J
Jonathan Kirwan
Guest
On Tue, 05 Oct 2004 19:32:45 GMT, Rich Grise <null@example.net> wrote:
work. Not that it cannot be shown. But it would take a book (okay, a small
one) to detail the various alternate thoughts to investigate so that in a
comprehensive view you see the "difficulties."
A single expression would show you nothing. The methods applied to the data are
as important as the data, itself. And that requires understanding which is
acquired by study and working through the ideas and their equations.
Don had quipped:
You go on to add:
In the process, I believe, they've made significant progress with hybrid engines
-- but that's another thing. But they also haven't been designing for much of
anything else, either. It's just to go get the prize.
What Don was alluding to (hell, not alluding to but saying outright) is that you
use the "rocket equation" to see what the meaning of delta vee is:
dV = Exhaust velocity * ln ( initial mass / final mass )
Since it's kind of hard to wave a stick in the exhaust to measure its velocity,
ISP is usually used and is measured as (thrust/flow rate.) With that, you get:
dV = ISP * g * ln ( initial mass / final mass )
I've read various estimates on the ISP for Rutan's hybrid, but the number is
probably somewhere around 325 seconds. (He uses a solid butadiene with nitrous
oxide.)
Circular orbital velocity at 112.2km is:
Vc = SQRT( G * M / (R + h) ) = SQRT( 398603.2 / (6371 + 112.2) )
or about 7.84 km/s.
Skipping all that for a moment....
I seem to have read that this ship was traveling at Mach 3 (which varies a lot
based on the air so I really don't know what this means) when it started
coasting. I'm going to assume that mach 3 means what google says, namely
1020.87 m/s. Final height was 112.2 km and coasting time would be V(final)/g or
about 1020.87/9.82 or say 114 seconds. Rate of loss of velocity to zero is
constant (g), so we can use the midpoint as the average velocity or (1/2) of
1020.87 m/s, so the distance traveled is .5*1020.87*114 or about 58 km. (I've
neglected air resistance here.) Release height, I've read, was something on the
order of 15km, so the fuel was burning for the time from (112 - 58 - 15) or 39
km. I also read that the burn time was something like 70 seconds, so this means
a net acceleration of a=2*d/t^2 or about 15.92 m/s^2 (about 1.62 gee.)
Reasonableness check: 1.62 gee sounds reasonable? yes.
Now, I've also read that there was a planned up-angle of 84 degrees. No idea if
they followed through with that. But it would adjust some of these figures.
I'll leave it as an exercise.
Meanwhile, we've already figured that there was a vertical delta vee of about
1.02 km/s (going from zero to Mach 3 before coasting.) This compares with
needing yet another 7.84 km/s to go from about 0 to orbital velocity,
tangentially. Since energy increases by V^2, this suggests as much as 61 times
as much energy is needed. (This isn't strictly correct, as the rocket gets
lighter as it uses fuel, but it gets the idea across.)
Another way to look at this, though, for a single stage is to use the rocket
equation:
dV = ISP * g * ln ( initial mass / final mass )
We already know some of the factors or can estimate them. The dV is what's
required to "go up" (against gravity) plus what's required to "to tangential"
(orbital velocity) and some more for drag losses (and Earth's rotational
velocity one way or another.) Let's say it remains airplane launched from 15km
to keep it simpler. You'll need the 1.02 km/s plus the 7.84 km/s as your d-vee.
This means:
e^((dV/ISP)/g) = (initial mass / final mass) = about 16.06 : 1
This is ... tough to reach. Usually called... impractical.
Hmm. I wonder what his actual ratio was on this flight... Well, I've read
somewhere that it was about 1.6 : 1. About 10-fold less. Let's see what the
above estimates give us:
e^(1020.87/325/9.82) = about 1.38 : 1
Not too far from this suggestion I'd read somewhere. The differences here may
be from changes in various estimates, including the up-angle which may not have
been 90 degrees, but closer to 84 degrees.
If my numbers are (the 1.38:1 and the 16.06:1), then the fuel they used compared
to the fuel they'd need to use (assuming no change in payload and dead weight):
(16.06-1)/(1.38-1) = 39.6 times
This isn't so different from the 61X gross estimate arrived at from the earlier
V^2 estimate. So, about 40X or so.
In other words, they did about 2.5% of what they'd need to do, keeping the same
payload they currently have and launching from 15km up.
Anyway, all this gives a rough idea. This ship was carried aloft to over 15 km
before being released. In doing so, much of the drag effect was removed. It
actually ran powered for some 114 seconds total (less than two minutes) and
glided the rest of the way up. At this point, it canted back down but with
essentially zero (or close to it) horizontal velocity. All of that would need
to be made up if it were to go orbital, as well. And this would amount to many
times as much fuel -- perhaps 40 times, or so. In the rough ballpark, anyway.
Which gets right back to Don's broader point, I think.
used frequently enough (in the English system, anyway.)
Jon
I think he's suggesting that the understanding will only come from doing the
work. Not that it cannot be shown. But it would take a book (okay, a small
one) to detail the various alternate thoughts to investigate so that in a
comprehensive view you see the "difficulties."
A single expression would show you nothing. The methods applied to the data are
as important as the data, itself. And that requires understanding which is
acquired by study and working through the ideas and their equations.
Don had quipped:
and you took issue with this (out of ignorance?):
So what were you thinking here? From knowledge? Or not?
You go on to add:
Rutan designed this puppy for one thing and one thing only -- winning the prize.
In the process, I believe, they've made significant progress with hybrid engines
-- but that's another thing. But they also haven't been designing for much of
anything else, either. It's just to go get the prize.
What Don was alluding to (hell, not alluding to but saying outright) is that you
use the "rocket equation" to see what the meaning of delta vee is:
dV = Exhaust velocity * ln ( initial mass / final mass )
Since it's kind of hard to wave a stick in the exhaust to measure its velocity,
ISP is usually used and is measured as (thrust/flow rate.) With that, you get:
dV = ISP * g * ln ( initial mass / final mass )
I've read various estimates on the ISP for Rutan's hybrid, but the number is
probably somewhere around 325 seconds. (He uses a solid butadiene with nitrous
oxide.)
Circular orbital velocity at 112.2km is:
Vc = SQRT( G * M / (R + h) ) = SQRT( 398603.2 / (6371 + 112.2) )
or about 7.84 km/s.
Skipping all that for a moment....
I seem to have read that this ship was traveling at Mach 3 (which varies a lot
based on the air so I really don't know what this means) when it started
coasting. I'm going to assume that mach 3 means what google says, namely
1020.87 m/s. Final height was 112.2 km and coasting time would be V(final)/g or
about 1020.87/9.82 or say 114 seconds. Rate of loss of velocity to zero is
constant (g), so we can use the midpoint as the average velocity or (1/2) of
1020.87 m/s, so the distance traveled is .5*1020.87*114 or about 58 km. (I've
neglected air resistance here.) Release height, I've read, was something on the
order of 15km, so the fuel was burning for the time from (112 - 58 - 15) or 39
km. I also read that the burn time was something like 70 seconds, so this means
a net acceleration of a=2*d/t^2 or about 15.92 m/s^2 (about 1.62 gee.)
Reasonableness check: 1.62 gee sounds reasonable? yes.
Now, I've also read that there was a planned up-angle of 84 degrees. No idea if
they followed through with that. But it would adjust some of these figures.
I'll leave it as an exercise.
Meanwhile, we've already figured that there was a vertical delta vee of about
1.02 km/s (going from zero to Mach 3 before coasting.) This compares with
needing yet another 7.84 km/s to go from about 0 to orbital velocity,
tangentially. Since energy increases by V^2, this suggests as much as 61 times
as much energy is needed. (This isn't strictly correct, as the rocket gets
lighter as it uses fuel, but it gets the idea across.)
Another way to look at this, though, for a single stage is to use the rocket
equation:
dV = ISP * g * ln ( initial mass / final mass )
We already know some of the factors or can estimate them. The dV is what's
required to "go up" (against gravity) plus what's required to "to tangential"
(orbital velocity) and some more for drag losses (and Earth's rotational
velocity one way or another.) Let's say it remains airplane launched from 15km
to keep it simpler. You'll need the 1.02 km/s plus the 7.84 km/s as your d-vee.
This means:
e^((dV/ISP)/g) = (initial mass / final mass) = about 16.06 : 1
This is ... tough to reach. Usually called... impractical.
Hmm. I wonder what his actual ratio was on this flight... Well, I've read
somewhere that it was about 1.6 : 1. About 10-fold less. Let's see what the
above estimates give us:
e^(1020.87/325/9.82) = about 1.38 : 1
Not too far from this suggestion I'd read somewhere. The differences here may
be from changes in various estimates, including the up-angle which may not have
been 90 degrees, but closer to 84 degrees.
If my numbers are (the 1.38:1 and the 16.06:1), then the fuel they used compared
to the fuel they'd need to use (assuming no change in payload and dead weight):
(16.06-1)/(1.38-1) = 39.6 times
This isn't so different from the 61X gross estimate arrived at from the earlier
V^2 estimate. So, about 40X or so.
In other words, they did about 2.5% of what they'd need to do, keeping the same
payload they currently have and launching from 15km up.
Anyway, all this gives a rough idea. This ship was carried aloft to over 15 km
before being released. In doing so, much of the drag effect was removed. It
actually ran powered for some 114 seconds total (less than two minutes) and
glided the rest of the way up. At this point, it canted back down but with
essentially zero (or close to it) horizontal velocity. All of that would need
to be made up if it were to go orbital, as well. And this would amount to many
times as much fuel -- perhaps 40 times, or so. In the rough ballpark, anyway.
Which gets right back to Don's broader point, I think.
Interesting play on words, because a 'slug' is actually a unit that rocketeers
used frequently enough (in the English system, anyway.)
Jon
about 'Bravo to the