will it take off?

[HTWT]

I want to know how to find the three points of intersection of the 2 equations

1. Y= 2^x
2. Y= X^2

Two solutions are easy by inspection but what about the negative value for X.

 

[patrick_g]

Tom, My A/C all use lubricants in the wheel bearings. Besides friction is a minor contributor compared to the rotating mass energy storage that alows the MCB to accelerate at a constant rate to produce the force equivalent to the propulsion systems thrust. Folks just kept on focusing on friction, due to familiarity I suppose.

If you are in for a penny you are in for a pound. If the bird in flight and at rest show the same weight on the scales...

Lets make it a very tall hour glass with very course sand. Will a sensitive scale show the big grains hitting the bottom? Either group the must fly crowd or the heredical no fly bunch would realize that the hour glass would weigh the same after the sands of time had shifted all to the bottom no matter if the red end or the black one was on bottom when the sand stopped. The question is does the weight change while the grains are trickling down? At the start with a freshly inverted hour glass does it weigh more, less, or the same. What becomes of the potential energy which the elevated sand had. Does the sensitive scale show any variatioin as grains free fall and then come to rest?

Can a man in an enclosed elevator car tell if the elevator is accelerating upward as opposed to a large mass appearing under the car?

Can there be a doubter doubting that there could be a doubter doubting?

Can two mediteranean larks working in tandom carry a cocoanut? (appologies to Monty Python)

Time and about everything else looks smoothly continuous when viewed from the right distance. Conversely, just about everything including time starts looking as granular as a low res computer graphic on sufficiently close inspection. "Great fleas have little fleas upon their backs to bite 'em,: And little fleas have lesser fleas, and so ad infinitum.

Pat

 

[MossRoad]

So the plane powers up and moves forward. The MCB powers up and moves backwards. At this point the plane isn't standing still, is it? ...

Yes.

At the very moment the plane tries to move forward .00000001mm, the conveyor moves backwards .00000001mm, right? Now instead of .00000001mm run this decimal place out to a bazillion places to the right. Every atom width movement forward would be countered by an atom width movement backwards. If the conveyor could truly match the forward speed of the plane in the opposite direction, you would hear the engine run up but would not see the plane or the conveyor move, just as if the brakes were on.

 

[Tom_Veatch]

I want to know how to find the three points of intersection of the 2 equations

1. Y= 2^x
2. Y= X^2

Two solutions are easy by inspection but what about the negative value for X.

Since you are talking about points of intersection, I assume we are working in 2-space with the 3rd coordinate held to a constant value. IOW, an X-Y plane. I further assume we are working in the real instead of the complex number system.

That being the case, I only see one obvious solution, the point (2,4).

I question there being three points of intersection under the assumption stated above.

Eq. 1 is an exponential curve asymptotic to the lines X= +inf. and Y= 0. Equation 2 is a parabola, concave upward with a vertex at (0,0), asymptotic to the lines X = +inf and X= -inf.

There are only two points short of the asymptotes at which these curves can intersect, the obvious point (2,4) in the 1st quadrant and, in the 4th quadrant:

X = (2/ln(2)) * ln(abs(X)) where X < 0

This expression for the X-value of an intersection point can be obtained by simple algebraic manipulation but is a transcendental equation which cannot be solved by simple algebraic methods. An approximate numerical value can be obtained by straightforward numeric computational techniques but I suspect that an exact solution can only be expressed as a slowly converging infinite series similar to the case for other transcendental values and functions.

I suppose, since both curves are asymptotic in the 1st quadrant to the line X= +inf, one could claim (+inf, +inf) as a third intersection point. But that claim would be made in the same sense that parallel lines are claimed to intersect at infinity. This implies that "infinity" is an actual rather than a conceptual value, so I argue that there are only two actual intersection points on the real plane.

If you admit complex solutions, X = a + bj where j = sqrt(-1), then perhaps one, or even many, complex roots exist, but the half-life of my expertise with complex numbers has expired many times over so I'll pass on that one.

Now, to get on topic for this forum. I have a 22HP CUT. What size moldboard plow would be appropriate for that size tractor and how long do you think it might take to plow the 1st quadrant of the X-Y plane?

 

[Tom_Veatch]

Tom, My A/C all use lubricants in the wheel bearings. Besides friction is a minor contributor compared to the rotating mass energy storage that alows the MCB to accelerate at a constant rate to produce the force equivalent to the propulsion systems thrust. Folks just kept on focusing on friction, due to familiarity I suppose.

Yeah, I was just being a little too nit-picking. Besides, if there were no lubricants in the bearing, they would soon be lubricated by molten metal anyway.

Lets make it a very tall hour glass with very course sand. Will a sensitive scale show the big grains hitting the bottom? Either group the must fly crowd or the heredical no fly bunch would realize that the hour glass would weigh the same after the sands of time had shifted all to the bottom no matter if the red end or the black one was on bottom when the sand stopped. The question is does the weight change while the grains are trickling down? At the start with a freshly inverted hour glass does it weigh more, less, or the same. What becomes of the potential energy which the elevated sand had. Does the sensitive scale show any variatioin as grains free fall and then come to rest?

Let's simplify the configuration a little. Assume there is only one grain of sand in the hourglass. If we can conclude something about that configuration, the conclusion extends to the case of many grains. Further assume that a perfect vacuum exists within the hourglass so we can neglect any aerodynamic influences inside the hourglass. Agreed?

At the start of the experiment, the grain, which weighs G pounds, is somehow restrained in the upper chamber. Perhaps the opening is slightly too small to allow passage. The hourglass, containing the grain, is resting on a sensitive, instantaneously registering scale. The hourglass alone, without the grain, has been previously determined to weigh H pounds. Thus the total system, hourglass plus restrained grain weighs H + G pounds.

The experiment begins by, say heating the hourglass so that the opening expands enough to allow the grain to pass through and fall to the bottom of the lower chamber. As soon as the grain begins to fall, the weight of the grain is no longer being supported by the hourglass and measured on the scale. Therefore, during the period of free fall the scale will register only the weight of the hourglass, H pounds.

The potential energy, PE = Gh, of the grain in the upper chamber is converted to kinetic energy, KE = 0.5 (G/g) * V^2. The PE decreases as the height (h) of the grain above the bottom of the hourglass decreases. The KE increases as the square of the speed of the grain's fall increases. At any point during the fall, the sum, PE + KE, will be constant and equal to the initial potential energy.

When the grain impacts the bottom of the hourglass, the weight of the grain will again be supported by the hourglass and the scale will again register H + G. So the scale will register H+G before the fall, H during the fall, and H + G once the grain comes to rest at the bottom of the fall.

So what happened to the kinetic energy of the grain that it had accumulated at the instant prior to impact. Any number of things. But if we assume neither the hourglass or the grain was physically deformed that energy will heat both grain and hourglass so that both are very slightly warmer than they were at the instant before the fall. If we allow deformation of either grain or hourglass bottom, some of the energy will be used by the work to accomplish that deformation and the amount of heating will be slightly less.

I believe that touches all the points in your question except to say that what happens to one grain happens to all grains in the falling stream. So, once the stream of sand starts falling, the scale will detect only the weight of the hourglass plus only that sand that is not falling.

Can a man in an enclosed elevator car tell if the elevator is accelerating upward as opposed to a large mass appearing under the car?

According to General Relativity an observer cannot distinguish between the two conditions in an inertial reference frame. However, the surface of the earth is a rotating sphere and is not a true inertial reference system. Close enough to be considered one in cases of airplanes flying off conveyor belts, but not in the strictest sense. That fact has to be taken into account for every launch from Vandenburg or Kennedy, or for any launch from the Earth's surface except an exactly vertical launch from either of the Earth's poles.

Therefore, the elevator moving upward perpendicular to the Earth's surface experiences a very slight Coriolis acceleration in a Westerly direction. That acceleration would be absent if the increase in apparent weight were due to the sudden appearance of a large concentrated mass below the elevator. Therefore, with sufficiently sensitive instrumentation, the passenger could determine the difference between the two cases.

Can there be a doubter doubting that there could be a doubter doubting?

Can two mediteranean larks working in tandom carry a cocoanut? (appologies to Monty Python)

Wow! This is beginning to remind me of the my Master's Degree Orals. I'm going to pass on those two.

Time and about everything else looks smoothly continuous when viewed from the right distance. Conversely, just about everything including time starts looking as granular as a low res computer graphic on sufficiently close inspection. "Great fleas have little fleas upon their backs to bite 'em,: And little fleas have lesser fleas, and so ad infinitum.

Pat

From my readings several years ago, I knew there were some theoretical physicists working on one or more of the GUTs who were hypothesizing a granularity or quantum nature of time. I had not heard or read that those hypotheses were in general acceptance or had reached the status of theory. I wonder how it could be verified experimentally.

Well, to get back on topic, will allowing my brush-hog to free fall from a full elevation on my 3pt hitch relieve some of the load on my rear tires? Or will the Coriolis forces during the fall bend the lift arms if the tractor happens to be facing North or South when I do it?

 

[Tom_Veatch]

Yes.

At the very moment the plane tries to move forward .00000001mm, the conveyor moves backwards .00000001mm, right? Now instead of .00000001mm run this decimal place out to a bazillion places to the right. Every atom width movement forward would be countered by an atom width movement backwards.
...

What is your explanation for the method by which the movement of the conveyor is transmitted to the airframe?

 

[MossRoad]

What is your explanation for the method by which the movement of the conveyor is transmitted to the airframe?

The entire weight of the aiplane is on the little patch of rubber tires that is contacting the conveyor. None of it will ever get transferred to the wings because it will never move forward one spec.

Put a brick on a treadmill and slowly turn on the treadmill. The brick moves backwards. Now put a roller skate on the treadmill and slowly turn it on. The roller skate moves backwards. Now try and push the roller skate forward. Under normal circumstances, you could push it forward. But if at the very time you attempted to push it forward(much like a propeller screwing into the air), for each tiny, tiny increase in movement you tried to move it forward the conveyor increases it's speed the same amount in the opposite direction, instantly(since the problem said it could match the speed). You would never be able to break the friction of the wheels from the conveyor to get them to start rolling. You would see no visible movement in the conveyor or the wheels. It would be like trying to push two extremely powerful magnets together when the poles were not opposite. You know that feeling? If I can just push them harder, I can overcome the force. But it doesn't work. The harder you push, the harder the conveyor pushes back without moving. The entire will-fly team will drop their jaws, hang their heads in disbelief, tuck their helmets under their arms and walk off the field in disbelief. We are good winners, so we will buy them a root-beer.

 

[Egon]

This aircraft can not fly at the present time. It's been on the darned belt so long that now it's in the shop having the engine rebuilt!

Whatever happens in sealed truck box full of flying sparrows if the truck comes to an full emergency stop??

 

[MossRoad]

This aircraft can not fly at the present time. It's been on the darned belt so long that now it's in the shop having the engine rebuilt!

Whatever happens in sealed truck box full of flying sparrows if the truck comes to an full emergency stop??

I agree. The paperwork on this thing is going to be a nightmare.

 

[patrick_g]

Tom, You assume the elevator was in the Northern hemisphere, a fact not in evidence, but well done!

My FIL was, for a time, a rocket scientist and one of his papers was "Precision Azimuth Allignment of Ballistic Missiles" He worked directly for Werner VB and was liaison to a lot of the "imported German talent." He was a fun guy to banter with about stuff of physics (THE FATHER SCIENCE.) He didn't let you get away with sloppy thinking.

Now MOSS... Making the "will fly" assumtion on the speed of the MCB where the speed of the MCB exactly equals the speed of the plane... Don't get into Zeno's paradox subdividing speeds into infintessimals. The concepts of calculus will handle those. as well as a very simple arrangement described just a tad down the page.

The plane will start moving down the conveyor pretty normally except for the fact that the wheels will always be spinning at twice the airplanes speed, i.e. as the plane accelerates through any given speed however small or large prior to lifting off. The MCB will MATCH the planes speed. The plane will have slightly less takeoff performance since it will have to provide the extra energy to account for the fact that the wheels will accelerate at twice the acceleration of the plane.

If the bearings don't fail, a reasonable assumption, and the tires don't fly apart at twice the normal rotational speed, I'm not sure about the safety margin on aircraft tires, the plane will take off almost as nicely as if the MCB were switched off. For the purposes of this exercise it would be fair to assume the bearings and tires survive as their performance is not the central issue of the mental exercise.

The plane takes off at x MPH while the MCB is producing x MPH and the wheels are rotating at a rate to make 2xMPH.

If you assume low or no friction in the design of the MCB and a low or no weight in its moving parts you could attach a rope with a hook on it to the airplane and through a turning block and drum let the plane's forward motion supply the motion of the MCB. Then there would be no question about the MCB dectecting and responding to infintessimal motions and such distractions.

Every inch forward the plane moves the rope and pulley pulls the MCB an inch in the oppoosite direction at the same speed the plane is moving however small or large (since our rope is a low stretch polyester.)

For safety reasons the rope is attached to the drum like a non-recoil starter rope is attached to the flywheel of an engine so after the end of the rope is reached it disconnects from the drum. after liftoff the pilot releases the mechanical connection of the hook on the plane (like a glider's tow plane.)

Anyway this rope and pulley exactly fits the "WILL FLY" argument while it simplifies the MCB to just CB (not much magic in a rope and pulley.)

The somewhat more fun and considerably more interesting (but a bit less defensible in its enabling assumptions) is the "NO-FLY" contention. If for whatever reason you allow the MCB to run at whatever speed is required to prevent the plane's forward motion and there are no material failures the pllane will run out of gas sitting stationery
on the MCB unless the wind generated by the MCB will lift the plane. If it does lift off it may stall leaving the region where the wind is being generated if the MCB is not sufficiently long.

I assert that even in theory there is no way to attach the rope of the above example to a mechanical mechanism that will cause the MCB to experience an accelerated motion sufficient to cause the plane to remain statioinery since if the plane is stationery there is no motion of the rope.

Pat