Firing the payload to space with cannon

[Eagle9]

Good day
As you know there are a lot of projects for launching the payload into space by means of huge cannons, for example the http://quicklaunchinc.com/sharp" project.
Imagine that such cannon is ready to be used and can launch the payload with initial speed let’s say 8 km/sec, this is a bit more than value of Orbital Velocity at Low Earth Orbit. The cannon may be tilted towards Earth’s surface at various angles, such as 30º, 45º, 60º and etc:

Is it possible to calculate what will be payload’s speed when it leaves atmosphere at the altitude of 100 km? Of course if initial angle is 30º then the payload will have to cover more distance in air and loss in speed will be more, if that angle is 45º than the loss will be less….so, is it possible to calculate?

 

[berkeman]

Good day
As you know there are a lot of projects for launching the payload into space by means of huge cannons, for example the http://quicklaunchinc.com/sharp" project.
Imagine that such cannon is ready to be used and can launch the payload with initial speed let’s say 8 km/sec, this is a bit more than value of Orbital Velocity at Low Earth Orbit. The cannon may be tilted towards Earth’s surface at various angles, such as 30º, 45º, 60º and etc:

Is it possible to calculate what will be payload’s speed when it leaves atmosphere at the altitude of 100 km? Of course if initial angle is 30º then the payload will have to cover more distance in air and loss in speed will be more, if that angle is 45º than the loss will be less….so, is it possible to calculate?

It would seem to depend a great deal on the aerodynamics of the payload "shell". What assumptions are you going to make about the size and shape of the projectile?

http://en.wikipedia.org/wiki/Air_resistance

.

 

[maburne2]

Actually I believe the minimum escape velocity for an object is 11 km/sec. And just off the top of my head you could do energy conservation to calculate its final velocity after air and gravity have performed work on the payload.

 

[Eagle9]

berkeman

It would seem to depend a great deal on the aerodynamics of the payload "shell".

Yes, of course I know that aerodynamics is very important

What assumptions are you going to make about the size and shape of the projectile?

Shape: as I know the supersonic aircrafts/rockets (should) have narrow/tapered shape like cone, so the payload will have the same like this:
http://img841.imageshack.us/img841/9012/formf.jpg

As for size……is it important? In my case the diameter will be several meters, for example 5 meters, length-twice more, about 10 meters then what? Can we calculate something important based on these sizes?
maburne2

Actually I believe the minimum escape velocity for an object is 11 km/sec. And just off the top of my head you could do energy conservation to calculate its final velocity after air and gravity have performed work on the payload.

Well, in this case we need to put the payload on the orbit around the Earth so, the trajectory will be tilted and not vertical.

 

[russ_watters]

Well, in this case we need to put the payload on the orbit around the Earth so, the trajectory will be tilted and not vertical.

A payload launched from Earth has an orbit that intersects with the surface of the earth. How do you propose to correct the orbit once the projectile is at altitude?

Anyway, I'd start with conservation of energy and the assumption that the projectile isn't in the atmosphere long enough for substantial drag-induced speed reduction. Unless you're at a steep angle, you're essentially out of the atmosphere in a few seconds. Even if you decelerated at 100g for 10 seconds, you'd only drop to 7 km/sec.

 

[pallidin]

A payload launched from Earth has an orbit that intersects with the surface of the earth. How do you propose to correct the orbit once the projectile is at altitude?

"Houston, we have a problem..."

 

[Nik_2213]

"...diameter will be several metres, for example 5 metres, length-twice more,"

I take it you are using a magnetic cannon or such, as building a 'conventional' gun barrel that diameter would be very difficult...

With the exception of short-barrelled mortars, the biggest guns fielded were ~ 80cm diameter and wore out after ~300 shots...
http://en.wikipedia.org/wiki/Schwerer_Gustav

 

[maburne2]

"Houston, we have a problem..."

...not necessarily, we must not forget that the Earth itself is moving with a speed of 29.78 km/s around the sun. So if a projectile is launched from the surface of the Earth traveling in the opposite heading of the Earth it is possible to find a point SOMEWHERE above the Earth where the payload will be bound by an effective potential that does not include crashing into the earth.

 

[russ_watters]

No, sorry, the projectile is also orbiting the sun so the orbital speed of Earth has no effect at all on the trajectory.

 

[maburne2]

Lame, this problem is like inception...you can always go deeper.

 

[berkeman]

...not necessarily, we must not forget that the Earth itself is moving with a speed of 29.78 km/s around the sun. So if a projectile is launched from the surface of the Earth traveling in the opposite heading of the Earth it is possible to find a point SOMEWHERE above the Earth where the payload will be bound by an effective potential that does not include crashing into the earth.

Lame, this problem is like inception...you can always go deeper.

Unless you can post your calculations that show what you are claiming, your posts will be deleted as misinformation.

 

[Eagle9]

russ_watters

A payload launched from Earth has an orbit that intersects with the surface of the earth. How do you propose to correct the orbit once the projectile is at altitude?

By means of its own rocket engines, they will fire in space

Anyway, I'd start with conservation of energy and the assumption that the projectile isn't in the atmosphere long enough for substantial drag-induced speed reduction.

Yes, if the initial speed is high (10 km/sec or so) then the payload will leave the atmosphere soon, but still its speed will be reduced and I want to know how it is possible to calculate this loss, at least approximately.

Unless you're at a steep angle, you're essentially out of the atmosphere in a few seconds. Even if you decelerated at 100g for 10 seconds, you'd only drop to 7 km/sec.

From 8 km/sec? How did you calculate this?

Nik_2213

I take it you are using a magnetic cannon or such, as building a 'conventional' gun barrel that diameter would be very difficult...

With the exception of short-barrelled mortars, the biggest guns fielded were ~ 80cm diameter and wore out after ~300 shots...

Well, this is absolutely different matter; of course I know that in reality firing “my” payload will be extremely difficult by means of cannon, but imagine that this is possible and that is it. I want to clarify how to calculate the speed loss only

 

[maburne2]

Unless you can post your calculations that show what you are claiming, your posts will be deleted as misinformation.

Challenge accepted.

 

[berkeman]

Challenge accepted.

Great! And...

 

[maburne2]

...dude give me some time, trying to compensate for drag due to humidity and temp, cross-sectional drag, effective potentials of the earth, sun, and moon ( and now that I think about it, maybe mars) is not trivial. Then I have to find the apogee and perihelion of the orbit of the payload to make sure it does not collide with the surface of the earth. Should have an answer by tomorrow, I made sufficient headway for tonight and will continue tomorrow, so until then.

 

[maburne2]

I meant aphelion, sorry.

 

[unedjoocated]

Giant catapult would work better. Read "The Moon is a Harsh Mistress" by Robert Heinlein. :)

 

[Eagle9]

...dude give me some time, trying to compensate for drag due to humidity and temp, cross-sectional drag, effective potentials of the earth, sun, and moon ( and now that I think about it, maybe mars) is not trivial. Then I have to find the apogee and perihelion of the orbit of the payload to make sure it does not collide with the surface of the earth. Should have an answer by tomorrow, I made sufficient headway for tonight and will continue tomorrow, so until then.

Well, to be perfectly honest I think that the payload WILL collide with Earth unless it changes its trajectory by means of rocket engines for now I need to know how it is possible to calculate the loss in speed at 100 km altitude when the payload is fired at some certain angle from the Earth, for example 45º

 

[Eagle9]

I read the following iformation at this site:
http://colonyfund.com/Reading/papers/phys_econ_leo.html

The net result is that an effective penalty of about 15 to 25 percent is added to the basic orbital velocity requirement for a real-world launch to LEO [Ref. 3]. As shown in Table 1, a practical speed change capability (Delta-V) of about 9 to 10 kilometers per second is required for a launch vehicle to deliver a payload to LEO

Does it mean that for reaching the 7.7 km/sec at 150 km altitude the cannon (or railgun or any other device-never mind) should give 9.24 km/sec (this is by 20 % more than the 7.7 km/sec) velocity? Or perhaps for rockets and for cannon (railgun and etc.) this is calculated in a different ways?

ground launch adds an additional Delta-V burden of between 100 and 160 meters per second for air drag, and between 1,100 to 1,500 meters per second for gravitational losses.

The same question: for reaching the 7.7 km/sec at 150 km altitude the cannon (or railgun or any other device-never mind) should give about 9360 km/sec to payload?