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#102
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![]() "cjcampbell" wrote in message oups.com... Saw this question on "The Straight Dope" and I thought it was amusing. http://www.straightdope.com/columns/060203.html The question goes like this: "An airplane on a runway sits on a conveyer belt that moves in the opposite direction at exactly the speed that the airplane is moving forward. Does the airplane take off?" (Assuming the tires hold out, of course.) Cecil Adams (world's smartest human being) says that it will take off normally. When this question was raised on Sci.aeronautics it lasted 14 replies, (as archived by Google groups, but I think that some other posts were dropped by the news servers.) http://groups.google.ca/group/sci.aeronautics/browse_thread/thread/e344b133e57880a0/57a2a6789dd92bf6?lnk=st&q=(treadmill)+group%3Asci. aeronautics&rnum=1&hl=en#57a2a6789dd92bf6 One of the posters raised the issue of what would happen when the aircraft returned for landing. we are up to about 110 replies here on r.a.s, r.a.p. A similar question was posed in the Airliners.net forums, what a great discussion it was! See he http://www.airliners.net/discussions...d.main/136068/ over 400 replies! I am quite surprised that this question has received the amount of consideration that it has. Must be a slow news month or maybe is an indication that there are more people thinking about flying than there are actually doing it. |
#103
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In article , rmoore16
@tampabay.rr.com says... BillJ wrote The earth is a treadmill. Goes about about 900 knots (at equator). Does that bother your takeoff? Suppore treadmill stopped (rotation stopped). Takeoffs any different? You forgot one major difference....in the case of the earth, the airmass is travelling at the same 900kts, ignoring any localized wind effect....not so in the treadmill case. So when a treadmill runs the wind starts blowing ? OK... wot are yer saying then? -- Duncan |
#104
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![]() "Dave Doe" wrote in message . nz... In article , says... "Dave Doe" wrote in message . nz... In article , says... At the point where the tire contacts the ground, it's speed is zero. 180° away, at the top, it is moving forward at twice the speed of the car. Negative - yer forgetting centripetal force. ? Negative what? Talking about a point on the surface of the tire, not the wheel as a whole. Centripital force has nothing to do with the forward velocity of that point (how it travels in one axis). Are you talking about a round tire or not What other kind of tire is there? .. Sorry bud, can't make the initial assumption that's been made - I'm not your bud,. And what assumption are you talking about? as it's on a tire, and yep, even that point, at that time - has the centripetal force. We aren't talking about the forces at work on the wheel or tire, we are talking about the forward velocity. I can see this concept is lost on you. -- Duncan |
#105
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![]() "cjcampbell" wrote The question goes like this: "An airplane on a runway sits on a conveyer belt that moves in the opposite direction at exactly the speed that the airplane is moving forward. Does the airplane take off?" (Assuming the tires hold out, of course.) Cecil Adams (world's smartest human being) says that it will take off normally. Maybe he's not so smart after all :) On a calm day you can run and feel a wind on your face because you are moving across the ground as well as through the air. But, if you run on a treadmill there will be no wind because you are not moving through the air - the air is calm so it has no relative motion with respect to the ground. Neither do you when you run on a treadmill. Assume the airplane is on the conveyor and there is a 10 kt headwind, and assume we need 60 kts for takeoff. The only way to generate the additional 50 kts of airspeed is by moving across the ground at 50 kts. If the airplane is standing still because the conveyor is moving backwards at the same speed that the airplane is moving across the ground at, then the airspeed will still be 10 kts. If the conveyor keeps the airplane standing still relative to the ground, then it cannot take off. If it could, then we'd all have problems during run up because the brakes do the same thing that the theoretical conveyor does - prevent motion across the ground. BDS |
#106
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![]() "BDS" wrote in message t... Maybe he's not so smart after all :) On a calm day you can run and feel a wind on your face because you are moving across the ground as well as through the air. But, if you run on a treadmill there will be no wind because you are not moving through the air - the air is calm so it has no relative motion with respect to the ground. Neither do you when you run on a treadmill. Yes, but an airplane isn't propelled by its feet. |
#107
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"BDS" wrote in message
t... 50 kts of airspeed is by moving across the ground at 50 kts. If the airplane is standing still because the conveyor is moving backwards at the same speed that the airplane is moving across the ground at, then the airspeed will still be 10 kts. But the airplane *isn't* standing still (as others have pointed out). Since the wheels are able to spin freely, the reverse thrust by the conveyor belt is *not* transferred past the wheels to the plane's fuselage, and so the only remaining applied force is the unbalanced forward force of the thrust of the engine. The conveyor *doesn't* "drag" the plane backwards to compensate for the engine's thrust. All it does (as others have pointed out) is spin the wheels faster. If the conveyor keeps the airplane standing still relative to the ground, It doesn't. (Don't worry... I didn't understand it at first either.) BDS Jeff Shirton (PP-ASEL) |
#108
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"One of the posters raised the issue of what would happen when the
aircraft returned for landing." If the belt were moving backwards at the speed of the aircraft when it touched down, it would be similar to landing with that much tailwind, basically, your ground speed would double you airspeed at touch down. Jester |
#109
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Dave Doe wrote:
Try this for a brain scrambler. Think about a tire on your car, driving down the highway. At the point where the tire contacts the ground, it's speed is zero. 180° away, at the top, it is moving forward at twice the speed of the car. Negative - yer forgetting centripetal force. http://www.wordiq.com/definition/Centripetal Well, I'm impressed that you know of the existence of centripetal force. But in what possible way do you think it negates the comment about the speeds (relative to the ground) of points at the top and bottom of the tire on a moving car? -- Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently. |
#110
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"Peter Duniho" wrote:
"alexy" wrote in message .. . Yes, the problem could have been made uninteresting by removing any ambiguity. But as stated, it is very common (almost universal) to speak of movement of a terrestrial object with respect to the surface of the earth. If another frame of reference is intended, it is almost always specified. Very amusing. According to you: On the one hand, the problem is uninteresting if one removes the ambiguity in the phrasing. On the other hand, there is no ambiguity, because if a different frame of reference were intended, "it is almost always specified". So, the logical conclusion you arrive it in your post is that the problem is uninteresting. For an uninteresting problem, it sure generated a lot of traffic. True. Which really surprised me. When I first saw CJ's post, I thought it was too obvious to draw in this kind of activity. -- Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently. |
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