4 ms·

Rocketry is just plain difficult and expensive. Consider the "Tyranny of Rockets" problem: if you want to send a rocket up 1 km, you need X fuel. But to get to

by mabbo 7y ago

Rocketry is just plain difficult and expensive. Consider the "Tyranny of Rockets" problem: if you want to send a rocket up 1 km, you need X fuel. But to get to 2 km, you need way more than 2X fuel- because you first have to carry all that extra fuel up 1 km, which takes more energy/fuel, before you can use it to go the other km. And if you want 3 km up... well, you get the idea. It's exponential in cost. Then the problem of the fuel itself. It has to be something super-energy-dense: lots of energy (velocity) for the least mass. The most super energy dense substances are usually used to make bombs, so basically you're building a metal tube with explosives inside and hoping that you can direct the explosion correctly such that your rocket goes up. And then you have to make sure your payload- in this case some sensors to prove you actually went up that high- have to be lifted (mass) and not break. In the case of this team, their previous rocket probably did reach space but the sensors weren't working so they have no evidence! If you want to really learn this stuff, play the game "Kerbal Space Program". It's not accurate in any sense, but it gives you the instincts of why and how rocketry and orbital mechanics work.

superkuh 7y ago
>The most super energy dense substances are usually used to make bombs, By volume or per mass high explosives usually don't have much energy density compared to things like liquid fuels (ie, gasoline) or even solid fuels. This holds true for all high explosives (substances used to make bombs). They're all pretty low energy density. They're just high power because of the low time to release the energy.
theLotusGambit 7y ago
>Consider the "Tyranny of Rockets" problem: if you want to send a rocket up 1 km, you need X fuel. But to get to 2 km, you need way more than 2X fuel- because you first have to carry all that extra fuel up 1 km, which takes more energy/fuel, before you can use it to go the other km. And if you want 3 km up... well, you get the idea. It's exponential in cost. Is this correct? I know how the Tyranny of the Rocket Equation relates to mass, but I've never heard it used in terms of altitude before. Using the kinematic equations, it seems the initial velocity required to reach height 2X would actually be less than double of that for just X. However, I'm not sure if that also applies to rocket launches and if it does how it relates to fuel requirements. Feel free to correct me if I'm on the completely wrong track here.
- CydeWeys 7y ago
  It's not correct. The rocket equation is exponential for delta-v, not for altitude. Getting to 2km could easily be free once you've hit 1km, if your rocket is traveling fast enough at 1km when it cuts out to coast much higher. It's getting to, say, 2 km/s that takes more than double the fuel of getting to 1 km/s. It's not a simple relationship though (like say the inverse-square law); it's related to propellant velocity, which for chemical rockets is in the neighborhood of 4 km/s. Reaching velocities greater than your propellant velocity is where the exponential ramp really starts to take off. And no, it's not just mass either. If you have a rocket that sends mass M to velocity V, then double the size of the rocket and it'll send mass 2M to velocity V. The rocket equation tells you the ratio of fuel mass to payload mass required to reach any given delta-v. That's what grows exponentially as the desired delta-v grows.
deepnotderp 7y ago
That's not at all true, the tsiolkovsky rocket equation relates to delta-v (change in speed) and mass fraction, NOT altitude.
- chucksmash 7y ago
  Delta-v is not change in speed here: Delta-v, as used in spacecraft flight dynamics is a measure of the impulse that is needed to perform a maneuver such as launch from, or landing on a planet or moon, or in-space orbital maneuver. It is a scalar that has the units of speed. >>As used in this context, it is not the same as the physical change in velocity of the vehicle<<.[1] [1]: https://en.wikipedia.org/wiki/Delta-v https://en.wikipedia.org/wiki/Delta-v
  - EForEndeavour 7y ago
    To clarify: is the distinction between delta-v and change in re velocity that thrust could be applied in any direction, including in the braking direction, so a rocket applying a maneuver of given delta-v could end up with an increased or decreased (or zero) final speed? If the above distinction is correct, then in general delta-v is not coupled to the physical change in the rocket's velocity. But in the case of a conventional rocket with a fixed thrust vector launching from Earth's surface (as in OP's comment), isn't it perfectly true that delta-v is equivalent to change in velocity (barring air resistance)?
    - deepnotderp 7y ago
      It's also that you can go "backwards", e.g. slow down to land on another planet.
  - deepnotderp 7y ago
    Yes, you're right, I was talking about the restricted case of straight up as the Karman line would likely be.
daniel-cussen 7y ago
It's not exponential in cost. It's just not linear in cost. It must be polynomial then.
- ddingus 7y ago
  Added complexity likely provides a nice bump.

by mabbo 7y ago

superkuh 7y ago

>The most super energy dense substances are usually used to make bombs, By volume or per mass high explosives usually don't have much energy density compared to things like liquid fuels (ie, gasoline) or even solid fuels. This holds true for all high explosives (substances used to make bombs). They're all pretty low energy density. They're just high power because of the low time to release the energy.

theLotusGambit 7y ago

>Consider the "Tyranny of Rockets" problem: if you want to send a rocket up 1 km, you need X fuel. But to get to 2 km, you need way more than 2X fuel- because you first have to carry all that extra fuel up 1 km, which takes more energy/fuel, before you can use it to go the other km. And if you want 3 km up... well, you get the idea. It's exponential in cost. Is this correct? I know how the Tyranny of the Rocket Equation relates to mass, but I've never heard it used in terms of altitude before. Using the kinematic equations, it seems the initial velocity required to reach height 2X would actually be less than double of that for just X. However, I'm not sure if that also applies to rocket launches and if it does how it relates to fuel requirements. Feel free to correct me if I'm on the completely wrong track here.

CydeWeys 7y ago
It's not correct. The rocket equation is exponential for delta-v, not for altitude. Getting to 2km could easily be free once you've hit 1km, if your rocket is traveling fast enough at 1km when it cuts out to coast much higher. It's getting to, say, 2 km/s that takes more than double the fuel of getting to 1 km/s. It's not a simple relationship though (like say the inverse-square law); it's related to propellant velocity, which for chemical rockets is in the neighborhood of 4 km/s. Reaching velocities greater than your propellant velocity is where the exponential ramp really starts to take off. And no, it's not just mass either. If you have a rocket that sends mass M to velocity V, then double the size of the rocket and it'll send mass 2M to velocity V. The rocket equation tells you the ratio of fuel mass to payload mass required to reach any given delta-v. That's what grows exponentially as the desired delta-v grows.

deepnotderp 7y ago

That's not at all true, the tsiolkovsky rocket equation relates to delta-v (change in speed) and mass fraction, NOT altitude.

chucksmash 7y ago
Delta-v is not change in speed here: Delta-v, as used in spacecraft flight dynamics is a measure of the impulse that is needed to perform a maneuver such as launch from, or landing on a planet or moon, or in-space orbital maneuver. It is a scalar that has the units of speed. >>As used in this context, it is not the same as the physical change in velocity of the vehicle<<.[1] [1]: https://en.wikipedia.org/wiki/Delta-v https://en.wikipedia.org/wiki/Delta-v
- EForEndeavour 7y ago
  To clarify: is the distinction between delta-v and change in re velocity that thrust could be applied in any direction, including in the braking direction, so a rocket applying a maneuver of given delta-v could end up with an increased or decreased (or zero) final speed? If the above distinction is correct, then in general delta-v is not coupled to the physical change in the rocket's velocity. But in the case of a conventional rocket with a fixed thrust vector launching from Earth's surface (as in OP's comment), isn't it perfectly true that delta-v is equivalent to change in velocity (barring air resistance)?
  - deepnotderp 7y ago
    It's also that you can go "backwards", e.g. slow down to land on another planet.
- deepnotderp 7y ago
  Yes, you're right, I was talking about the restricted case of straight up as the Karman line would likely be.

daniel-cussen 7y ago

It's not exponential in cost. It's just not linear in cost. It must be polynomial then.

ddingus 7y ago
Added complexity likely provides a nice bump.