Brute force re-entry

In article ,
Allen Meece wrote:
They could send up some solid fuel boosters that the shuttle could dock
with
and use for braking. How how long would it take to brake from 18k mph to
about 200mph without exceeding 2 G's?

That would be insane. You'd need "solid fuel boosters" with very nearly as
much energy as the SRB's and the ET used to *launch* the shuttle in the
first place

This is the second poster who erroneously contends that it takes as much to
brake an empty shuttle out of orbit as it did to get the fully fueled orbiter
into orbit.

C'mon guys, it only takes a fraction of the launch fuel to brake the empty
craft.

A fraction of the fuel? Let's look at some numbers.

From the Rogers Commission report at
http://history.nasa.gov/rogersrep/v3appoe8.htm we find the following
weights for Challenger:

Wet-with crew and cargo - 268,829 lbs
wet-with crew, w/o cargo - 216,494 lbs
Dry-with crew, w/o cargo - 177,372 lbs

The dry weight is completely dry- no fuel, no liquid oxygen for life
support, no fuel to power the APUs to power the aerosurfaces.

The wet weight without cargo is a reasonable number for an "empty"
shuttle returning from orbit. As you can see, "empty" isn't all that
"empty."

The returning weight is about 80% of the takeoff weight.
So we need 80% of the energy used to put the shuttle in orbit to
stop it in place. That's not a particularly small fraction.

But there's no need to "send up" an SRB to brake it, just hang onto the ET
and use its spare liquid fuel for braking and then cut it loose.

Sorry, no, the spare liquid fuel in the ET after the main engine burns
is something like 2000 lbs of fuel. That's 2000 lbs out of 1.6 million
pounds of fuel. So the spare fuel, even if you could use it, would
give you about one tenth of one percent of the energy we need. What
you would need is more than 1.3 million pounds of fuel and oxidizer in
orbit.

cheers,

Steven

--


"M-Theory is the unifying pachyderm of the five string theories."
- Brian Greene, _The Elegant Universe_

(Allen Meece) wrote in message ...
This is the second poster who erroneously contends that it takes as much to
brake an empty shuttle out of orbit as it did to get the fully fueled orbiter
into orbit.

Simple question: how much delta-V are you expecting the rockets to
brake the shuttle? By 100mph or the full 17500mph?

C'mon guys, it only takes a fraction of the launch fuel to brake the empty
craft.

The rockets that launch the shuttle essentially put an empty vehicle
into orbit, give or take 20 tons of payload and OMS fuel. A fully
empty shuttle (~90 tons) would take ~1800 tons of fuel to brake from
orbital velocity to 0mph.

Mike Miller, Materials Engineer

(Allen Meece) wrote in
:

Incorrect. The space shuttle re-enters at a flight path angle
typically
between -1 and -1.5 degrees. That's hardly "steep".
Also incorrect. The shuttle's flight path angle in the atmosphere
is 20 --
25 degrees.
You're thinking about the initial orbital departure angle, not the
average
sink rate along the entire descent.

No, the number I quoted was for EI, not "initial orbital departure angle."
The average sink rate along the *entire* descent is irrelevant if we're
talking about entry, which is only the first part of the descent. During
*entry* the shuttle's flight path angle remains quite shallow, close to the
EI flight path angle I quoted. The flight path angle doesn't come anywhere
close to -20 until TAEM, which is well after entry.

The shuttle's reentry path IS STEEP. It is a high wing-loading,
controlled
brute force reentry. Hardly as graceful as a good lifting body like
the cancelled [!?] X-38 would do.

Incorrect. The shuttle has a much higher hypersonic L/D (4.0) than the X-38
(1.4).


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brute force reentry. Hardly as graceful as a good lifting body like
the cancelled [!?] X-38 would do.

Incorrect. The shuttle has a much higher hypersonic L/D (4.0) than the X-38
(1.4).
Also incorrect. The X-38 would have been more graceful than the brute force
reentry of the shuttle because the shuttle is so darn heavy that its higher
lift ratio would not prevent it from sinking faster than the lighter X-38.

The returning weight is about 80% of the takeoff weight. So we need 80% of
the energy used to put the shuttle in orbit to
stop it in place. That's not a particularly small fraction.
Wheeewww. Not when you realize that most of the weight of the shuttle
*system* was in the ET and the SRB's.
Without them, the orbiter weighs a *fraction* of the complete STS weight and
it won't take *anything* like an SRB to brake the shuttle into a slow speed
reentry. Probalby the ullage left in the ET would be plenty of fuel for
sufficient braking..
^
//^\\
~~~ near space elevator ~~~~
~~~members.aol.com/beanstalkr/~~~

Allen Meece wrote:
The returning weight is about 80% of the takeoff weight. So we need 80%
of the energy used to put the shuttle in orbit to
stop it in place. That's not a particularly small fraction.
Wheeewww. Not when you realize that most of the weight of the shuttle
*system* was in the ET and the SRB's.
Without them, the orbiter weighs a *fraction* of the complete STS
weight and
it won't take *anything* like an SRB to brake the shuttle into a slow
speed reentry. Probalby the ullage left in the ET would be plenty of fuel
for sufficient braking..

Yeah, but the shuttle's the only important bit. That vast weight of the ET
and the SRBs is there for the sole purpose of accelerating the shuttle into
orbit. Think of them just as fuel; you burn it during launch.

In order to get it *out* of orbit you're going to need the same delta-vee.
Since the shuttle's mass hasn't changed much, you're going to need the same
amount of fuel, i.e. an ET and two SRBs.

Getting down is exactly as hard as it is to get up, unless you can cheat by
using the atmosphere.

--
+- David Given --McQ-+ "A character is considered to be a letter if and
| | only if it is a letter or digit (§20.5.16) but is
| ) | not a digit (§20.5.14)." --- SMSDN Java
+- www.cowlark.com --+ documentation

Allen Meece wrote:
The returning weight is about 80% of the takeoff weight. So we need 80%
of the energy used to put the shuttle in orbit to
stop it in place. That's not a particularly small fraction.
Wheeewww. Not when you realize that most of the weight of the shuttle
*system* was in the ET and the SRB's.
Without them, the orbiter weighs a *fraction* of the complete STS
weight and
it won't take *anything* like an SRB to brake the shuttle into a slow
speed reentry. Probalby the ullage left in the ET would be plenty of fuel
for sufficient braking..

Yeah, but the shuttle's the only important bit. That vast weight of the ET
and the SRBs is there for the sole purpose of accelerating the shuttle into
orbit. Think of them just as fuel; you burn it during launch.

In order to get it *out* of orbit you're going to need the same delta-vee.
Since the shuttle's mass hasn't changed much, you're going to need the same
amount of fuel, i.e. an ET and two SRBs.

Getting down is exactly as hard as it is to get up, unless you can cheat by
using the atmosphere.

--
+- David Given --McQ-+ "A character is considered to be a letter if and
| | only if it is a letter or digit (§20.5.16) but is
| ) | not a digit (§20.5.14)." --- SMSDN Java
+- www.cowlark.com --+ documentation

Suppose the entire 25 tonnes payload of the shuttle orbiter was LH2 and LOX,
and the dry mass of the orbiter is 75 tonnes. By the rocket equation v/ve =
ln((F+P)/P), where v is the change in velocity, ve is exhaust velocity for
LH2/LOX (4500 m/s), P is dry mass of shuttle, and F is mass of LH2/LOX...

v = 4500 * ln(100/75) = 4500 * 0.288 = 1296 m/s

Since orbital velocity is 8000 m/s, sacrificing the entire shuttle payload
to carry fuel would only allow the shuttle orbiter to decelerate by 16% of
the amount needed to come to rest. The only result would be to enter the
atmosphere at a much steeper angle and thus suffer much greater reentry
forces and heat.

Since the entire purpose of the shuttle is to place payload into orbit, I
can see why braking beyond the amount needed to just intersect the
atmosphere has never been seriously considered.

"Allen Meece" wrote in message
...
The returning weight is about 80% of the takeoff weight. So we need 80%
of
the energy used to put the shuttle in orbit to
stop it in place. That's not a particularly small fraction.
Wheeewww. Not when you realize that most of the weight of the shuttle
*system* was in the ET and the SRB's.
Without them, the orbiter weighs a *fraction* of the complete STS
weight and
it won't take *anything* like an SRB to brake the shuttle into a slow
speed
reentry. Probalby the ullage left in the ET would be plenty of fuel for
sufficient braking..
^
//^\\
~~~ near space elevator ~~~~
~~~members.aol.com/beanstalkr/~~~

(Allen Meece) wrote in message ...
Probalby the ullage left in the ET would be plenty of fuel for
sufficient braking..

Heh. Do the math. This is an easy application of the rocket equation
(now someone correct my math errors).

Orbiter with typical ET residuals (15,000lbs): 200,000lbs
Orbiter after expending residuals: 185,000lbs
Specific impulse of the SSMEs in vacuum: 455 sec^-1.

(455 sec^-1) x (9.8m/s/s) x Natural Log (200,000lbs / 185,000lbs) =
347 m/s.

With those 15,000lbs of residuals, you'll slow the orbiter by 779
miles per hour. That's not much out of the 17500mph orbital velocity.

That assumes you get the SSMEs to restart in orbit and that the
shuttle orbiter carries the residuals internally, rather than in the
external tank.

Mike Miller, Materials Engineer

Jorge Incorrect. The shuttle has a much higher hypersonic L/D (4.0) than
Jorge the X-38 (1.4).

Whoa! L/D = 4 hypersonic is pretty good! Where'd you get that number?