible materials can handle them, so the deceleration system can be enclosed in a relatively small container until needed, and its total mass is less than that of a system that must handle more challenging heating environments.



        NASA and collaborators are still looking for a material that is light enough, flexible enough, and sufficiently resistant to heating that you can build such a ballute that actually saves mass. For a while they were considering polybenzoxazole but found there were problems in manufacturing thin, flexible, uniform sheets. The search continues.






        share|improve this answer









        $endgroup$



        I've done a lot of work on this subject with researchers and engineers at JPL, NASA Langley, and NASA Ames. There are some interesting things that come out of high-fidelity CFM (Computational Fluid Mechanics) modeling of entries or re-entries, and also from flight experience. This FAA tutorial segment is a good general reference for the principles involved.



        There is no single entry circumstance (atmosphere-relative entry velocity, entry flight path angle, atmospheric temperature profile, atmospheric composition, etc.) that tells unambiguously what the entry conditions (heat flux profile, maximum heat flux, integrated heat load, inertial force profile ["g-load" profile], stagnation pressure profile, etc.) will be. It's a multivariate problem.



        One of the important entry circumstances is the vehicle's ballistic coefficient, the vehicle's mass divided by its effective area, i.e. the actual frontal area times the drag coefficient. This online paper by Dinesh Prabhu (one of my Ames colleagues) shows a type of chart they use a lot, this one specific to an Earth entry and heat shield geometry. The axes are two entry circumstances: entry flight path angle and ballistic coefficient $beta$. The three sets of curves you can read off that chart are entry conditions: peak heating rate, integrated heat load, and peak deceleration rate in g's.



        One interesting result is that for an exponential atmosphere without huge deviations from the typical temperature profile, the deceleration profile is only mildly sensitive to the ballistic coefficient! "How can that be??" you say. As @Uwe pointed out, a larger surface area for roughly the same mass (and hence lower ballistic coefficient) will produce a larger drag force, thus a larger deceleration rate.



        If the velocities and air densities are the same.



        But they're not. The vehicle with the lower ballistic coefficient starts seeing non-trivial deceleration at lower atmospheric densities, and thus higher altitudes. By the time it gets down to the altitude where the higher-$beta$ vehicle is seeing non-trivial deceleration, the lower-$beta$ vehicle is now traveling slower.



        The net result is that for the lower-$beta$ vehicle the time profile of deceleration looks very much like that of the higher-$beta$ vehicle, just earlier and at higher altitudes. This means that the lower-$beta$ vehicle's deceleration is, on average, in less dense atmosphere, so the heating rate and integrated heat load per unit area do indeed decrease.



        This fact has led to the "ballute" concept for atmospheric entry and aerocapture (example papers here, here, and here). The fundamental concept is that you use a balloon-like structure to create a huge surface area, so the major part of deceleration occurs at very high altitudes and at very low atmospheric densities, so heating rates and heat loads are much lower. If they get low enough, flexible materials can handle them, so the deceleration system can be enclosed in a relatively small container until needed, and its total mass is less than that of a system that must handle more challenging heating environments.



        NASA and collaborators are still looking for a material that is light enough, flexible enough, and sufficiently resistant to heating that you can build such a ballute that actually saves mass. For a while they were considering polybenzoxazole but found there were problems in manufacturing thin, flexible, uniform sheets. The search continues.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Mar 27 at 4:27









        Tom SpilkerTom Spilker

        11.5k2957




        11.5k2957





















            10












            $begingroup$

            There is a fixed amount of energy which has to be dissipated. You can, to some extent, choose how fast this is done -- more air resistance (either by getting into thicker air or having a bigger surface) dissipates it faster, with higher g forces. Less air resistance dissipates it slower, but you do have to make sure to get rid of it all before you hit the ground. Choosing a slower (or at least not faster) dissipation with a bigger area (eg a pointy vehicle with wings that spreads the load out along the sides and wings) will mean you can use a lighter heat shield per unit area, but, you need more of it, so it might be heavier overall. Finally, aerodyanmic stability across a huge range of speeds may limit the shapes you can use.






            share|improve this answer









            $endgroup$








            • 4




              $begingroup$
              I suppose you don't HAVE to get rid of all the energy before you hit the ground, but the people inside would really appreciate it if you did.
              $endgroup$
              – corsiKa
              Mar 27 at 18:20










            • $begingroup$
              Afaik the possibility of hitting the ground with a significant speed (coming from the orbital kinetical energy and not from the low athmosphere freefall) is practically impossible. 0.001 bar athmosphere could kill the Columbia, I don't think that anything man-made could exist what could reach the ground with living humans therein. I think most of the deceleration should happen in $approx$ 40-60 km altitude.
              $endgroup$
              – peterh
              Mar 27 at 18:50










            • $begingroup$
              @corsiKa well technically you don't have to get rid of ALL of the energy, considering some energy is transferred to the earth or ocean on touchdown in anything other than a retropropulsive hoverslam landing. Still, the energy left is several orders of magnitude smaller than before re-entry.
              $endgroup$
              – Sdarb
              Mar 27 at 21:40










            • $begingroup$
              If you've got a wheeled gliding vehicle, you could tolerate quite a bit of residual horizontal speed (up to several hundred meters per second), the exact amount depending on the maximum certificated tyre speed and the heat tolerance of the tyres and brake assemblies.
              $endgroup$
              – Sean
              Mar 27 at 22:33






            • 1




              $begingroup$
              @Sean: So basically the Space Shuttle. Still, the problem is that shedding 99% of the speed means shedding 99.99% of the kinetic energy. And those are realistic figures - starting at Mach 25 and landing at Mach 0.25.
              $endgroup$
              – MSalters
              Mar 28 at 11:07















            10












            $begingroup$

            There is a fixed amount of energy which has to be dissipated. You can, to some extent, choose how fast this is done -- more air resistance (either by getting into thicker air or having a bigger surface) dissipates it faster, with higher g forces. Less air resistance dissipates it slower, but you do have to make sure to get rid of it all before you hit the ground. Choosing a slower (or at least not faster) dissipation with a bigger area (eg a pointy vehicle with wings that spreads the load out along the sides and wings) will mean you can use a lighter heat shield per unit area, but, you need more of it, so it might be heavier overall. Finally, aerodyanmic stability across a huge range of speeds may limit the shapes you can use.






            share|improve this answer









            $endgroup$








            • 4




              $begingroup$
              I suppose you don't HAVE to get rid of all the energy before you hit the ground, but the people inside would really appreciate it if you did.
              $endgroup$
              – corsiKa
              Mar 27 at 18:20










            • $begingroup$
              Afaik the possibility of hitting the ground with a significant speed (coming from the orbital kinetical energy and not from the low athmosphere freefall) is practically impossible. 0.001 bar athmosphere could kill the Columbia, I don't think that anything man-made could exist what could reach the ground with living humans therein. I think most of the deceleration should happen in $approx$ 40-60 km altitude.
              $endgroup$
              – peterh
              Mar 27 at 18:50










            • $begingroup$
              @corsiKa well technically you don't have to get rid of ALL of the energy, considering some energy is transferred to the earth or ocean on touchdown in anything other than a retropropulsive hoverslam landing. Still, the energy left is several orders of magnitude smaller than before re-entry.
              $endgroup$
              – Sdarb
              Mar 27 at 21:40










            • $begingroup$
              If you've got a wheeled gliding vehicle, you could tolerate quite a bit of residual horizontal speed (up to several hundred meters per second), the exact amount depending on the maximum certificated tyre speed and the heat tolerance of the tyres and brake assemblies.
              $endgroup$
              – Sean
              Mar 27 at 22:33






            • 1




              $begingroup$
              @Sean: So basically the Space Shuttle. Still, the problem is that shedding 99% of the speed means shedding 99.99% of the kinetic energy. And those are realistic figures - starting at Mach 25 and landing at Mach 0.25.
              $endgroup$
              – MSalters
              Mar 28 at 11:07













            10












            10








            10





            $begingroup$

            There is a fixed amount of energy which has to be dissipated. You can, to some extent, choose how fast this is done -- more air resistance (either by getting into thicker air or having a bigger surface) dissipates it faster, with higher g forces. Less air resistance dissipates it slower, but you do have to make sure to get rid of it all before you hit the ground. Choosing a slower (or at least not faster) dissipation with a bigger area (eg a pointy vehicle with wings that spreads the load out along the sides and wings) will mean you can use a lighter heat shield per unit area, but, you need more of it, so it might be heavier overall. Finally, aerodyanmic stability across a huge range of speeds may limit the shapes you can use.






            share|improve this answer









            $endgroup$



            There is a fixed amount of energy which has to be dissipated. You can, to some extent, choose how fast this is done -- more air resistance (either by getting into thicker air or having a bigger surface) dissipates it faster, with higher g forces. Less air resistance dissipates it slower, but you do have to make sure to get rid of it all before you hit the ground. Choosing a slower (or at least not faster) dissipation with a bigger area (eg a pointy vehicle with wings that spreads the load out along the sides and wings) will mean you can use a lighter heat shield per unit area, but, you need more of it, so it might be heavier overall. Finally, aerodyanmic stability across a huge range of speeds may limit the shapes you can use.







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Mar 26 at 21:54









            Steve LintonSteve Linton

            9,30612451




            9,30612451







            • 4




              $begingroup$
              I suppose you don't HAVE to get rid of all the energy before you hit the ground, but the people inside would really appreciate it if you did.
              $endgroup$
              – corsiKa
              Mar 27 at 18:20










            • $begingroup$
              Afaik the possibility of hitting the ground with a significant speed (coming from the orbital kinetical energy and not from the low athmosphere freefall) is practically impossible. 0.001 bar athmosphere could kill the Columbia, I don't think that anything man-made could exist what could reach the ground with living humans therein. I think most of the deceleration should happen in $approx$ 40-60 km altitude.
              $endgroup$
              – peterh
              Mar 27 at 18:50










            • $begingroup$
              @corsiKa well technically you don't have to get rid of ALL of the energy, considering some energy is transferred to the earth or ocean on touchdown in anything other than a retropropulsive hoverslam landing. Still, the energy left is several orders of magnitude smaller than before re-entry.
              $endgroup$
              – Sdarb
              Mar 27 at 21:40










            • $begingroup$
              If you've got a wheeled gliding vehicle, you could tolerate quite a bit of residual horizontal speed (up to several hundred meters per second), the exact amount depending on the maximum certificated tyre speed and the heat tolerance of the tyres and brake assemblies.
              $endgroup$
              – Sean
              Mar 27 at 22:33






            • 1




              $begingroup$
              @Sean: So basically the Space Shuttle. Still, the problem is that shedding 99% of the speed means shedding 99.99% of the kinetic energy. And those are realistic figures - starting at Mach 25 and landing at Mach 0.25.
              $endgroup$
              – MSalters
              Mar 28 at 11:07












            • 4




              $begingroup$
              I suppose you don't HAVE to get rid of all the energy before you hit the ground, but the people inside would really appreciate it if you did.
              $endgroup$
              – corsiKa
              Mar 27 at 18:20










            • $begingroup$
              Afaik the possibility of hitting the ground with a significant speed (coming from the orbital kinetical energy and not from the low athmosphere freefall) is practically impossible. 0.001 bar athmosphere could kill the Columbia, I don't think that anything man-made could exist what could reach the ground with living humans therein. I think most of the deceleration should happen in $approx$ 40-60 km altitude.
              $endgroup$
              – peterh
              Mar 27 at 18:50










            • $begingroup$
              @corsiKa well technically you don't have to get rid of ALL of the energy, considering some energy is transferred to the earth or ocean on touchdown in anything other than a retropropulsive hoverslam landing. Still, the energy left is several orders of magnitude smaller than before re-entry.
              $endgroup$
              – Sdarb
              Mar 27 at 21:40










            • $begingroup$
              If you've got a wheeled gliding vehicle, you could tolerate quite a bit of residual horizontal speed (up to several hundred meters per second), the exact amount depending on the maximum certificated tyre speed and the heat tolerance of the tyres and brake assemblies.
              $endgroup$
              – Sean
              Mar 27 at 22:33






            • 1




              $begingroup$
              @Sean: So basically the Space Shuttle. Still, the problem is that shedding 99% of the speed means shedding 99.99% of the kinetic energy. And those are realistic figures - starting at Mach 25 and landing at Mach 0.25.
              $endgroup$
              – MSalters
              Mar 28 at 11:07







            4




            4




            $begingroup$
            I suppose you don't HAVE to get rid of all the energy before you hit the ground, but the people inside would really appreciate it if you did.
            $endgroup$
            – corsiKa
            Mar 27 at 18:20




            $begingroup$
            I suppose you don't HAVE to get rid of all the energy before you hit the ground, but the people inside would really appreciate it if you did.
            $endgroup$
            – corsiKa
            Mar 27 at 18:20












            $begingroup$
            Afaik the possibility of hitting the ground with a significant speed (coming from the orbital kinetical energy and not from the low athmosphere freefall) is practically impossible. 0.001 bar athmosphere could kill the Columbia, I don't think that anything man-made could exist what could reach the ground with living humans therein. I think most of the deceleration should happen in $approx$ 40-60 km altitude.
            $endgroup$
            – peterh
            Mar 27 at 18:50




            $begingroup$
            Afaik the possibility of hitting the ground with a significant speed (coming from the orbital kinetical energy and not from the low athmosphere freefall) is practically impossible. 0.001 bar athmosphere could kill the Columbia, I don't think that anything man-made could exist what could reach the ground with living humans therein. I think most of the deceleration should happen in $approx$ 40-60 km altitude.
            $endgroup$
            – peterh
            Mar 27 at 18:50












            $begingroup$
            @corsiKa well technically you don't have to get rid of ALL of the energy, considering some energy is transferred to the earth or ocean on touchdown in anything other than a retropropulsive hoverslam landing. Still, the energy left is several orders of magnitude smaller than before re-entry.
            $endgroup$
            – Sdarb
            Mar 27 at 21:40




            $begingroup$
            @corsiKa well technically you don't have to get rid of ALL of the energy, considering some energy is transferred to the earth or ocean on touchdown in anything other than a retropropulsive hoverslam landing. Still, the energy left is several orders of magnitude smaller than before re-entry.
            $endgroup$
            – Sdarb
            Mar 27 at 21:40












            $begingroup$
            If you've got a wheeled gliding vehicle, you could tolerate quite a bit of residual horizontal speed (up to several hundred meters per second), the exact amount depending on the maximum certificated tyre speed and the heat tolerance of the tyres and brake assemblies.
            $endgroup$
            – Sean
            Mar 27 at 22:33




            $begingroup$
            If you've got a wheeled gliding vehicle, you could tolerate quite a bit of residual horizontal speed (up to several hundred meters per second), the exact amount depending on the maximum certificated tyre speed and the heat tolerance of the tyres and brake assemblies.
            $endgroup$
            – Sean
            Mar 27 at 22:33




            1




            1




            $begingroup$
            @Sean: So basically the Space Shuttle. Still, the problem is that shedding 99% of the speed means shedding 99.99% of the kinetic energy. And those are realistic figures - starting at Mach 25 and landing at Mach 0.25.
            $endgroup$
            – MSalters
            Mar 28 at 11:07




            $begingroup$
            @Sean: So basically the Space Shuttle. Still, the problem is that shedding 99% of the speed means shedding 99.99% of the kinetic energy. And those are realistic figures - starting at Mach 25 and landing at Mach 0.25.
            $endgroup$
            – MSalters
            Mar 28 at 11:07

















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