Bravo to the SpaceShipOne team!

[Rich Grise]

On Thursday 07 October 2004 08:48 am, Rolavine did deign to grace us with
the following:

Thanks to all of you, we have some really smart people here, and your
instruction on rocket effiency was great.

Can anyone explain how SpaceShipOne's feather works? I think this the
reconfiguration of the ship for reentry, and something about that prevents
it from buring up?

The thing about reentry that prevents it from burning up is that when
it starts reentry, it's going at approximately zero miles an hour. ;-)

This has been explained to me in great detail, by a couple of generous
folks, whom I might not have thanked properly. Thanks!

Well, it's going maybe 100 or so MPH - that's within three significant
digits of zero percent of orbital velocity, anyway. ;-)

Now, if it'd been shot up there with a proper mass driver -

I've been doing some fantasizing about a 15,000 foot LIM up the side
of some handy mountain, with enough power to it to maintain 1G all
the way to the end.

s = (a/2) * t^2
t^2 = s / (a/2) = 2 * s / a, which is associative.

So, if I assign 32 ft./sec/sec as a, and 15,000 feet as s, then
t is 21 seconds? And you're only going 472 MPH? Hmmmm.... how about
1.5G....

But the only limit on how much gets to that height and speed is
engineering. Remember, it's going 472 MPH at 30,000 or so feet. Cargo
could handle, say, 4G .... it looks like it'd make 15Kft in just under
11 sec, with a final speed of 944.75 MPH.

Of course, the whole mass driver would have to be enclosed in an
evacuated tube.

Thanks!
Rich
Of course, if my math is from planet Neptune, please enlighten me, thanks.

 

[Charles Edmondson]

Rich Grise wrote:

On Thursday 07 October 2004 08:48 am, Rolavine did deign to grace us with
the following:


Thanks to all of you, we have some really smart people here, and your
instruction on rocket effiency was great.

Can anyone explain how SpaceShipOne's feather works? I think this the
reconfiguration of the ship for reentry, and something about that prevents
it from buring up?



The thing about reentry that prevents it from burning up is that when
it starts reentry, it's going at approximately zero miles an hour. ;-)

This has been explained to me in great detail, by a couple of generous
folks, whom I might not have thanked properly. Thanks!

Well, it's going maybe 100 or so MPH - that's within three significant
digits of zero percent of orbital velocity, anyway. ;-)

Now, if it'd been shot up there with a proper mass driver -

I've been doing some fantasizing about a 15,000 foot LIM up the side
of some handy mountain, with enough power to it to maintain 1G all
the way to the end.

s = (a/2) * t^2
t^2 = s / (a/2) = 2 * s / a, which is associative.

So, if I assign 32 ft./sec/sec as a, and 15,000 feet as s, then
t is 21 seconds? And you're only going 472 MPH? Hmmmm.... how about
1.5G....

But the only limit on how much gets to that height and speed is
engineering. Remember, it's going 472 MPH at 30,000 or so feet. Cargo
could handle, say, 4G .... it looks like it'd make 15Kft in just under
11 sec, with a final speed of 944.75 MPH.

Of course, the whole mass driver would have to be enclosed in an
evacuated tube.

Thanks!
Rich
Of course, if my math is from planet Neptune, please enlighten me, thanks.

Hi Rich,

Win and those guys know a lot more about turning large inductive loads
on and off quickly, but as I understand it, that is where the real
problem with a linear accelerator comes from. As you reach higher
velocities, you have to get those coild to turn on, and then turn off
more quickly than our present technology would allow. Even on the moon,
where the delta v required is a lot less, and you are already in a
vacumn, we still can't handle the large currents needed...
--
Charlie
--
Edmondson Engineering
Unique Solutions to Unusual Problems

 

[Dirk Bruere at Neopax]

Charles Edmondson wrote:

Rich Grise wrote:

On Thursday 07 October 2004 08:48 am, Rolavine did deign to grace us with
the following:


Thanks to all of you, we have some really smart people here, and your
instruction on rocket effiency was great.

Can anyone explain how SpaceShipOne's feather works? I think this the
reconfiguration of the ship for reentry, and something about that
prevents
it from buring up?



The thing about reentry that prevents it from burning up is that when
it starts reentry, it's going at approximately zero miles an hour. ;-)

This has been explained to me in great detail, by a couple of generous
folks, whom I might not have thanked properly. Thanks!

Well, it's going maybe 100 or so MPH - that's within three significant
digits of zero percent of orbital velocity, anyway. ;-)

Now, if it'd been shot up there with a proper mass driver -

I've been doing some fantasizing about a 15,000 foot LIM up the side
of some handy mountain, with enough power to it to maintain 1G all
the way to the end.

s = (a/2) * t^2
t^2 = s / (a/2) = 2 * s / a, which is associative.

So, if I assign 32 ft./sec/sec as a, and 15,000 feet as s, then t is
21 seconds? And you're only going 472 MPH? Hmmmm.... how about
1.5G....

But the only limit on how much gets to that height and speed is
engineering. Remember, it's going 472 MPH at 30,000 or so feet. Cargo
could handle, say, 4G .... it looks like it'd make 15Kft in just under
11 sec, with a final speed of 944.75 MPH.

Of course, the whole mass driver would have to be enclosed in an
evacuated tube.

Thanks!
Rich
Of course, if my math is from planet Neptune, please enlighten me,
thanks.

Hi Rich,
Win and those guys know a lot more about turning large inductive loads
on and off quickly, but as I understand it, that is where the real
problem with a linear accelerator comes from. As you reach higher
velocities, you have to get those coild to turn on, and then turn off
more quickly than our present technology would allow. Even on the moon,
where the delta v required is a lot less, and you are already in a
vacumn, we still can't handle the large currents needed...

You make the coils longer at the fast end.

--
Dirk

The Consensus:-
The political party for the new millenium
http://www.theconsensus.org

 

[Nicholas O. Lindan]

"Charles Edmondson" <edmondson@ieee.org> wrote

Win and those guys know a lot more about turning large inductive loads
on and off quickly, but as I understand it, that is where the real
problem with a linear accelerator comes from.

Not so much turning on quickly as turning off: though there really is not
a whole lot of difference between the two. It's all about volt-seconds.

--
Nicholas O. Lindan, Cleveland, Ohio
Consulting Engineer: Electronics; Informatics; Photonics.
Remove spaces etc. to reply: n o lindan at net com dot com
psst.. want to buy an f-stop timer? nolindan.com/da/fstop/

 

[Rolavine]

I asked for an explanation of the 'feather' move used in reentry.

I found this
"
Here's another quote, from an article
(http://www.aiaa.org/aerospace/Article.cfm?issuetocid=446&ArchiveIssueID=46) in
Aerospace America: SpaceShipOne, the rocket element of Tier One, is designed to
reenter like a stable shuttlecock, then glide and land like an airplane. The
wings, with an ultralow aspect ratio of 1.7, span 16.4 ft. Their size is based
on the requirement to provide enough lift to rotate the vehicle into its ascent
attitude after horizontal launch, and to permit conventional gliding approaches
and landings. At the top of the climb, the rear part of the wing and the
tailbooms—still known collectively as the “feather”—hinge upwards. As
the spacecraft starts to reenter the atmosphere, the feather stabilizes it in a
flat attitude with the slab-like wings at right angles to the airflow. This
creates so much drag in relation to the vehicle’s weight (without fuel) that
peak heating is moderate."

So, it is based on not going very fast by having lots of resistance.

Rocky

 

[Rich Grise]

On Wednesday 13 October 2004 10:50 am, Nicholas O. Lindan did deign to grace
us with the following:

"Charles Edmondson" <edmondson@ieee.org> wrote

Win and those guys know a lot more about turning large inductive loads
on and off quickly, but as I understand it, that is where the real
problem with a linear accelerator comes from.

Not so much turning on quickly as turning off: though there really is not
a whole lot of difference between the two. It's all about volt-seconds.

Well, I blame the gullibility of a teenage sci-fi fan for my vast, in-depth

knowledge of the workings of mass drivers. ;-)

Thanks!
Rich