Re: Airliner 1/4 mile ETs and MPH?? :-)

From:         "David K. Cornutt" <cornutt@hiwaay.net>
Organization: Residential Engineering
Date:         13 Sep 96 03:03:36 
References:   1
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In article <airliners.1996.1782@ohare.Chicago.COM> Ed Treijs,
transam@idirect.com writes:
> ET> I'm wondering if anyone has any "drag racing" times for
> ET> airplanes--let's talk commercial, from turboprop commuter specials to
> ET> whatever the "fastest" jet (I assume it would be a jet) is.  Yeah,
> ET> jet fighters might be able to hit 70,000 feet 3:42 after starting
> ET> takeoff roll, but that's not quite in the ballpark.

Not quite what you asked for, but related...  Here's a little
bit of calculation that I wrote up a few years ago.  (Keep
this in mind when you read the comments in the text about
NHRA classes and records.)  It was originally posted to
sci.space.shuttle:

Space Shuttle Acceleration (1/5/94)

Someone recently posted a query on how long it takes the Shuttle
to fly a quarter mile from a dead stop.  Well, I haven't sat down
and done anything mathematical in a while, so I regurgitated as
much of my high-school physics as I could remember, and sat
down and worked it out.  You might be disappointed, though.
(Explanation later)

Since all the data I had available were the Shuttle mass and
thrust (from a KSC pamphlet), I started with force = mass times
acceleration and went from there.  The pamphlet gives the weight
of the Shuttle stack on the pad (fully fueled) as being about
2 million kilograms.  (I'm going to do this in metric, both
as an exercise to myself, and to avoid getting wrapped around
the axle of lbs-mass vs. lbs-force, which always confuses the
hell out of me when I try to work out the units.)  It gives
the combined thrust of the SSMEs and SRBs at liftoff as being
about 28.9 million newtons.  (The newton is the metric unit
of force, equivalent to an accleration of 1 meter/sec/sec on
a mass of one kilogram, leading to the units being m*kg/sec/sec.
One pound of force is about 4.5 newtons.)

The first thing we need to do is find the acceleration, since
that's what we need to derive distance and velocity.  We
know the upward force from the engines and boosters; however,
part of this force is offset by the downward force of gravity
acting on the Shuttle's mass.  To find this force, we use
F=MA.  The acceleration of gravity is about 9.75 m/sec/sec.
(This is burned into my brain, from the gruesome chant we
used in eighth grade: "Thirty-two feet per second per second,
until you hit the ground." :-)  Plugging in, we get:

                 9.75M m
Force = 2M kg * -----------
                   sec^2

or, 19.5 M newtons downward.  Subtracting this from the 28.9 M
newtons of force from the engines produces a net upward force
at liftoff of 9.4 M newtons.

Now, we can plug this back into F=MA and solve for acceleration:


                9.4 M m*Kg    /
Acceleration = ------------  /    2M Kg
                  sec^2     /

             = 4.7 m/sec/sec

Now, we know that velocity = acceleration * time (plus initial
velocity, but that's 0, so we can ignore it).  Integrating
this, we also know that distance = .5 * acceleration * time^2.
To find the time it takes to do the quarter mile from a
standing start, we have to solve the last equation for time:

time^2 = 2 * distance / accelaration, or

time = SQRT (2 * distance / accelaration)

So, given a distance of a quarter mile (which we'll round off
to 400 meters), and the acceleration above, we can solve
for time:

            {             /  4.7 m    }
time = SQRT { 2 * 400 m  /  --------  }
            {           /   sec ^2    }

It works out to 13 seconds.  (You can do the math.)  To find
the velocity, we plug the time back into V=AT, and get a
terminal velocity after 13 seconds of 61.1 m/sec.  Converting
this back to English units, the velocity works out to 137 mph
at MET 00:13.

Sure doesn't seem like much, does it?  Lots of cars can reach
137 mph in less than 13 seconds.  There are three points to
remember here:

1.  Acceleration is a little like compound interest (psycho-
logically, not mathematically) in that it doesn't start
to do surprising things until it's been going for a while,
but once it gets going, it really gets going.  No automobile
can sustain this kind of acceleration for several minutes,
but the Shuttle can.

2.  The Shuttle's mass decreases as it goes.  Fuel is depleted,
reducing the tank & SRB mass, and there's a big step decrease when
the SRBs drop off.  Also, the SSMEs produce about 25% more
thrust in a vacuum then they do at sea level.

3.  For most of the flight profile, the Shuttle isn't going
straight up (it would never achieve orbit if it did).  The
force of gravity that offsets the thrust is reduced (at least
for the purpose of gaining velocity) as the pitch angle
decreases.  (At MECO, the velocity vector is at a fairly
low angle [I think about 15 degrees] to the earth's surface.)

This last brings up an interesting point.  What if one could
put the Shuttle on wheels and drag race it?  What could it
do in the quarter mile?  Let's find out:

Traveling horizontally, the Shuttle's engines wouldn't have
to fight the Earth's gravity, so the entire thrust would
be available for lateral acceleration.  Using the entire
28.9 M newtons in the F=MA equation and solving for
acceleration produces:

                28.9M m*Kg    /
Acceleration = ------------  /    2M Kg
                  sec^2     /

             = 14.4 m/sec/sec

Putting this into the time equation gives us:

            {             /  14.4 m   }
time = SQRT { 2 * 400 m  /  --------  }
            {           /   sec ^2    }

     = 7.4 sec

Now, computing the terminal velocity gives:

velocity = 14.4 m/sec * 7.4 sec

         = 106 m/sec

Converting this to English units gives 239 mph.  Since the "top
speed" in drag racing is an average over a short distance which
is mainly ahead of the finish line, the Shuttle would probably
turn in a quarter mile of 7.4 @ 230.  That might be a respectable
time in the NHRA Pro Stock class.  However, the current NHRA
world record for a Top Fuel dragster is about 4.88 @ 304 mph.
So you can forget about seeing Kenny Bernstein or John Force
racing a Shuttle anytime soon.  (Then again, Top Fuel dragsters
weigh about 0.03% as much as a Shuttle.  And they don't fly,
at least not intentionally. :-)

Of course, this ignores a lot of real-world stuff, like wind
resistance and losses due to offset thrust.  But what's
interesting about the little drag-racing calculation above
is that the acceleration value that we started with, 14.4
m/sec/sec, is roughly the mid value between the liftoff
acceleration (which we computed to be 4.7 m/sec/sec), and
the maximum accleration of 3G (about 29 m/sec/sec), which
is always reached around MET 7:00.  If I remember right,
MECO is timed to occur when a velocity of about
23,000 feet/sec, or 7000 m/sec, is reached.  If one plugs
this back into the velocity equation, it yields a MECO time
of 8:06, which is not very far off the actual MECO time of
around 8:30.

---
David K. Cornutt, Residentially Engineered, Huntsville, AL
email: cornutt@hiwaay.net
I'm a rocket scientist.  Don't tell me what TV I must see.