From:"David K. Cornutt" <cornutt@hiwaay.net>Organization:Residential EngineeringDate:13 Sep 96 03:03:36References: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.