Quote from: KILYAV on 10/25/2021 03:11 pmQuote from: eriblo on 10/25/2021 02:30 pmQuote from: KILYAV on 10/25/2021 09:35 amQuote from: Yggdrasill on 10/23/2021 10:49 amI can't see how that is accurate. This is assuming the Starship is aerodynamically controlled to follow a circular trajectory around Mars, and has enough control authority to do so. But that's not the case. It will be going through thorugh the atmosphere in a hyperbolic trajectory, which then degrades into a parabolic trajectory, which then degrades into a suborbital parabolic trajectory, right? Also you have to subtract the Mars gravitational acceleration.Moving at an acceleration of more than 8 G, the starship will be able to fly at the same height above the surface and decelerate as much as necessary.With 8G acceleration, the starship will be able to "follow" the "bend" of the planet all the time, remaining at the same altitude, which will allow it to decelerate as much as necessary.The problem is that for a L/D ratio of 0.3 you need 3.3 g of drag for every 1 g of lift. So the 8 g circular trajectory implies a total of ~28 g... Fortunately, the peak is much less due to the lower but still significant drag as it descends into the atmosphere.You are using too “perfect” aerodynamic bodies. Other body shapes can give much better drag coefficients.For example, a triangle with angles 45-90-45. Front bottom corner 45, rear bottom corner 90, rear top corner 45. At hypersonic speeds, such a figure should reflect most of the atmospheric flow upward, with the deceleration being equal to centrifugal acceleration.The magnitude of the centrifugal acceleration decreases very quickly and is equal to 8G only at a speed of 16.5 km / s, after the speed drops to 11.7 km / s, the acceleration drops to 4G.I do not quite follow...? Higher lift is certainly possible with a more aerodynamic shape but I thought we were discussing Starship?I took the lift and drag values at an angle of attack (AoA) of 70° modeled here: https://www.researchgate.net/publication/334611503_3DoF_simulation_model_and_specific_aerodynamic_control_capabilities_for_a_SpaceX%27s_Starship-like_atmospheric_reentry_vehicleThus is for an older version of Starship but the values are probably similar as most of the drag and lift are generated by the body. Starship is certainly able to generate more lift at a lower AoA but I have not seen any mention of this despite the fact that it would greatly help with g-forces and heat flux. This suggest that they are limited by stability or local heating to an AoA around 70° and a L/D of about 0.3.
Quote from: eriblo on 10/25/2021 02:30 pmQuote from: KILYAV on 10/25/2021 09:35 amQuote from: Yggdrasill on 10/23/2021 10:49 amI can't see how that is accurate. This is assuming the Starship is aerodynamically controlled to follow a circular trajectory around Mars, and has enough control authority to do so. But that's not the case. It will be going through thorugh the atmosphere in a hyperbolic trajectory, which then degrades into a parabolic trajectory, which then degrades into a suborbital parabolic trajectory, right? Also you have to subtract the Mars gravitational acceleration.Moving at an acceleration of more than 8 G, the starship will be able to fly at the same height above the surface and decelerate as much as necessary.With 8G acceleration, the starship will be able to "follow" the "bend" of the planet all the time, remaining at the same altitude, which will allow it to decelerate as much as necessary.The problem is that for a L/D ratio of 0.3 you need 3.3 g of drag for every 1 g of lift. So the 8 g circular trajectory implies a total of ~28 g... Fortunately, the peak is much less due to the lower but still significant drag as it descends into the atmosphere.You are using too “perfect” aerodynamic bodies. Other body shapes can give much better drag coefficients.For example, a triangle with angles 45-90-45. Front bottom corner 45, rear bottom corner 90, rear top corner 45. At hypersonic speeds, such a figure should reflect most of the atmospheric flow upward, with the deceleration being equal to centrifugal acceleration.The magnitude of the centrifugal acceleration decreases very quickly and is equal to 8G only at a speed of 16.5 km / s, after the speed drops to 11.7 km / s, the acceleration drops to 4G.
Quote from: KILYAV on 10/25/2021 09:35 amQuote from: Yggdrasill on 10/23/2021 10:49 amI can't see how that is accurate. This is assuming the Starship is aerodynamically controlled to follow a circular trajectory around Mars, and has enough control authority to do so. But that's not the case. It will be going through thorugh the atmosphere in a hyperbolic trajectory, which then degrades into a parabolic trajectory, which then degrades into a suborbital parabolic trajectory, right? Also you have to subtract the Mars gravitational acceleration.Moving at an acceleration of more than 8 G, the starship will be able to fly at the same height above the surface and decelerate as much as necessary.With 8G acceleration, the starship will be able to "follow" the "bend" of the planet all the time, remaining at the same altitude, which will allow it to decelerate as much as necessary.The problem is that for a L/D ratio of 0.3 you need 3.3 g of drag for every 1 g of lift. So the 8 g circular trajectory implies a total of ~28 g... Fortunately, the peak is much less due to the lower but still significant drag as it descends into the atmosphere.
Quote from: Yggdrasill on 10/23/2021 10:49 amI can't see how that is accurate. This is assuming the Starship is aerodynamically controlled to follow a circular trajectory around Mars, and has enough control authority to do so. But that's not the case. It will be going through thorugh the atmosphere in a hyperbolic trajectory, which then degrades into a parabolic trajectory, which then degrades into a suborbital parabolic trajectory, right? Also you have to subtract the Mars gravitational acceleration.Moving at an acceleration of more than 8 G, the starship will be able to fly at the same height above the surface and decelerate as much as necessary.With 8G acceleration, the starship will be able to "follow" the "bend" of the planet all the time, remaining at the same altitude, which will allow it to decelerate as much as necessary.
I can't see how that is accurate. This is assuming the Starship is aerodynamically controlled to follow a circular trajectory around Mars, and has enough control authority to do so. But that's not the case. It will be going through thorugh the atmosphere in a hyperbolic trajectory, which then degrades into a parabolic trajectory, which then degrades into a suborbital parabolic trajectory, right? Also you have to subtract the Mars gravitational acceleration.
You are correct, I assume that the "Martian Express" in its shape will be much more like a Boeing X-51 than a starship.The form of the "express" should not create lift, but press it to the surface so that it does not spill out of the atmosphere ahead of time, before it has time to decelerate to the "10 o'clock" orbit.
The form of the "express" should not create lift, but press it to the surface
It might be more correct to say that the angle of attack will be minus 70 degrees.
