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Fay Slag
3rd Dec 2004, 01:32
Dear fellow PPruners,

Myself and my colleagues have been wondering about the following conundrum, and your help would be greatly appreciated...

Picture this - A Bell 206 in a huge cardboard box. The box, with the aircraft inside it is sitting on a massive pair of bathroom scales which show the weight of the helicopter plus that of the box. Now, the pilot is already inside the aircraft inside the box and lifts into a hover.

What do the bathroom scales show now???

We really want to know!

After several discussions with various people, we cannot come to a conclusion. Some people think the scales would show the same weight, others think they would show less.

Regards,

Fay-Slag

Flingwing207
3rd Dec 2004, 01:46
Assuming the box was airtight, then the weight of the box would include, the box, the Jet Ranger, and the air. It would still contain all of the above whether the JR was on the skids or in the air, ergo the weight would not change.

Fay Slag
3rd Dec 2004, 02:03
but why will the scales read the same? - is it because the downwash equals the weight of the helicopter...or is it because the helicopter sucks the roof of the box down equivalent to the weight of the helicopter.....


or is it a bit of both?

many thanks,
Fay Slag

Flingwing207
3rd Dec 2004, 05:16
Helicopter on skids = weight of helicopter supported by bottom of box.

Helicopter in air = Helicopter supported by moving an equal mass of air downward.

Scale doesn't know or care how the helicopter is being supported or if air in box is being moved.

Hope I'm making sense to someone but me!:8

Gomer Pylot
3rd Dec 2004, 05:28
Why would the scales change? It's the same question as the trucker with the load of canaries. When he pulls onto the scales, he's overweight whether the canaries are flying or not.

csmorris
3rd Dec 2004, 07:37
Warning: rambling answer from an insomniac physics major contained within.

If the box is sealed, the helicopter and box form a closed system. Physically speaking, the question becomes not what the system weighs, but what the scale will measure.

If the helicopter is resting on the bottom of the box or hovering, the center of mass of the box + helicopter system is stationary. Thus, there is no net force and the force on the box from the scale is equal to the weight of the system. If the helicopter is descending, the center of mass is as well. The net force is downward (the gravitational force is greater than the upward force from the scale), and the scale will read low. The opposite is true if the helicopter is ascending.

The trucker with the canaries, however, will find his weight changing as they take off and land (unless his truck is sealed, and then the canaries won't be flying very long!).

I'll stop there because I should be studying for my thermo and electro exams instead of reading pprune :\

Dave_Jackson
3rd Dec 2004, 08:56
Perhaps people should be 'thinking outside the box'. :D

boomerangben
3rd Dec 2004, 10:47
What on earth has cog got to do with it? If CoG affects the reading on the scales, does that mean that the scales would read different if the pilot were sat down or stood up?

In a steady hover, the air is accelerated by the blades and the mass of this accelerated air causes a reaction force equal to the mass of the helicopter. The bottom of the box decelerates this air, causing a reaction which is measured on the scales. I agree that the reading on the scales will change if the aircraft climbs or descends within the box, since the aircraft will be accelerating, but once in a steady climb, the scales will return to a steady reading. In other words, the scales would read the mass of the helicopter plus box, unless the aircraft was starting/ stopping a climb or decent. What is more, the hover height will not affect the scales (ie independant of ground affect).

I wish Santa would bring me a Jet Banger in a box, or even better a twin squirrel....... he can keep the scales!

csmorris
3rd Dec 2004, 11:17
The center of mass of a closed system can be considered a point where all mass is concentrated. Any external forces that act on the system (like the scale pushing up from the earth in opposition to the weight of the box) effectively act upon the center of mass. The position of the CoM within the system (which is not necessarily the CoG, though on Earth I suppose it would be) is irrelevant in this case. What counts is the vertical component of its velocity :)

The rotor downwash has no measurable effect on the outside of the box, since energy cannot enter or leave a closed system.

Can't say I'd mind a heli for Christmas either, though :ok:

Flingwing207
3rd Dec 2004, 12:32
C of G could change - if you moved the helicopter from one end of the box to the other, but the total weight on both scales wouldn't.

If you entered autorotation or did a low-G pushover, for a moment the box would show less weight.

If you flew long enough, the box would show less total weight, as some of the mass of your fuel would have been converted into energy (heat and noise). Probably wouldn't be able to measure it on a bathroom scale though.

What if the box were on the back of a truck and you drove it through a TFR while hovering the helicopter inside the box? Is it a bust? :E

SASless
3rd Dec 2004, 13:01
As to the truck driver....if he let some air out of the tires....would that not reduce the air pressure being exerted upon the exterior of the tire thereby reducing the weight on the scale....in that the scale measures the pressure exerted against the platform?:confused:

Genghis the Engineer
3rd Dec 2004, 13:02
Just for fun, why not take a radio controlled helicopter, put it in a large box on some scales (or probably easier to hang from a spring balance) and try.

G

mattsmith678
3rd Dec 2004, 13:38
Hey all,
I'm new to this forum, i'm an air cadet and fly with a Volunteer Gliding School at RAF Halton. I wondered if any of you know of similar problem solving questions like this one which i could give to the cadets at my squadron so as to get them thinking! Any help would be appreciated!
Thanks Matt

SASless
3rd Dec 2004, 14:04
Find your offshore oil rig....

Bell 296B-III helicopter

Leg length: 90nm

Indicated Airspeed: 90 knots

Track: 180 degrees magnetic

Variation: 5 degrees west

Deviation: +5 Degrees for 180 degrees heading

Wind: 270/10

Visibility: 5nm


Remarks....No GPS, No NDB on rig, only fuel is on the rig. Mag compass and clock for navigation.

Weather is forecast to remain VFR for the duration, no frontal passage forecast. Sky clear. Departure point is the shoreline...rig is in open water.

How would you plan the flight???

Banjo
3rd Dec 2004, 17:01
Not very PC but all this talk of boxes reminded me of a question.

If with a male crew an aircraft has a cockpit would it with an all female crew be called a box room?

Red Wine
3rd Dec 2004, 17:42
Hey Sassy!!

Is that one of those new 3 engined Bells?

SASless
3rd Dec 2004, 17:52
OOPS!

One cup of coffee shy of enough this morning....sorry! I meant the old proverbial Jetranger Short. Freudian slip I fear!


Per the request....all female crew sits in the "Box Office".

csmorris
3rd Dec 2004, 20:19
Matt,

Are you looking for aviation specific stuff, or just physics riddles in general? The question in this thread is typically posed in physics classes as the "bird in a bottle experiment," so you could try googling that for a starting point.

Here's another one that gets people (Robert Oppenheimer famously embarassed himself with this one). Imagine you're in a small boat in a swimming pool. With you in the boat is a large boulder. You toss the boulder out of the boat and into the pool. What happens to the water level in the pool? :8

Fay Slag
3rd Dec 2004, 20:47
Ok, how about this then..............


A hot air baloon, in the same kind of box (somewhat taller though). The air baloon is FLOATING inside the box with neither an ascent or descent.

What do the scales weigh, and why?

Cheers,

Fay

Barotrauma
3rd Dec 2004, 21:10
don't know what the scales will show but I do know that the value of the box will go up when the balloon rises....
due to inflation of course!

I thank you!

ShyTorque
3rd Dec 2004, 21:22
The box will weigh the same as if it were full of air. The balloon displaces a similar mass of air as itself and has no effect on weight.

Fay Slag
3rd Dec 2004, 21:23
Do some work Barotrauma!

Fay

boomerangben
3rd Dec 2004, 22:21
Just worked out the boat and rock one. Those in small boats should not throw rocks - the CoG will go up and the boat will capsize :E

Flingwing207
4th Dec 2004, 03:55
if a boodle beetle battles on a bottle with a noodle

Gomer Pylot
4th Dec 2004, 07:47
How would you plan the flight??? I wouldn't. It's pointless. The dispatcher will call with a change of destination anyway, so why bother to plan. I flew on a contract for a drilling company for about 2 years, and when I started I would call up each day and find out where I was going so I could plan fuel stops, etc. I never, not even once, flew what I had planned. The only thing to do is load on all the fuel you can, (it will fly over gross, but not without fuel) and head south, and wait for the radio call. :mad:

Thomas coupling
4th Dec 2004, 07:57
Are we talking 'MASS' or 'WEIGHT'??

The mass of the total body will never change whether it is in flight or in space, but the weight will??

In this case, the 'weight' (because of 'G') of the helo displaces that same amount of air downward and exerts a force on the scales. Hence no 'weight' change in readout.

If the air could escape thru cracks in the box, then the 'weight' would indicate less.

The mass never changes.

TeeS
4th Dec 2004, 09:36
Well this one got me thinking, in the first few minutes I managed to convince myself that the scales would measure the weight of the box only and then I convinced myself that they would measure both the box and helicopter, then I wasn’t sure Ahhhhhh!!!

However several hours later, I think this explanation answers it for me: -

As TC stated, the downwash will strike the bottom of the box and exert a force on it. However, that is an oversimplification because the air will now dissipate sideways, strike the side of the box deflect upwards hit the top of the box and then get sucked back down through the rotor system. At this point there will be a reduction in pressure on the ‘inside’ of the top of the box and an increase in pressure on the base of the box, hence there will be a greater force exerted by the air on the bottom of the box than on the top of the box. The resultant of these two forces will be a force downward (onto the scales) equal to the weight of the helicopter.

The scales will show the weight of the helicopter and the box.

That’s my theory for what it is worth.

Cheers

TeeS

Helisurfer
5th Dec 2004, 04:29
When the helicopter is not running and sitting on the floor of the box, all the force, or potential energy, is on the floor of the box and being read by the scales.

When the aircraft is hovered the air is being forced down to the floor then out to the walls of the box and up to the roof to be recirculated. It would make sense to me to say that now some of this force, or energy, is being aplied sideways to the walls of the box and therefore would not bear pressure down on the scales thus the scales would read lighter in the hover.

Now though, what effect, if any, would changing the size of the box so that now non of the sideways air movement ever reaches the sides of the box?

Now my head hurts.

Bomber ARIS
5th Dec 2004, 18:25
............and what if the box were tall enough to allow the helicopter to hover OGE :E

ShyTorque
5th Dec 2004, 19:37
Or transitions into forward flight?

I think the scales won't weigh the same figure in either case as some of the rotor thrust will cause vortices to form, which won't bear down on the scales.

Anyone got a big box? Now don't be rude.....

Flingwing207
5th Dec 2004, 22:36
You guys crack me up - you really do.

I'm going to make a statement: the box is airtight.

So it doesn't matter whether the box has a helicopter, a balloon or a duck in it, nor does it matter whether whatever is in the box is being supported by skids, rotors, heated air or flapping wings.

Whatever mass you put into the box before you sealed it will remain there. The mass of the air will remain unchanged whether it is still or being swirled or heated. Since weight is a function of gravity acting on mass, if the mass stays the same, the weight stays the same.

Actually, I crack myself up - I can't believe I'm actually getting concerned about this! Time for a nice hot cuppa...

SilsoeSid
5th Dec 2004, 22:54
Mmmm;
The box is airtight.
The box is cardboard.

From the start point of the experiment, a vacuum will be created due to the engine and pilot consuming the fixed volume of air.

This vacuum will increase in strength ultimately causing the box to collapse in on itself, resulting in the weight of the wreckage + box + pilot being the same as at the start of the experiment.

However, assuming the integrity of said box has not been compromised, ie is still airtight, the resultant carbon dioxide that was produced, being heavier than the air it has replaced, would cause the resultant wreckage on the scales to weigh more than at the start.


Just a thought.

Sir HC
5th Dec 2004, 23:07
Sid, You should know better than that, certainly, a vacuum will be created due to the enignes intake however this will be offset due to the pressure of the exhaust gases. Like Flingwing said, the mass will remain unchanged. The box will not collapse. The first law of thermodynamics, Energy cannot be created nor destroyed. This does not answer the original question however it may point you in the right direction.

Gaseous
5th Dec 2004, 23:11
Sounds like a recipe for vortex ring and carbon monoxide poisoning to me.
:}

You actually dont need the box for this scenario.

Does the world weigh the same when my Enstrom takes off????

SilsoeSid
5th Dec 2004, 23:50
Sir HC,

The reason my vacuum is caused, is by the burning of the air by the engines and by the pilots lungs, (albeit minutely compared to the engines consumption).

Nothing to do with the complicated dynamic properties associated with a jet engine, but simply the conversion of fossil fuels to carbon dioxide.

Much the same as the old candle in an up turned jam jar.
As the flame burns the air, the candle gradually goes out, as there is no further air to burn.
The smoke produced follows the draught created by the carbon dioxide falling to the lowest point.
Should the jar be sat upon a dish with water in, the effects of the vacuum can be seen by the rise in water level within the jar.

OK, so in reality the engines will flame out and the pilot will die before any reasonable vacuum is produced, but the conversion of the air to a heavier than air gas will cause the scales to show, if sensitive enough, an increase in the weight of the total contents of the box.


edited to avoid the use of atomic weights

csmorris
6th Dec 2004, 00:47
Einstein would like a word with you :E

Neither mass nor energy can be created or destroyed. They can be exchanged for one another, but this typically happens (on a macroscopic scale) inside a nuclear bomb and not a helicopter. The helicopter and human expel the same amount of gas over time that they ingest. It might be a different gas, but weight is a function of mass and not volume. The energy comes from chemical reactions and not breakdowns on the atomic level :)

The scale only cares about the direct force that it has to apply to keep the box from accelerating downward. In this case, the force depends on the weight of the box and whether or not the helicopter (and thus the system, in physical terms) is accelerating up or down.

A more abstract explanation of the experiment can be found here (http://physics.about.com/cs/gravity/f/birdinabottle.htm).

NickLappos
6th Dec 2004, 01:43
csmorris,

In your experiment with a rock in a boat then thrown into the pool, the water level in the pool will go down. This is because the boat had displaced the water's volume equivalent to the weight of the rock, which is more volume than the dense rock will displace when tossed out of the boat. When the rock is tossed into the water, the boat floated up to stop displacing the water weight of the rock, and the rock displaced less water when it sank to the bottom. In fact the pool water went down by an amount equal to the difference between the specific gravity of the rock and the SG of the water times the weight of the rock.


Your points about the box are right on! Guys, regarding the box experiment, remember that the helo flies by exchanging momentum with the earth system. Inside the box, the rotor allows the helo to fly, but the entire sealed box/helo system is constant weight, forever. Even as the engine and pilot consume the oxygen, it is mearly recombined in the H2O and CO2 expelled in the exhaust. The weight of the entire system stays constant (conservation of energy and mass.)

A great science show 20 years ago performed this exact experiment. "The Ring of Truth" with Philip Morrison weighed the box and then ran little cars, rockets and such inside the box, then re-weighed it, to the gram. Here is a reference to the video tape, which some libraries might have (there are 13 copies in Connecticut, for example):
Ring of Truth, Morrison, Program 2: Change
Summary: Using everyday examples that are dramatic, exciting & entertaining, Philip Morrison devotes an entire program to this subject, demonstrating in simple experiments that it is true that matter is equivalent to energy. 060 Min. VIDEO 1987 Subject: PHYSICS, ASTRONOMY Series: RING OF TRUTH, THE

chopperdude
6th Dec 2004, 03:18
Hey all, lets keep it simple. Please remember the question, and I quote:

"Picture this - A Bell 206 in a huge cardboard box. The box, with the aircraft inside it is sitting on a massive pair of bathroom scales which show the weight of the helicopter plus that of the box. Now, the pilot is already inside the aircraft inside the box and lifts into a hover.

What do the bathroom scales show now??? "

I am not going back on the threads, but someone did mention earlier about flight planning off shore. I agree with you. Nothing is confirmed until you land back home.

Fay Slag: You have brought up a topic, that I have heard of, and been asked many times before. On one flight off shore we had a fly in the flight deck, and started an hours discussion on exactly the same topic. Everything has mass, no matter dimension. What you are asking in simple terms is weight. Now, in a sealed box, for a simple explanation, should be the same as mass. Mass has many factors affecting, weight fewer. If anybody can remember when we start our flight training, Newtons law: For every action, there will be an equal and opposite reaction. The aircrafts weight is displacing air to counter its weight to overcome gravity. Mass is involved, but not considered in this discussion. In theory, the weight indicated on the scale should remain the same, right??? I still don't know, but think about it. We don't want to know about exhaust gases, and vortex ring, and, and ,and... Simply, lift, opposes weight, the same as thrust is opposite to drag. In a 'sealed' box, with no other influences, nothing should change.

I would like to see how far this is gonna go. Keep it up all of you, some good thinking going on here. Once we answer this one, we can solve anything, maybe time travel.... This is not an easy one.

NickLappos
6th Dec 2004, 04:38
chopperdude,

This is dirt simple and about 20 people have answered it. The scale does not change. See my post just above your last.

Try to imagine this one as a swimming pool:

You are sitting on the bottom of a swimming pool wearing scuba gear, slightly negitively bouyant so you must paddle to stay off the bottom. The pool is mounted on a big scale. The water, pool and you are all registered on the scale. Now you swin lazily off the bottom and paddle around for a while. While off the bottom, does the scale show less total weight?

This is exactly the same question as the helicopter question, but the answer is more obvious. The scale does not change in either case.

SilsoeSid
6th Dec 2004, 08:40
"Even as the engine and pilot consume the oxygen, it is mearly recombined in the H2O and CO2 expelled in the exhaust."

But surely the newly produced elements of Hydrogen and Carbon add their weight to the volume of the box.

Atomic weights of ;
Hydrogen = 1.00794
Carbon = 12.011
Oxygen = 15.9994

Let's say,
A 'unit' of oxygen = 16
A 'unit' of H20 = 1+1+16=18
A 'unit' of CO2 = 12+16+16=48

Total exhausted gases = 18+48=66

Doesn't the weight of exhausted gases therefore exceed original gases weight?

:confused:

Gomer Pylot
6th Dec 2004, 09:42
Sid, you need a course in basic chemistry. Carbon and oxygen aren't produced, they're merely combined in different configurations. It's impossible to manufacture atoms of any element from nothing. Elements cannot just suddenly appear, they're all there all the time in a different form. The fuel is converted from a liquid to a gas, and the carbon and hydrogen in it are recombined with the oxygen in the air. The only thing that changes is the temperature, which changes the pressure, which is the cause of the water rising in your previous example.

mattsmith678
6th Dec 2004, 10:02
csmorris.

Absolutely anything aviation related is helpful! Anything that would get them thinking! Thanks everyone for your input!

Matt

SilsoeSid
6th Dec 2004, 10:54
"The only thing that changes is the temperature."

Thank you GP, at last. Like pulling teeth! :rolleyes:
So the temperature has risen and as we all know hot air rises therefore the scales will indicate less. :ok:

Got there eventually.


psst, how about this;

Does the TRT in the hover, actually EQUAL the a/c weight, or does it have to be ever so slightly greater in order to compensate for gravitational acceleration, due to momentary lapses in constant lift throughout the rotor disc area??

NickLappos
6th Dec 2004, 11:39
Silsoesid,

The scale will read higher when the aircraft is lifting up, since the aircraft is accelerating upward briefly (just as the scale bounces higher when you step on it) but when you are in a steady hover, the weight reading is the same as when the helo was sitting on the ground.

The high temp gasses rsing is a slight bouyant effect but does not change this discussion (it is an interesting side discussion, though!)

6th Dec 2004, 12:31
How is the helicopter actually going to hover when it is in the middle of recirculation - as fast as it displaces air downwards (accelerating it) the air is deflected round the box and enters the disc from above with a higher velocity - therefore more pitch is required to accelerate the air which then goes round the box faster and enters the disc faster than the last time it went through - eventually the blades are having to work at such high AoA that they stall - so how is it going to get off the bottom of the box. There is no constant supply of slow moving or static air for the rotor to accelerate and therefore it will not lift the helicopter.

PhilJ
6th Dec 2004, 12:52
It is a very big box

TeeS
6th Dec 2004, 13:29
"So the temperature has risen and as we all know hot air rises therefore the scales will indicate less."

SilsoeSid - Sorry but the hot gasses are not going to make the box appear lighter because they cannot escape the sealed system. The mass of the system has remained exactly the same (let's not get onto the subject of energy escaping the system in the form of heat!!!)

Cheers

TeeS

boomerangben
6th Dec 2004, 13:35
If the box is airtight and rigid, the gas mixture will not be able to expand so the pressure will increase. The overall mass of the box plus B206, fuel, pilot and air/exhaust mixture would remain constant, as would the volume of the box. Therefore the density of the box would remain constant and there would be no buoyancy effect.

From the pilot's point of view, it would be dark, noisy, smelly, hot and your altimeter would be lying. No change there then!

Beat me to it Tees!

SilsoeSid
6th Dec 2004, 15:28
If he put his landing light on it wouldn't be dark any more. ;)

Besides, in this very strong cardboard box :rolleyes: , if the pressure increases, would there be a point at which, if the door of the box was to suddenly be opened during this experiment, the pilot would suffer from the bends? (I'm not a scuba diver by the way so, :confused: )

csmorris
6th Dec 2004, 19:27
Not likely. You need a pretty large pressure differential to experience the bends. For every 30 ft a diver descends, he experiences an additional 1 atmosphere of pressure (actually a little less but eh). A diver at 90 feet is therefore experiencing 4 atm, or 58.8 psi.

On the other hand, the difference between sea level and outer space is only 1 atm. Good thing too, or you'd get the bends while driving in the mountains. I haven't done the math, but I imagine the temperature required to create that much pressure would cook the pilot long before he'd be subject to the bends :}

Flingwing207
6th Dec 2004, 20:16
Hi Sid,

If you heat air but do not allow it to expand, it does not get less dense.

If you take a fixed mass oxygen and hydrocarbons and rearrange the structure of their respective electrons, the total mass does not change (as you already saw).

However, the question remains - will NR go up or down as the CO2 content increases?:E

SilsoeSid
6th Dec 2004, 21:04
FW207, surely with a govenor and fuel controller fitted, the Nr would remain the same. (N2 = 100% and N1 altered to maintain constant Nr, unless of course the N2 govenor goes haywire)

Although as the air content was replaced with CO2, wouldn't the T4 soar, as the flame wouldn't be cooled as easily?

If we emptied the box and instead of the Jetranger and pilot placed in it an electric hot plate and a pot containing 1 litre of water. As the water reached and maintained boiling point, would the weight of the box plus contents remain the same?

Gomer Pylot
6th Dec 2004, 23:25
If the box is sealed, yes. If the box is open, no. There is a big difference between a closed and an open system.

boomerangben
7th Dec 2004, 09:17
In a sealed box, the fuel will be converted to gas and the ambient temperature will increase. This will result in an increase in pressure and density (more gas in the system as the liquid fuel is converted to gas). There would be an increase in drag and providing the govenor is functional, Nr would be constant, but Tq will increase. Whether this increase would be seen on a torque meter is questionable. The T4 would increase and the efficiency of the engine would reduce since T1 increases. I would imagine this would cause an increase in Ng.

Bomber ARIS
7th Dec 2004, 09:42
the fuel will be converted to gas and the ambient temperature will increase :confused:

Surely the ambient temperature would have to DECREASE to provide the energy to convert the liquid fuel to its vapour form, etc., etc., ......................

TeeS
7th Dec 2004, 10:21
Boomerangben - But surely as the drag increased due to more dense air so would the lift, lower the lever to prevent it climbing and the drag will reduce!

Bomber ARIS - I think he meant from an Exothermic chemical reaction not evaporation!

(Note to self - TeeS get yourself a life!)

boomerangben
7th Dec 2004, 11:35
TeeS,

Agreed, in flight you would be able to reduce pitch. I was thinking about the flat pitch, ie burning and turning with skids on the bottom of the box.

Bommber,

A gas turbine does not vapourise the fuel, it atomises it (converts it to a very fine spray). In fact as the temperture goes up the fuel temp increases and viscosity will go down and the engine driven fuel pump will have to work less hard...........

Bollox to this, off to get myself a life.