3 ms·

Indeed! Assuming that your centripital accel. numbers are right, I wonder why not even just use linear accel. Assuming that you want to exit the tube/system a

by toss1 8y ago

Indeed! Assuming that your centripital accel. numbers are right, I wonder why not even just use linear accel. Assuming that you want to exit the tube/system at 5000mph, you could keep acceleration to 9g (tolerable for a human)with less than a 9 mile long acclerator and 18 seconds. If you were ok with 40g, you're down under 4miles and 6 seconds. (I'm not estimating the deceleration Gs from suddenly hitting MaxQ on exiting into the lower atmosphere. So perhaps that shock puts us up into a high G situation anyway, so might as well go for the smaller real-estate circular acceleration option?) edit: "tolerable for a human" might better read "survivable for trained humans"

AlanSE 8y ago
I don't see the term "MaxQ" very applicable here. The idea of Max Q is that you start at low aerodynamic forces and then accelerate via rocket thrust, with dynamic pressure increasing until the atmosphere density causes it to fall. For a projectile from this launcher, MaxQ is the second it opens the airlock of the vacuum chamber and releases it to atmosphere. Then it hits a wall of air, and dynamic pressure spikes to the maximum instantly. It is true that you get rid of the centripetal acceleration upon release, so to a degree you trade the centripetal acceleration for drag acceleration.
nickparker 8y ago
Yeah I'm not up on my supersonic aero anywhere near enough to work out max-Q for their very pointy projectile, but I'll point out that 'suddenly hitting MaxQ' is a jerk problem, not an acceleration one. MaxQ is MaxQ, whether you've been cruising at that speed and air density for hours or just hit it out of a near-vacuum. I think centripetal / launch acceleration dramatically outweighs air drag though, by analogy to the SR-71. The max thrust divided by its dry mass gives a maximum thrust/weight of about 1g, and that thing could cruise at Mach 3+ which is about halfway to Spinlaunch's 5000 mph. Their little teardrop projectiles have got to be way lower drag than anything with wings and intakes, and it's probably much heavier for its size than the SR-71 as well, so I can't imagine MaxQ on these things gets past 10-15g.

by toss1 8y ago

AlanSE 8y ago

I don't see the term "MaxQ" very applicable here. The idea of Max Q is that you start at low aerodynamic forces and then accelerate via rocket thrust, with dynamic pressure increasing until the atmosphere density causes it to fall. For a projectile from this launcher, MaxQ is the second it opens the airlock of the vacuum chamber and releases it to atmosphere. Then it hits a wall of air, and dynamic pressure spikes to the maximum instantly. It is true that you get rid of the centripetal acceleration upon release, so to a degree you trade the centripetal acceleration for drag acceleration.

nickparker 8y ago

Yeah I'm not up on my supersonic aero anywhere near enough to work out max-Q for their very pointy projectile, but I'll point out that 'suddenly hitting MaxQ' is a jerk problem, not an acceleration one. MaxQ is MaxQ, whether you've been cruising at that speed and air density for hours or just hit it out of a near-vacuum. I think centripetal / launch acceleration dramatically outweighs air drag though, by analogy to the SR-71. The max thrust divided by its dry mass gives a maximum thrust/weight of about 1g, and that thing could cruise at Mach 3+ which is about halfway to Spinlaunch's 5000 mph. Their little teardrop projectiles have got to be way lower drag than anything with wings and intakes, and it's probably much heavier for its size than the SR-71 as well, so I can't imagine MaxQ on these things gets past 10-15g.