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Conveyor-belt runway

More time wasting puzzles

Okay, it looks like the moving runway has been put to bed, so here's one with no tricky words and the chance of actually trying it at home.

You've got an empty mayonaise jar sitting on your fancy electronic postal scale and all is working fine. You notice a fly taking a rest on the bottom of the jar. The fly now lifts off and hovers inside the jar. Does the scale register a change?

If the scale does read lighter now, where did the weight of the fly go? Would it matter whether the jar had a lid?

If, on the other hand, the scale still indicates the combined weight, how high can the fly go? What if the jar suddenly vanished? Would the reading drop by only the weight of the jar and still register the weight of the hovering fly?

When you feel good about your answer, ask yourself how things would differ if the jar were full of water and a fish.

No obfuscation on this one; just plain old physics -- I think.
 
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Agree to agree and disagree

gmcjetpilot said:
Yes you are right and wrong. :D

No its not just interpretation, and yes the plane will fly. I'll take the stand that there is only one correct interpretation and solution.

I agree interpretation can be an issue in Puzzles. Some have no solution and are more philosophical than analytical (does a tree falling in a forest make a sound), but most riddles/puzzles have one intended solution, albeit with miss leading or non-relevant statements to "de-clarify", trying to lead you to the wrong conclusion.

I agree people interpret words differently and that's why there are lawyers, but I the real reason for disagreement is analytical.

The main reason some folks came to the "plane won't fly" conclusion is they did not understand forces involved with bodies in motion (balanced forces, free body equations, summation of forces ...... thrust, friction, lift, gravity, relative motion, degrees of freedom, reference systems and Newton's laws of motion Force = mass x acceleration).

Yes it's common to obfuscate and "de-clarify" to miss lead like:

What is heaver, a pound of feathers or a pound of gold?
Feathers are lighter so GOLD are heaver!​

Well of course GOLD IS heavier!!!!! Well the answer is neither is heaver. A pound of anything is a pound. Again yes one might be miss lead by intuition or lack of careful reading, but they are also miss lead by assumptions of weight, mass and density. That is why there are nerdy engineers. Some people no matter what you say will tell you a pound of feathers is lighter than a pound of gold.

Before anyone gets pissed and takes insult, please don't. The intent of riddles is fun. The whole moving runway scenario is not realistic or practical...... Puzzles and riddles and solving them is an acquired taste and you get better with practice.


If the plane was tied down or could not move, it would never takeoff despite the tire speed, thrust or conveyor speed. Of course this would be a trivial situation. The "puzzle" only said the conveyor moved inversely proportional to the aircraft (fwd) speed. People assume that means the plane can't move forward. Clearly the puzzle never SAID it can't move, but people ASSUME it can't if the runway moved opposite. THIS IS THE GOTCH-A.

A free body diagram (engineering tool to analyze the forces on an object) clears the issue up. If the plane has thrust and is NORMAL in every way (degrees freedom of motion, e.g. not tied down), except it's on a weird moving runway.

The OTHER INFORAMATION that they included to throw you off is the conveyor can SENSE the aircraft speed and move OPPOSITE at the SAME speed. Again like feathers vs. gold, you assume the WORDS opposite and same speed means the plane will not move forward.


The take off distance may be a few feet longer on the moving runway, but in the end excess thrust will overcome the small extra drag, which allows the plane to accelerate (until thrust = drag). The first point in time where the thrust and drag (gravity and lift) are all balanced after starting the "take-off roll" is after lift-off, established in a steady state speed climb.


A backwards moving runway will not stop the plane from moving forward.

So the question is clear. Will it take off or will it not?

The answer is yes IT will. G :D
Click to expand...
George, it looks like we agree that the solution will change depending on interpretation, but we will have to disagree that there is only one solution. Maybe I'm a boring engineer by believing that the riddle was poorly written rather than purposefully declairified. One of my many pet peeves is when relative terms are not properly defined. The speed of the plane relative to the belt, the ground, another plane in the air and Haley's comet are all very different.

I don't see how the gold and feathers analogy is comparible as that is simply a trick question and it really isn't left up to interpretation.

I think the reason that I initially (and many others) thought that the riddle was referring to the speed of the plane relative to the belt was because it is a really easy puzzle otherwise (i.e. the plane will obviously fly with 2X wheel speed) and not very amusing (at least not to me).

All of that said, after re-reading it yet again, if I had to pick one, I would guess that the intention of the author was that the belt speed would match the plane speed relative to the ground. This would result in a flying aircraft.

It's possible that the original author was a diabolical genius, but I think that it is more likely that he/she was rather clueless, just my opinion.
 
A hint of truth and a dash of bogus

ericwolf said:
because it is a really easy puzzle otherwise (i.e. the plane will obviously fly with 2X wheel speed) and not very amusing (at least not to me).
Click to expand...
Apparently it was not that easy. :D Like all good puzzles there is just hint of truth and a dash of bogus. There are some folks that still swear a pound of gold weighs more than the pound of feathers.

Truth be told, the moving runway thing is silly and almost pegs the bogosity meter, but if you replaced the airplane with a car, the car would obviously remain stationary to the "global" reference system and the answer changes, SO it is not that easy.

YOU can win a new RV-10, which is behind one of three hanger doors. The winning prize is a brand new RV-10, full panel and new IO540, leather interior. If you pick the hanger door with the RV-10 behind it, you will win it. Pick one of the doors, 1, 2 or 3. Lets say you pick door 3, just for an example. Now one of the other two doors, door 1 or 2 is reviled, and when opened it shows that the new RV-10 is not there. Just for example lets say door 1 is opened. You are offered to keep door 3, the first door you picked, or you could switch to door 2 if you want?

Will switching doors increase your chances of winning the RV-10?
Will you be better off by keeping the door you first picked?
Will switching doors make no difference in your chances?​

You will always have free choice of any of the 3 doors initially; After choosing a door you will always be shown one of the other doors which reviles a non-prize (since Bob Barker knows where the plane is), and you are always given the chance to change to the other door that is still closed or stick to your original door. On average what is the best tactic, stay, switch or makes no difference.

IF you know, post it. The answer might surprise you. I'll post the answer tomorrow or day after. G

(please no post that its not RV related, hey I am talking about winning a RV-10. :D )
 
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On average, my first pick will be wrong 2/3 of the time. Switching in this case gets me the plane. Therefore, the switching strategy will produce a winner 2/3 of the time. Not switching is only desirable if I've already chosen correctly, which happens 1/3 of the time. Thus the switcher will win twice as often as the non-switcher. Most counter-intuitive!!

So what about my bug in a jar? What if the jar was huge and there was an RV4 doing cubans inside?
 
Here we go again...

How 'bout this one....

A cargo plane is flying along hauling 10,000 Canaries in the cargo hold. All at once all the Canaries decided to start flying inside of the cargo plane. Does the weight of the plane change? If so would it cause the plane to suddenly climb or decend?

:rolleyes:
 
We have a winner, but your new RV-10 is sized for a fly in a Jar

szicree said:
On average, my first pick will be wrong 2/3 of the time. Switching in this case gets me the plane. Therefore, the switching strategy will produce a winner 2/3 of the time. Not switching is only desirable if I've already chosen correctly, which happens 1/3 of the time. Thus the switcher will win twice as often as the non-switcher. Most counter-intuitive!!

So what about my bug in a jar? What if the jar was huge and there was an RV4 doing cubans inside?
Click to expand...
We have a WINNER!!!

The first choice is 33.33%, even after they take one of the three doors away, its still the 33.33%. The second door, that you switch to is 50% chance, or switching is 16.7% more likely to win the RV-10 than if you kept the original door. So if you are the "price is right" switch doors. George

I can't answer the bug and the RV-4 in a jar doing Cubans question, because it only works if the pilot is a tiny Cuban or the fly is smoking a Cuban cigar, clearly a trick question.
 
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Same answer for a car

gmcjetpilot said:
Apparently it was not that easy. :D Like all good puzzles there is just hint of truth and a dash of bogus. There are some folks that still swear a pound of gold weighs more than the pound of feathers.

Truth be told, the moving runway thing is silly and almost pegs the bogosity meter, but if you replaced the airplane with a car, the car would obviously remain stationary to the "global" reference system and the answer changes, SO it is not that easy.
Click to expand...

What can I say about the people who believe that a pound gold is heavier than a pound of feathers, they just need a good dope slap.

I agree with Steve's analysis of the game to get the RV-10. I've seen the puzzle before and it was a game show a long time ago. Let it be known, however, that I would rather it be a light and simple RV-8A with sheep skin seats. :D

I agree that the moving runway riddle is silly, but I have to once again disagree with you. If it were a car rather than an airplane, the car would move just the same assuming that the speed is referring to the speed of the car relative to the ground. You said it yourself that nowhere does it say that the car cannot move (relative to the ground). The belt will be moving at the same speed as the car relative to the ground. For example, let's take a car moving forward relative to the ground at 20mph. By the rules, the belt will have a speed of 20mph backwards. Then the speed of the car relative to the belt (and the reading on the speedometer) will be 40mph. The only difference in the answer is that the car will not fly regardless of speed. :D It is obvious that the car will not be stationary and it is that easy.
 
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Let's make a price is right deal

I don't know much 'bout science and stuff but I do know that the game where you picked what was behind door #3 was 'Let's Make Deal' hosted by Monty Hall. If you wanted to score an RV10 from Bob Barker you'd have to know the retail value of the RV10 and/or have your dog/cat spayed/neutered or you'd have to be a big boobed game show hottie.

FWIW though I agree with Eric. According to the letter of the riddle, a car must move forward in order for the belt to move backward so the car would in fact have to move.

Also I don't have clue about the fly/jar riddle or the cannary riddle assuming they both refer to the same principal but I'd love to know the answer.

If I'd read books even 10% of the time in my youth that I wasted watching Bob Barker I wouldn't need to build an RV because I'd be retired and flying myself around the planet in a Citation right now.
 
szicree said:
So what about my bug in a jar? What if the jar was huge and there was an RV4 doing cubans inside?
Click to expand...
The jar is a boundary of a system and the system mass does not change so the weight will be constant whether the plane/bug is flying or not.

Another proposition to this type of question occurred while I was in submarines.

Suppose you have a 500 pound canary in the bow compartment and the canary starts flying but stays in the bow compartment. Will the boat need to be retrimmed to maintain zero bubble?

-mike
 
szicree said:
Okay, it looks like the moving runway has been put to bed, so here's one with no tricky words and the chance of actually trying it at home.

You've got an empty mayonaise jar sitting on your fancy electronic postal scale and all is working fine. You notice a fly taking a rest on the bottom of the jar. The fly now lifts off and hovers inside the jar. Does the scale register a change?

If the scale does read lighter now, where did the weight of the fly go? Would it matter whether the jar had a lid?

If, on the other hand, the scale still indicates the combined weight, how high can the fly go? What if the jar suddenly vanished? Would the reading drop by only the weight of the jar and still register the weight of the hovering fly?

When you feel good about your answer, ask yourself how things would differ if the jar were full of water and a fish.

No obfuscation on this one; just plain old physics -- I think.
Click to expand...

The jar is almost irrelavant in this case. The basic physics of why a plane, or ANYTHING, flys is that it pushes down on something (in this case air). Without getting too technical, the plane stays level because it pushing hard enough on enough air molecules to to keep from falling out of the sky, much like a snowmobile being able to glide across a lake if you get it going fast enough (actually, it doesn't need to "push" on the air, but it does need to give enough "down" momentum to enough air molecules to keep flying...this is a much more accurate explanation).

In order for the fly to fly, it needs to push on the surrounding air. As the fly is taking off, the "weight" of the system will go up as measured by the scale (since the fly is accelerating up). If it's hovering, it'll be the same weight, and if the fly is going down, the weight will be less. The jar being sealed is irrelavant, and if the measuring device is close enough to and much bigger than the thingy flying, the jar is irrelavant also.

The only reason you normally don't notice this effect is that in terms of psi, when a plane (or fly or whatever) passes overhead, the pressure is spread over an enourmous area and the change in psi is pratically unmeasurable. Also, as you get further away from the flying thingy, the momentum you gave to the air molecules very rapidily gets dissipated as heat (the end result of pratically everything). The effect is obvious, however, if you've ever been near a helicopter that's landing.

When it comes to flying, Bernoulli's law has about as much to do with why an airplane flys as transporting propane has to do with cooking a steak on the BBQ. The link is there, but it's very subtle.

Think about this one: Let's say the jar is mounted on an ultra low friction bearing, and the fly starts to fly in a circle inside the jar, much like when you'd make a whirlpool in the pool when you where a kid. What happens to the jar? On argument is that it will go in the opposite direction of the fly because the fly is "pushing" against the jar. Another argument is that it will go in the same direction of the fly because of friction between the air and the jar. A third argument is that the jar will stay put because....well, just because :D
 
John,

I'm liking this explanation, but it sounds like this would mean that as the fly goes higher and higher, the indicated weight would read lower. Is this correct?
 
Win a ten

well, in the first door selection, you have a one in three chance, then you eliminate one door, and you have a 50-50 chance of having picked the correct one in the first place, or if you switch, you still have a 50-50 chance.

eliminating the first door changes the set you are dealing with, and the 1/3 chance no longer applies.

Now on to a bigger question---------if the plane on the runway belt is a "-a" model, when will it flip over?????

Mike
 
szicree said:
John,

I'm liking this explanation, but it sounds like this would mean that as the fly goes higher and higher, the indicated weight would read lower. Is this correct?
Click to expand...

As the fly went higher and higher, part of the downwash from the wings would sorta spread out, and the sides of the jar would become more important. You'd start getting an effect from the friction of the air on the sides of the jar. Ultimately, though, the jar/air/fly system has to push against gravity so everthing will still weigh the same.

If the jar didn't exist, though, then as the fly went higher, the scale would measure less. This is just because the scale is only so big and the downwash spreads out. In this case you would have an "open" system. If you forget about things being sealed or unsealed (what most people think of for open vs. closed systems) and think about closed vs. open as "I can measure EVERYTHING" vs. "There are something I can't measure because they slip by", then it's easy. The scale would read less because if the downwash from the fly is partially pushing on a tree 2 feet away from the scale, then the scale's not measuring everything. It's open :) Make the scale big enough (like, wrap the whole earth in a tremendous pressure sensor) and the height of the fly becomes irrelavant again. You can now measure everything and the system's "closed".

So the bottom line is that the jar doesn't really matter at all, or rather only matters as much as the scale. What makes the jar interesting is that it's rigid and resting solely on the scale so that all of the effects of the fly are transfered directly to the scale (making the system closed). Make the scale big enough (or the fly small enough and close enough) and the jar can go away and you can still consider the system closed for a practical measurement. Alternately, make the jar so big that it doesn't fit on the scale i.e. part of the jar is resting on a table and part is resting on the scale, then you have to start considering again where the fly is hovering because you're allowing at least part of the "push" to transfer to something other than the scale (a table in this case). The jar can still be sealed, but the system is effectively open because you've lost energy through the table and have no way of measuring how much. Put 4 additional scales under the table legs, add them all up, and you're back in business. :)

Holy cow...that was a totally long-winded blowhard explanation to a simple question. LOL.
 
Nope, close but no prize

Mike S said:
well, in the first door selection, you have a one in three chance, then you eliminate one door, and you have a 50-50 chance of having picked the correct one in the first place, or if you switch, you still have a 50-50 chance.

eliminating the first door changes the set you are dealing with, and the 1/3 chance no longer applies.

Now on to a bigger question---------if the plane on the runway belt is a "-a" model, when will it flip over?????

Mike
Click to expand...
Negative: Prove it to yourself:

You agree that if given three doors, picking one results in a statistical 1 in 3 or 33.33% chance. (no debate I hope, keep in mind statistical and totally random, not actual.)

You agree that if you are given two doors, picking one results in a statistical 1 in 2 or 50% chance.


You can flip a coin 100 times and get heads every time, but statistically you will get a 50/50 split given a large sample. Variation in coin weight or centroid is ignored; Roulette wheels are known not to be random, and if you crack the characteristic of that wheel you can improve you chances. However we are talking total random prize location.


Now lets go back to the first example or the case in point, 3 doors.

"eliminating the first door changes the set you are dealing with, and the 1/3 chance no longer applies."


You assumed incorrectly, it does apply. ONCE YOU MAKE YOUR CHOICE, you are stuck with the 1 and 3 chance population. End of story, you have set you chance with a "33%" door and revealing a door does not change that 33.33%.

However after one door is eliminated you get the benefit of the 50/50 chance, but only if switching to a new "50%" door. You could switch and loose, but in the long run you will win more. How can we prove it. Well like flipping a coin you need to take a sample. If the sample is large enough and truly random, the theory proves correct.

Let me take an extreme example, say 100 doors. Pick ONE. Now "Bob Barker" eliminates 98 doors, leaving only two, the one you picked and one remaining closed door, which "Bob" offers to you instead of your original choice. Keep in mind one door, the original choice or the one remaining closed door has the RV-10. What do you think? One has the RV-10 for sure. What was the chance you picked the RV-10 hanger out of the 100 original hanger doors? Oh about 1 in 100. Do you think by doing nothing you go from 1% to 50%? It would be a good move to switch to the other hanger door. It will improve your chance from (1 in 100) to (1 in 2), an improvement of 1% to 99% (corrected was 49%)

In our case same thing, we go from a (1 in 3) 33% chance to a (1 in 2) 50% chance by switching. Statistically you will improve your chances by 16.66% (0.50-0.333) by switching after one door is eliminated.

Its statistically, you may looses every time, switch doors or not, but if you take enough random samples you will find your chances improve. Gamblers loose all the time even if the percentages are in their favor, that's why they call it gambling. BUT GOOD NEWS!! Even though you did not win the RV-10 :( you won a big jar with a bug that can do aerobatics inside the jar. :)

Again intuition is trumped by math. If you set up a truly random number generator (1,2,3) and test the theory; you will win more switching, post door reveal. Cheers George
 
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gmcjetpilot said:
Let me take an extreme example, say 100 doors. Pick ONE. Now "Bob Barker" eliminates 98 doors, leaving only two, the one you picked and one remaining closed door, which "Bob" offers to you instead of your original choice. Keep in mind one door, the original choice or the one remaining closed door has the RV-10. What do you think? One has the RV-10 for sure. What was the chance you picked the RV-10 hanger out of the 100 original hanger doors? Oh about 1 in 100. Do you think by doing nothing you go from 1% to 50%? It would be a good move to switch to the other hanger door. It will improve your chance from (1 in 100) to (1 in 2), an improvement of 1% to 49%
Click to expand...

I have to disagree here. In the case of 100 doors and one RV10, the switching strategy will actually produce a win 99% of the time, whereas the staying put strategy will only win 1% of the time.
 
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Corrected

szicree said:
I have to disagree here. In the case of 100 doors and one RV10, the switching strategy will actually produce a win 99% of the time, whereas the staying put strategy will only win 1% of the time.
Click to expand...
Just coming back to correct the typo. It is like picking one in 100, and than being offered to keep that one door or trade your ONE door pick for all the other 99 hanger doors, 99%. G
 
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Monty Hall three door problem

I've read a lot of explanations of this problem, but they always left something out - action.

Everyone agrees that if you are presented with three options, and you can choose one, then you have a 1/3 chance of winning every time. Pretty simple.

However, this is not a 1 out of 3 problem, this is a 2 out of 3 problem. You know in advance that he will eliminate one of the options. The difference between the two is to switch doors. If you switch doors, you have taken advantage of the 2/3 situation. If you don't, you have stayed with the 1/3 situation. Just like in life, you have to take action to increase your odds. :)
 
GMCJETPILOT is right with the 100 door extrapolation. Here's the actual way to best think of it. Forget all about Bob Barker showing you that one (or more) doors are empty. You already know that.
1. You pick one of three doors.
2. Bob offers to you this: You can trade your current ONE Door choice for BOTH OF the other two doors.
Obviously your 1/3 change has gone to 2/3. Even though you know that 2 of the 3 doors are empty.

If there are 100 doors, you pick a door.
Bob offers to you this: You can trade your ONE DOOR choice for ALL OF the 99 other doors.
Obviously the right thing to do, even though you know 99 total doors are empty.
See?


And oh - of course the plane flies.
Turn on the conveyer belt to 500 mph. Far faster than the max speed of the plane.
So the plane is sitting there, engine just above idle, with the wheels spinning at 500 mph. (Plane has Unobtanium tires)
The slightly above idle engine is counteracting the conveyor belt friction.
Advance the throttle and fly away.
 
And oh - of course the plane flies.
Turn on the conveyer belt to 500 mph. Far faster than the max speed of the plane.
So the plane is sitting there, engine just above idle, with the wheels spinning at 500 mph. (Plane has Unobtanium tires)
The slightly above idle engine is counteracting the conveyor belt friction.
Advance the throttle and fly away.[/QUOTE]



An airplane will fly just fine in a wind tunnel but not so well on a conveyor belt. Airspeed not groundspeed is what makes it fly. Think about the old airshow act with the Cub landing on a truck rolling down the runway. If the truck were sitting stationary on a moving conveyor, though the speedo was showing 50 mph, would the piper be able to land safely on it?

Put a really big fan in front of the airplane and it will lift. Watch out for the wind shear as you exit the fan's influence, though.

Ed
 
So on a very fast conveyor belt, with the wheels spinning, you could just apply the brakes to gain airspeed...IF the conveyor belt is running "backwards," i.e., the surface of the belt is going in the forward direction of the airplane. :)
 
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rv7boy said:
So on a very fast conveyor belt, with the wheels spinning, you could just apply the brakes to gain airspeed...IF the conveyor belt is running "backwards," i.e., the surface of the belt is going in the forward direction of the airplane. :)
Click to expand...

The Navy already does that! ;)
 
Bicyclops said:
And oh - of course the plane flies.
Turn on the conveyer belt to 500 mph. Far faster than the max speed of the plane.
So the plane is sitting there, engine just above idle, with the wheels spinning at 500 mph. (Plane has Unobtanium tires)
The slightly above idle engine is counteracting the conveyor belt friction.
Advance the throttle and fly away.



An airplane will fly just fine in a wind tunnel but not so well on a conveyor belt. Airspeed not groundspeed is what makes it fly. Think about the old airshow act with the Cub landing on a truck rolling down the runway. If the truck were sitting stationary on a moving conveyor, though the speedo was showing 50 mph, would the piper be able to land safely on it?

Put a really big fan in front of the airplane and it will lift. Watch out for the wind shear as you exit the fan's influence, though.

Ed
Click to expand...
Click to expand...

No one said the plane was to remain motion-less. The original question asked "can the plane take off" if the conveyor belt matches the speed of the plane in the opposite direction. Presumably, you can apply full throttle on the plane and take off normally (but with double the wheel velocity of normal)...

This would sort of be like taking off with a 60 knot tail wind, with the plane starting out at 60 knots ground speed...
 
Make it an absolutely ideal model:

Tires never slip on the belt, no matter how fast the two are moving...they're always rolling
Frictionless bearings in the wheels

What you have now is a plane which feels no resistance to its movement in the forward direction from the wheels (which would be really cool if it existed). Thrust > drag + (now nonexistent) friction -> flight.
 
bruceh said:
Done and BUSTED by Mythbusters

https://www.youtube.com/watch?v=YORCk1BN7QY

Plane takes off just fine.
Click to expand...
Because the plane was doing 35mph forward motion relative to the wind and take off speed for that airplane was 25mph. If the belt was doing the same speed as the plane then no forward motion relative to the wind and he would go nowhere fast. How do you think helicopters take off with no forward motion? Aerodynamics 101.

:cool:
 
The Myth busters you tube is a little misleading due to the fact that the plane actually has forward motion to the ground(or air) . All that means is that the wheels were able to go faster than they normally do ( the forward speed of the plane plus the speed of the conveyor belt ) and the prop don't care about wheel speed only velocity thru the air! Its only velocity over the wings that creates lift plus a little help from the prop wash over the wing.
 
I just find it hard to believe this discussion could actually take place on this forum, where one would think most participants were intelligent.

If the airplanes wheels provided the forward movement there would be something to discuss......
 
Also, the conveyor belt itself might generate enough air moving backward to generate a wee bit o' lift beneath the wings, actually helping the plane get airborne.
 
Veetail88 said:
I just find it hard to believe this discussion could actually take place on this forum, where one would think most participants were intelligent.
Click to expand...

One would think so, but, have you seen what some folks use for primer? Don't get me started ...
 
RV7A Flyer said:
Make it an absolutely ideal model:

Tires never slip on the belt, no matter how fast the two are moving...they're always rolling
Frictionless bearings in the wheels

What you have now is a plane which feels no resistance to its movement in the forward direction from the wheels (which would be really cool if it existed). Thrust > drag + (now nonexistent) friction -> flight.
Click to expand...

Not quite true. As long as the belt is increasing in speed, it applies a force to make the wheel speed increase. The airplane feels the same (aft direction) force; however, its acceleration will be much less than the wheels', by the ratio of the wheels' mass to the airplane's mass. Because this ratio is so small, the force applied to the wheels and airframe is relatively small, and easily overcome by the engine/propeller. e.g., the plane takes off, although it needs a slightly longer time compared to a runway.
 
Run on a treadmill. You will get hot because of a distinct lack of wind. No relative wind, no takeoff.
That's what I think.
DaveH
 
I'm surprised that no-one has mentioned the effect of the boundary-layer on the (conveyor-belt) runway : Given that air is viscous (..it's why wings lift, after all), then if the (conveyor-belt) runway was long enough, the movement of the conveyor belt, itself, would result in a relative airflow over the wing!
...so if the (conveyor-belt) runway was long enough, and going substantially faster than Vs, then not only will the 'plane fly... it will fly with the engine off !! (..well... briefly, anyway.... before drag takes over, and it falls (backwards !) back to earth !!):)
 
Everyone uses the man on a treadmill anology....so imagine you are running on a treadmill and can't move forward right? Now grab the bar in front of you and pull yourself forward. Just like the prop would pull you forward.
 
I can't believe I'm getting into this...

Picture a hovercraft on a treadmill. The treadmill could be going 500mph backwards, and the hovercraft could remain stationary with minimal thrust to overcome any incidental drag between the skirt and the treadmill. This is because the thrust of the hovercraft is decoupled from the ground.

This is the same as an airplane. There is some slight friction in the bearings, but it's negligible. So you could picture a plane with the wheels replaced by 3 small hovercraft.

It wouldn't matter how fast the treadmill would go, just minimal thrust would overcome the drag from the treadmill and the plane would move forward, eventually hitting its minimum speed for flight and taking off.

THE PLANE WOULD FLY.
 
Shimoda said:
I'm surprised that no-one has mentioned the effect of the boundary-layer on the (conveyor-belt) runway : Given that air is viscous (..it's why wings lift, after all), then if the (conveyor-belt) runway was long enough, the movement of the conveyor belt, itself, would result in a relative airflow over the wing!
...so if the (conveyor-belt) runway was long enough, and going substantially faster than Vs, then not only will the 'plane fly... it will fly with the engine off !! (..well... briefly, anyway.... before drag takes over, and it falls (backwards !) back to earth !!):)
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Be surprised no more. Look at post No. 78! :)
 
I just have a FP, but if I upgrade to CS, oh my baby would leap off the belt. And ATC would say "wow, are you a Citation?" I would just answer with "No, just use 45 lbs in the tires." Of course I faced the shop heads all to the right and aft plus used rattle can primer. YMMV. :D
 
I think it's time for a lot of us, me included, to get back in the shop and do some productive things. If our RV is already flying, time would be better spent practicing takeoffs and landings in the pattern. This discussion is a "bunch of entropy." Lots of energy expended, but no net work. :eek:
 
How about somebody just get an RC plane, treadmill, and camera, and put this to bed once and for all, ala Mythbusters. Of course it will fly.

Chris
 
YellowJacket RV9 said:
How about somebody just get an RC plane, treadmill, and camera, and put this to bed once and for all, ala Mythbusters. Of course it will fly.

Chris
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Before doing it with the ultralight they did exactly that on the same episode of Mythbusters.

Spoiler Alert: the model plane flew too.
 
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