speed record set by scramjet - fair?

How fast was the rocket going when it released the record-setting
scramjet? If the rocket was going Mach 9 in the thin atmosphere at
100,000 feet and released a stone, for example, the stone would travel
several seconds at close to Mach 9. I assume that the rocket was not
going Mach 9, but I haven't seen any information on how fast it was
going.

Regardless, it seems to me that the rocket's speed has to be
subtracted from the jet's speed to arrive at the actual jet speed when
you talk about the world's record for speed of a jet plane.

-- Don French

"Don French" wrote in message

Regardless, it seems to me that the rocket's speed has to be
subtracted from the jet's speed to arrive at the actual jet speed when
you talk about the world's record for speed of a jet plane.

Hmm. Would you say the same for Yeager and the X-1, it having been dropped
from the belly of another aircraft, or is your particular question related
just to the rocket?

Would this same sort of criteria apply to the X-prize given that Space Ship
One was given a lift to an intermediate altitide?

Interesting.
-c

In article ,
"gatt" wrote:

Would this same sort of criteria apply to the X-prize given that Space Ship
One was given a lift to an intermediate altitide?

You could interpret the White Knight as a manned first stage of the
launch system. Since it's also reusable within the prize parameters, the
complete launch system would still qualify.

Whether or not you subtract the rocket's speed from the X-43, it's still
quite an accomplishment to have an air-breathing engine running at Mach
10. This is the proof of concept stage, right? Now we know that the
scramjet design doesn't just blow itself out.

- Nathan

The X-prize is different. They did what was required to win the
prize, which was get someone into space and return. Well, maybe the
scramjet did what was required to set the world's speed record too,
but it fails to impress since it wasn't the jet's engine that got it
going that fast. The jet only contributed the last few pounds of
thrust required to defeat air friction to keep it going at that speed
and maybe a few more to accelerate a half Mach or so.

I did a calculation that makes some assumptions that may or may not be
completely accurate. I am neither an aeronautical engineer or a fluid
dynamics expert, but I still made a go at trying to compute how
difficult it was for the scramjet to accelerate from Mach 9.5 to Mach
10.

Newton's first law of motion tells us that a plane released from a
rocket at Mach 10 will, in the absence of air friction, continue at
that speed indefinitely (or until it encounters another object, like
the Earth), and never have to turn on its engines to do so. The
scramjet only has to have enough power to overcome what little air
friction there is at 100,000 feet to maintain its release speed. The
question is how much air friction is there at Mach 10 at 100,000 feet.

Since I don't know how to compute the actual frictional effects at
that speed and altitude, it occured to me that maybe I can at least
compute the ratio between overcoming friction at that speed and
altitude and at say, Mach 1 at 5000 feet. That would provide a way of
making a comparison that makes sense to me.

My first assumption is that for the same air density, the friction is
directly proportional to the speed of the aircraft. If that is true,
the scramjet has to exert 10 times as much thrust to overcome friction
than a jet flying at Mach 1 for the same air density.

But the scramjet is flying at 20 times the altitude of the other jet,
and the air density is much lower up there. My second assumption is
that air resistance is directly proportional to air pressure. If this
is true, I can compute the relative ease of overcoming the friction by
simply computing relative air pressures. Air pressure decreases with
the square of the distance from Earth. So the difference between the
air pressure at 5000 feet and 100,000 feet is 1/20 squared, or 1/400.
And since air pressure and air density are proportional, there is
1/400 times as much air per cubic centimeter at 100,000 feet than
there is at 5,000 feet.

So, if all my assumptions are correct, then it is about 40 (400/10)
times as easy to maintain Mach 10 at 100,000 feet as it is to maintain
Mach 1 at 5000 feet. My final assumption is that this also means that
it takes about 1/40 the thrust to accelate from Mach 9.5 to Mach 10 at
100,000 feet as it does to accelerate from Mach Mach 1.0 to Mach 1.5
at 5000 feet. And if that is true, then it is also true that it takes
exactly the same amount of thrust to go from Mach 9.5 to Mach 10 at
100,000 feet as it takes to go from Mach 1 to Mach 1 plus 1/40 of a
half Mach. 1/40 of a half Mach is about 10 miles an hour increase in
speed.

Therefore, according to my calculations, if the scramjet accelerated
from Mach 9.5 to Mach 10, it took about as much thrust as for a jet
flying at Mach 1 at 5000 feet to increase its speed by 10 miles per
hour.

Like I said, I am neither an aeronautical engineer nor a fluid
dynamics expert, so consider the source. If there is an aeronautical
engineer or a fluid dynamics expert out there who can point out the
errors in these calculations, please do. Just leave out the flames,
OK? At least I made an attempt at reasoning through the problem and
realize my limitations.


-- Don French





Regardless, it seems to me that the rocket's speed has to be
subtracted from the jet's speed to arrive at the actual jet speed when
you talk about the world's record for speed of a jet plane.

Hmm. Would you say the same for Yeager and the X-1, it having been dropped
from the belly of another aircraft, or is your particular question related
just to the rocket?

Would this same sort of criteria apply to the X-prize given that Space Ship
One was given a lift to an intermediate altitide?

Interesting.
-c

My first assumption is that for the same air density, the friction is
directly proportional to the speed of the aircraft.

Nope. To oversimplify, it goes as the cube at subsonic speeds. Once supersonic other terms enter the equation. So at Mach 10 the scramjet would
have to exert more than 1000 times the thrust as for Mach 1 at the same altitude. And a scramjet can't run from a standing stop.

Jose
--
Freedom. It seemed like a good idea at the time.
for Email, make the obvious change in the address.

Jose wrote:

My first assumption is that for the same air density, the friction is
directly proportional to the speed of the aircraft.

Nope. To oversimplify, it goes as the cube at subsonic speeds. Once supersonic other terms enter the equation. So at Mach 10 the scramjet would
have to exert more than 1000 times the thrust as for Mach 1 at the same altitude. And a scramjet can't run from a standing stop.

Jose
Jose, hopefully someone will correct me if I'm wrong, but the drag
(and the thrust needed to overcome it) increase with the square of
velocity. It is the power needed that increases with v^3.
--
Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently.

Jose, hopefully someone will correct me if I'm wrong, but the drag
(and the thrust needed to overcome it) increase with the square of
velocity. It is the power needed that increases with v^3.

Yep. Misread it. But still far from linear.

Jose
--
Freedom. It seemed like a good idea at the time.
for Email, make the obvious change in the address.

On Fri, 19 Nov 2004 at 22:53:26 in message
, alexy
wrote:
Jose wrote:

My first assumption is that for the same air density, the friction is
directly proportional to the speed of the aircraft.

Nope. To oversimplify, it goes as the cube at subsonic speeds. Once
supersonic other terms enter the equation. So at Mach 10 the scramjet
would
have to exert more than 1000 times the thrust as for Mach 1 at the
same altitude. And a scramjet can't run from a standing stop.

Jose
Jose, hopefully someone will correct me if I'm wrong, but the drag
(and the thrust needed to overcome it) increase with the square of
velocity. It is the power needed that increases with v^3.

Perhaps it is time to insert the simple fundamental equation of
aerodynamic drag.

Drag = 0.5 * (air density)* ( a representative area - commonly wing
area )*(velocity squared)*( a 'constant that depends mostly on shape)

If using pounds divide the result by 'g': 32.2 ft/(second squared) for
an answer in pounds.

The speed of sound in air is proportional to the square root of the
absolute temperature of the air. Thus in the standard atmosphere the
speed of sound falls from 1,117 ft/sec at sea level to 968.5 ft sec at
36,000 ft. Above that it is constant up to around 80,000 ft when it
starts to rise again up to about 175,000 ft when it starts to fall
again!.

So the original simple calculation is wrong. It is more complex than I
have said as drag also depends on Mach number and a number called
Reynolds Number.

Altitude ft Speed of sound density
S.L. 1,117 ft/second 0.076475 lb/(cubic foot)
50,000 ft 968.5 ft/second 0.011642
100,000 ft 1003.2 ft/second 0.0010332

E&OE (' Errors And Omissions Excepted. That means I hope it's right but
I don't guarantee it! Check the tables yourself.)

I apologise that this is not metric units.
--
David CL Francis

(Don French) wrote:

The X-prize is different. They did what was required to win the
prize, which was get someone into space and return. Well, maybe the
scramjet did what was required to set the world's speed record too,
but it fails to impress since it wasn't the jet's engine that got it
going that fast. The jet only contributed the last few pounds of
thrust required to defeat air friction to keep it going at that speed
and maybe a few more to accelerate a half Mach or so.

I did a calculation that makes some assumptions that may or may not be
completely accurate. I am neither an aeronautical engineer or a fluid
dynamics expert, but I still made a go at trying to compute how
difficult it was for the scramjet to accelerate from Mach 9.5 to Mach
10.

Newton's first law of motion tells us that a plane released from a
rocket at Mach 10 will, in the absence of air friction, continue at
that speed indefinitely (or until it encounters another object, like
the Earth), and never have to turn on its engines to do so. The
scramjet only has to have enough power to overcome what little air
friction there is at 100,000 feet to maintain its release speed. The
question is how much air friction is there at Mach 10 at 100,000 feet.

Since I don't know how to compute the actual frictional effects at
that speed and altitude, it occured to me that maybe I can at least
compute the ratio between overcoming friction at that speed and
altitude and at say, Mach 1 at 5000 feet. That would provide a way of
making a comparison that makes sense to me.

My first assumption is that for the same air density, the friction is
directly proportional to the speed of the aircraft.
Bad assumption. I think that drag is proportional to the square of
speed.

If that is true,
the scramjet has to exert 10 times as much thrust to overcome friction
than a jet flying at Mach 1 for the same air density.
100 times if it is square.

But the scramjet is flying at 20 times the altitude of the other jet,
and the air density is much lower up there. My second assumption is
that air resistance is directly proportional to air pressure.
Intuitively that "feels" right. But I don't trust my intuition on
things like this.

If this
is true, I can compute the relative ease of overcoming the friction by
simply computing relative air pressures. Air pressure decreases with
the square of the distance from Earth.
Another assumption I would question. Please tell me you do not believe
that the air in New Orleans at 22'MSL is 57,000 time more dense and
higher pressured than the air in Denver!!!!

So the difference between the
air pressure at 5000 feet and 100,000 feet is 1/20 squared, or 1/400.
And since air pressure and air density are proportional, there is
1/400 times as much air per cubic centimeter at 100,000 feet than
there is at 5,000 feet.

So, if all my assumptions are correct, then it is about 40 (400/10)
times as easy to maintain Mach 10 at 100,000 feet as it is to maintain
Mach 1 at 5000 feet.
Highly suspect assumptions, even with sound reasoning, lead to highly
suspect conclusions.

My final assumption is that this also means that
it takes about 1/40 the thrust to accelate from Mach 9.5 to Mach 10 at
100,000 feet as it does to accelerate from Mach Mach 1.0 to Mach 1.5
at 5000 feet. And if that is true, then it is also true that it takes
exactly the same amount of thrust to go from Mach 9.5 to Mach 10 at
100,000 feet as it takes to go from Mach 1 to Mach 1 plus 1/40 of a
half Mach. 1/40 of a half Mach is about 10 miles an hour increase in
speed.
And you were complaining about the demise of physics education? g

Therefore, according to my calculations, if the scramjet accelerated
from Mach 9.5 to Mach 10, it took about as much thrust as for a jet
flying at Mach 1 at 5000 feet to increase its speed by 10 miles per
hour.

Like I said, I am neither an aeronautical engineer nor a fluid
dynamics expert
Glad you pointed that out!

, so consider the source. If there is an aeronautical
engineer or a fluid dynamics expert out there who can point out the
errors in these calculations, please do.
I'm neither.

Just leave out the flames,
OK? At least I made an attempt at reasoning through the problem and
realize my limitations.

Okay. Sorry for the sarcasm that crept in. But hopefully you see where
the assumptions look suspect to this layman.
--
Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently.

Are you sure that you are really not an aeronautical engineer? Are you a
high school graduate?

Mike
MU-2

"Don French" wrote in message
om...
The X-prize is different. They did what was required to win the
prize, which was get someone into space and return. Well, maybe the
scramjet did what was required to set the world's speed record too,
but it fails to impress since it wasn't the jet's engine that got it
going that fast. The jet only contributed the last few pounds of
thrust required to defeat air friction to keep it going at that speed
and maybe a few more to accelerate a half Mach or so.

I did a calculation that makes some assumptions that may or may not be
completely accurate. I am neither an aeronautical engineer or a fluid
dynamics expert, but I still made a go at trying to compute how
difficult it was for the scramjet to accelerate from Mach 9.5 to Mach
10.

Newton's first law of motion tells us that a plane released from a
rocket at Mach 10 will, in the absence of air friction, continue at
that speed indefinitely (or until it encounters another object, like
the Earth), and never have to turn on its engines to do so. The
scramjet only has to have enough power to overcome what little air
friction there is at 100,000 feet to maintain its release speed. The
question is how much air friction is there at Mach 10 at 100,000 feet.

Since I don't know how to compute the actual frictional effects at
that speed and altitude, it occured to me that maybe I can at least
compute the ratio between overcoming friction at that speed and
altitude and at say, Mach 1 at 5000 feet. That would provide a way of
making a comparison that makes sense to me.

My first assumption is that for the same air density, the friction is
directly proportional to the speed of the aircraft. If that is true,
the scramjet has to exert 10 times as much thrust to overcome friction
than a jet flying at Mach 1 for the same air density.

But the scramjet is flying at 20 times the altitude of the other jet,
and the air density is much lower up there. My second assumption is
that air resistance is directly proportional to air pressure. If this
is true, I can compute the relative ease of overcoming the friction by
simply computing relative air pressures. Air pressure decreases with
the square of the distance from Earth. So the difference between the
air pressure at 5000 feet and 100,000 feet is 1/20 squared, or 1/400.
And since air pressure and air density are proportional, there is
1/400 times as much air per cubic centimeter at 100,000 feet than
there is at 5,000 feet.

So, if all my assumptions are correct, then it is about 40 (400/10)
times as easy to maintain Mach 10 at 100,000 feet as it is to maintain
Mach 1 at 5000 feet. My final assumption is that this also means that
it takes about 1/40 the thrust to accelate from Mach 9.5 to Mach 10 at
100,000 feet as it does to accelerate from Mach Mach 1.0 to Mach 1.5
at 5000 feet. And if that is true, then it is also true that it takes
exactly the same amount of thrust to go from Mach 9.5 to Mach 10 at
100,000 feet as it takes to go from Mach 1 to Mach 1 plus 1/40 of a
half Mach. 1/40 of a half Mach is about 10 miles an hour increase in
speed.

Therefore, according to my calculations, if the scramjet accelerated
from Mach 9.5 to Mach 10, it took about as much thrust as for a jet
flying at Mach 1 at 5000 feet to increase its speed by 10 miles per
hour.

Like I said, I am neither an aeronautical engineer nor a fluid
dynamics expert, so consider the source. If there is an aeronautical
engineer or a fluid dynamics expert out there who can point out the
errors in these calculations, please do. Just leave out the flames,
OK? At least I made an attempt at reasoning through the problem and
realize my limitations.


-- Don French





Regardless, it seems to me that the rocket's speed has to be
subtracted from the jet's speed to arrive at the actual jet speed when
you talk about the world's record for speed of a jet plane.

Hmm. Would you say the same for Yeager and the X-1, it having been
dropped
from the belly of another aircraft, or is your particular question
related
just to the rocket?

Would this same sort of criteria apply to the X-prize given that Space
Ship
One was given a lift to an intermediate altitide?

Interesting.
-c