Physics Q&A Game: Calculate Minimum Power for Man-Powered Helicopter

[Gokul43201]

A Q&A game is simple: One person asks a relevant question (it can be research, calculation, off-the-top-of-the-head, anything as long as it is a physics question) and other people try to answer. The person who posts the first correct answer (as recognized by s/he who asked the question) gets to ask the next question, and so on.

I'll get this rolling with a simple back-of-the-envelope calculation question.

It is proposed to make a man powered helicopter, with a rotor 10 m in diameter. Assuming that the rotor blows a cylindrical column of air uniformly downwards, the cylinder diameter being the same as the rotor diameter, and the weight of the man plus machine is 200 kg, calculate the minimum mechanical power (in watts) that the man must generate, if he is to remain airborne. (take density of air ~ 1.2 kg/m3) Is the system feasible ?

(show, in a couple of lines, the essential steps in the calculation)

If a sufficient time passes with no correct answer, then the best attempt determines who goes next.

 

[Galileo]

Great initiative Gokul!

Here's my attempt. Call the radius of the rotor r.
The power generated is the kinetic energy transferred to the air per second.
The mass of the air column is the density of the air times it's volume, so

K=\frac{1}{2}Mv^2=\frac{1}{2}\rho \pi r^2 vt v^2
P=\frac{dK}{dt}=\frac{r^2\pi}{2}\rho v^3

The momentum imparted to the air is:

P_m=\rho \pi r^2 v^2t
so the force that keeps the copter floating and should equal mg is:
\rho \pi r^2 v^2=mg
which we can solve for the speed of the air particles.
We find for the power:

P=\frac{r^2\pi}{2}\rho \left(\frac{mg}{r^2\rho \pi}\right)^{3/2}
Plugging in the numbers, we find about 4,5 kW.
I had to look up some power consumption data. If a human runs at 24 km/h, the power consumption is about 1.7 kW. So it's not a feasible construction (not for me at least).

 

[Gokul43201]

Nice one Galileo !

World class sprinters and trained athletes can deliver close to 5 kW for no more than a minute. I certainly can not sustain more than about a kW for any reasonable length of time.

Your turn...

 

[jdavel]

Gokul and Galileo,

Notice though that the required power goes as 1/r. This makes sense because you're getting your momentum with more mass and less speed, and that reduces the energy which goes as speed squared.

So maybe with 50m rotors (made out something with a very high strength/weight ratio) you could do it. Of course your ideal model of the air column moving down monolithically will probably fail at low speed. Oh, well!

PS Gokul, be aware that your Q&A game could be abused by students wanting a quick solution to a homework without doing any work themselves! Come to think of it, maybe that's what you were doing.

 

[Gokul43201]

Gokul and Galileo,

Notice though that the required power goes as 1/r. This makes sense because you're getting your momentum with more mass and less speed, and that reduces the energy which goes as speed squared.

So maybe with 50m rotors (made out something with a very high strength/weight ratio) you could do it. Of course your ideal model of the air column moving down monolithically will probably fail at low speed. Oh, well!

PS Gokul, be aware that your Q&A game could be abused by students wanting a quick solution to a homework without doing any work themselves! Come to think of it, maybe that's what you were doing.

Point taken, on the abuse possibility (I'll not comment on the snide insinuation just yet). I'll leave it to the mentors to make that judgement.

As for the 50 m blades : you find me the material, and we're in business !

 

[Galileo]

I think homework questions posted here are easily detected. Such are usually not the kind of questions you want to post here.

As a result, it took me a little time to think up a nice question (that I have to solve myself), so I actually should have taken that into consideration before posting a solution , but anyhoo:

There's an astronaut going for a space walk and enjoying the awesome sight (who wouldn't).
Just when he's ready to return to his spaceship he notices the line that connected him to the spaceship has broken! This line was supposed to bring him back aboard the ship. The astronaut has only 5 minutes of oxygen left in his tank and he fears for his life. If something can be done, it has to be done quickly.
The astronaut realizes he has a small fire extinguisher on his spacesuit which he can use as a thrustrocket. He immediately grabs the extinguisher and wants to start spraying.

The astronaut is now 120 m away from his spaceship and has a neglegible velocity with respect to the ship. His mass (with suit, but without extinguisher) is 94.5 kg. De mass of the contents of the extinguisher is 5.0 kg and the mass of the empty container is 5.5 kg. The contents are sprayed outwards with a speed of 10 m/s, regardless of the remaining content. After 2 minutes of spraying, the container will be empty.

Will the astronaut arrive at the spaceship before his oxygen supply runs out?
Can he save his hide from this perilous predicament? Stay tuned!

 

[Gokul43201]

I think it may be important to specify that the extinguisher is attached to the spacesuit, and can not be removed. Else, there's an easy cheat.

 

[whozum]

v_f = v_i + v_{ex}*ln(m_i/m_f)

v_f = 10ln(\frac{105}{100}) = 0.487m/s

At t = 120s he will have traveled 120x0.487= 58.55m

At t = 600s when his gas is empty he will have traveled 600x0.487 = 292m.

He makes it with time to stop for donuts.

 

[Galileo]

v_f = v_i + v_{ex}*ln(m_i/m_f)

v_f = 10ln(\frac{105}{100}) = 0.487m/s

At t = 120s he will have traveled 120x0.487= 58.55m

At t = 600s when his gas is empty he will have traveled 600x0.487 = 292m.

He makes it with time to stop for donuts.

I`m sorry, your v_f is roughly equal to the final velocity of the astronaut, so you cannot use the distance formula for constant velocity.
Also, there is 5 minutes of oxygen left, not 10.

 

[Berislav]

I have an idea for the solution:

From the law of conservation of momentum

m_{e}v_e=(m_a+m_c)v

Deriving with respect to t we find:

\frac{d(m_e v_e)}{d t}=m_aa+\frac{dm_c}{d t}v+m_ca

We get an ODE which is easy to solve, since we know that

\frac{dm_e}{d t}=-\frac{dm_c}{d t}=5 kg/120s
v_e=10 m/s

The total time required to transverse distance s is

t=\frac{s-\int_{0}^{2min} v(t)dt}{v(120s)}+120s

P.S.
Sorry I didn't do any numerical calculations. My MS Calculator crashes everytime I try to start it, and I don't know where my pocket calculator is.

P.P.S.
I read your question again and made the necessary correction.

 

[Galileo]

Good try Berislav.

I can make the calculations for you if you tell me what numbers to plug in. Since you haven't got a formula for v(t) I can't solve your equation for t yet.

 

[Berislav]

For v(t) I get:

v(t)=\frac{5/120v_et}{-5/120t+10.5+m_a},

since m_c(t)=10.5 kg-5/120t

 

[Galileo]

For v(t) I get:

v(t)=\frac{5/120v_et}{-5/120t+10.5+m_a},

since m_c(t)=10.5 kg-5/120t

Ok, well you'll have to integrate it yourself. There are some internet tools to help you with it if your require it: http://integrals.wolfram.com/
Online calculators also exist, just goooogle it up.

 

[Galileo]

Anyone?

Too hard? Too easy? I`ll start posting mild hints tonight (on my clock) if no one answers.

 

[whozum]

Wouldn't average velocity for the first two minutes work?

v_avg = 0+0.487 / 2 = 0.2435/s

0.2435m/s * 120s = 29.22m plus

180 * 0.487 = 87.66m

= 116.88m. I'm not sure though, my algebra based physics skills are no match for you guys.

 

[Gokul43201]

Anyone?

Too hard? Too easy? I`ll start posting mild hints tonight (on my clock) if no one answers.

Can I suggest that you give it about 24 hrs before releasing hints ?

 

[Chronos]

Try using seconds as the time unit and see what you get. All the relevant constants defining work are so based.

 

[Berislav]

Integrator gives me:

s(t)=\frac{5}{120}v_et \log(10.5+m_a-0,0416667t)

EDIT:
No, scratch that! I meant at t=120s this is 100m. So,

t=\frac{s-\int_{0}^{2min} v(t)dt}{v(120s)}+120s

is 160 s. So, he makes it. It seems that mulitplication and division elude me.

P.S.

Sorry I posted this so late.

 

[Galileo]

Whozum. Taking the average velocity gives a good approximation, but from your anwer you cannot answer with certainty whether the astronaut makes it or not.

Berislav. You are close (i.e. on the right track), but there are some errors in your calculations.
(160s is off, it would mean he'd make it easily and I can guarantee the outcome will be a close call)

I wanted to post some hints at this time, since it's about two days after posting the problem, but at Gokul's request I'll wait a bit and post them in the morning (about 8:00 GMT) if no correct answer is in yet.

 

[whozum]

Can I waive my ability to solve this problem and request you message me the math needed to solve it?

 

[Galileo]

Can I waive my ability to solve this problem and request you message me the math needed to solve it?

That'd be a hint, wouldn't it?
You'll have to wait a little longer for that.

Don't forfeit your chance yet.

 

[whozum]

I'm dying here. Atleast let me know why average velocity wouldn't work? As far as I can think the acceleration is pretty much constant, no?

 

[Berislav]

Berislav. You are close (i.e. on the right track), but there are some errors in your calculations.

Yes, there are. And I think I found all of them. Apperantly, in Integrator c*x is not the same as cx.

I couldn't find any errors in my solution to the differential equation.

When I integrate v(t) from 0 to 120 s, I get 29.5121 m.

So t=300.9758 s. Close, indeed.

 

[Galileo]

Okay Berislav, you have basically solved the problem so you get to go next, although I would like you to give explicit answers to the actual question(s):
Does he make it before the oxygen supply runs out?
And: Can he (still) save himself (i.e. survive)?

Anyway, I wanted a question that was elementary on the surface, but not very easy. Also, I wanted one that could be solved by anyone, not just the 'elite'. You can answer this question using physical insight and very basic (Grade-K12) physics.
Notice the difficulty comes from the fact that the mass of the astronaut changes (the force acting on him is constant, but his acceleration isn't), but how big IS this effect? The guy starts out with 105 kg and ends with 100 kg, that's a small enough difference to consider the following: Can we solve it if we assume his mass is constant?
Ofcourse! It will be a simple linear acceleration problem. Moreover, if we assume his mass is and stays 100 kg, we can argue that he will arrive at the ship faster than in actuality. If he doesn't make it in this approximation, he surely won't make it in the real case. Similarly, if his mass stays 105 kg, you can use similar arguments.
Work it out yourself and you can easily estimate how long it will take the astronaut to reach the ship after his supply runs out. Then you can answer whether he can still save himself somehow (Gokul already hinted at an somewhat obvious idea).

In any case good job, Berislav. Your turn.

 

[Berislav]

Does he make it before the oxygen supply runs out?

No, he spends about a second is outerspace without oxygen.

Can he (still) save himself (i.e. survive)?

Since he is a astronut he's supposed to be at least in fair physical condition. So, he can go without oxygen for at least 30 s before falling unconcious. This is enough time repressurize a chamber in any modern spacecraft . All in all, he survives, although not a very pleasant experience.

Very good question.

In any case good job, Berislav. Your turn.

Thank you.

In spirit of your idea, my question can be answered with physical insight alone (no mathematics needed). Here it is:

In summer after a hot and dry period raindrops falling on wall made of red brick make a hissing sound. Explain the hissing sound.

NOTE: I am not the author of this question. I will post the reference per request.

 

[hemmul]

In spirit of your idea, my question can be answered with physical insight alone

outstanding!

In summer after a hot and dry period raindrops falling on wall made of red brick make a hissing sound. Explain the hissing sound.

if it was a very hot period (like at my location, in the summer the temperature of the surface placed under direct sunlight reaches really high levels) then maybe the small rain drop evaporates during the contact with the lots-of-heat-accumulated-red-brick? (like an oil on the pen)

 

[Berislav]

then maybe the small rain drop evaporates during the contact with the lots-of-heat-accumulated-red-brick?

It's not that simple. When you throw water drops on a very hot surface, like that of stove (I loved doing that as a young child), you will indeed hear a hissing sound, but a simple observation of a brick wall during summer rain and the case for the stove can tell you that the same process is not the reason for the hissing sound. I won't say anymore because that would be a hint.

 

[hemmul]

It's not that simple. When you throw water drops on a very hot surface, like that of stove (I loved doing that as a young child), you will indeed hear a hissing sound, but a simple observation of a brick wall during summer rain and the case for the stove can tell you that the same process is not the reason for the hissing sound. I won't say anymore because that would be a hint.

ah, ok... the other possibility is absorbtion - i shall think of the mechanism, but sometimes i noticed the hissing sound during the arbsorption of water by some materials... (maybe water acts like a catalyst in the unfinished reaction inside the material of a brick?)

 

[Berislav]

ah, ok... the other possibility is absorbtion - i shall think of the mechanism, but sometimes i noticed the hissing sound during the arbsorption of water by some materials... (maybe water acts like a catalyst in the unfinished reaction inside the material of a brick?)

The description of the mechanism is required, of course.

 

[jdavel]

As the water is absorbed it displaces air which then bubble out through the water, then the bubbles burst when they reach the outside surface of the water, creating the hissing sound?

 

[Berislav]

As the water is absorbed it displaces air which then bubble out through the water, then the bubbles burst when they reach the outside surface of the water, creating the hissing sound?

Yes, that is basically the correct answer. Although you neglected surface tension it is still satisfactory, in my opinion.

Good work. You can now ask the next question.

 

[jdavel]

Berislav,

I think hemmul deserves it; I wouldn't have gotten the answer without his "absorption" idea. As Newton said, I stood on the shoulders of those who went before me!

 

[Berislav]

I think hemmul deserves it; I wouldn't have gotten the answer without his "absorption" idea.

Maybe the OP can comment on what should be done in this case?

 

[Gokul43201]

While hemmul's answer was definitely noteworthy, it was not (as he said so himself) complete. Jdavel provided more of a complete answer.

In my opinion, however, the answer is still not satisfactorily answered as it does not explain the importance of the "hot and dry period".

I'd give the next turn to Jdavel.

 

[Berislav]

In my opinion, however, the answer is still not satisfactorily answered as it does not explain the importance of the "hot and dry period".

Yes, that's true. It would be nice of Jdavel to elaborate on his answer using the aditional information I supplied to explain why this is important before continuing with his question.

 

[jdavel]

Berislav,

This may be more of a guess than an "elaboration". As the bricks heat up in the sun water held in the bricks evaporates. The bricks can get much hotter than the air temperature, so in order for this vapor to leave the bricks, the air outside must have low relative humidity. So it needs to be a hot, dry day. Now there are empty spaces for water to enter, displacing the air through bubbles.

Just in case that was confusing, I'm not saying the bricks get hot enough to boil water (you said that doesn't happen) it's just that the vapor pressure of the water is very high inside the bricks.

?

Edit: Berislav, now I'm seeing that you said surface tension plays a role. I haven't figured that out yet, so I guess the game is still on.

 

[hemmul]

Edit: Berislav, now I'm seeing that you said surface tension plays a role. I haven't figured that out yet, so I guess the game is still on.

I believe the surface tension is responsible for the production of sound: when the bubble bursts the time during which the surface water will come together into single point depends on the radius of the bubble and surface tension (through the accelleration it provides).
The frequency of plop is inversely proportional to this time. And the plops from the vast number of bubbles produces hissing sound :) (like that in a hot cup of well-made coffee)

jdavel, "standing on the shoulders" and not falling down - is much more difficult than standing on the ground! YOU definitely do deserve the next turn :)

 

[Berislav]

Hemmul, you explained why we hear hissing instead of a 'plop'.

Jdavel, you are very close to the solution. If you want additional hints regarding surface tension, feel free to ask.

 

[Gokul43201]

jdavel, it is now your turn to ask a new question.

Berislav, feel free to post any additional details in the solution to your question. We shall consider the question answered for the purposes of the continuity of the game.

If jdavel does not come up with a question in 12 hrs, hemmul may go next. If neither asks a question within 24 hrs, I will.

 

[Berislav]

Very well.
The bricks have a porous structure. The pores can be viewed as interconnected by cavities. At the moment of impact when rain drops fall on a brick surface tension pulls the water down into the pores. This "pulling force" will be greater for water bubble on a smaller pore than on a larger one. Since there are more smaller pores than larger ones the air in the cavity is displaced until the large bubbles burst.

 

[jdavel]

"jdavel, it is now your turn to ask a new question."

How about this? Two 20-year-old twins live on the earth. One travels to Alpha Centauri and back at 90% the speed of light...JUST KIDDING!

Ok, here's my real question; it requires no calculation, and only high school level physics should be needed.

An astronaut is floating weightless in space when two wrist watches come drifting by. He reaches out and grabs them, sets them in front of him carefully so they don't drift away, and then observes them for a while. The first thing he notices is that they both have the same inscription on their faces, "Zurich, 1905". After watching them for a while, he notices they keep identical time, so he concludes (correctly) that these are identical watches. Then he puts one on.

Question: Do the identical watches continue to keep the same time while he is wearing one but not the other?. If so, explain. If not, explain.

 

[Crosson]

Question: Do the identical watches continue to keep the same time while he is wearing one but not the other?. If so, explain. If not, explain.

There are two effects, both of which are gravitational and minuscule, which effect the keeping of time in this situation.

Relative to the astronaut, the watch he is not wearing begins to "speed up", due to gravitational time dilation. The gravitational field is obviously stronger for the watch he is wearing.

Also, the gravitational field of the astronaut will effect the mechanical motion of the watches internal parts. Depending on the orientation of the astronaught to the watch he is wearing, this will either speed up of slow down the watch.

Despite all I have said, I conjecture that the clocks keep the same time to within one Planck time over the life of the universe (10^-52 seconds per year) under these vanishingly small gravitational differences. Therefore, the answer is yes, they continue to keep the same time.

 

[jdavel]

Crosson,

Your analysis is correct; there would be no way for the astronaut to detect the effect of his own gravitational field on how a clock on his wrist and one right in front of him keep time. I meant to convey that this is a problem in classical physics by saying that "only high school level physics should be needed". Sorry about the confusion.

That said, while your analysis was correct, it wasn't the right one and so your answer (they keep the same time) isn't necessarily correct.

 

[hemmul]

That said, while your analysis was correct, it wasn't the right one and so your answer 9they keep the same time) isn't necessarily correct.

getting back to high school physics course (luckily it is not that far!) the answer is no, however there are several points: how do we compare the time on two clocks? to be able to compare them we have to bring them to one and the same of space-time. Now, after the astronaut gets the watch on his hand his, and the other watch's world lines are different. Any watch shows its proper time.
Now, as the remained watch continues its inertial motion, let's consider the situation from it's RF. It is clear, that the astronaut is moving in the space (with some accelleration, searches the Earth, searches his spacecraft etc...). So his world line is curved, with respect to the linear line of the traveling watch. that's why when the astronaut stackles upon the traveling watch again - he will see the difference (his watch will show that less time passed)!

This is only in the case if the traveling watch is traveling at a constant velocity e.i. we can bind a single Inertial Reference Frame to it.

If no - (the other anstronaut grabs the remained watch) - then we have to integrate the proper time along the world lines for both astronauts, that is far beyond the high school level ;)

 

[jdavel]

hemmul,

You sure got further in high school physics than I did!

But again, to clarify, this is a problem in classical physics. That is, you can assume Galilean/Newtonian relativity (c is infinite) and you can assume zero uncertainty (h = 0).

 

[Gokul43201]

jdavel,

Could you clarify/describe what you mean by "then he puts one on" ? I'm imagining an astronaut in a bulky space suit. How on Earth (or in space) does he put on a wristwatch ? If this is an irrelevant detail, you may say so. But if there is any form of energy transfer between objects, this may be necessary to better estimate the sizes of effects.

Also, does it matter how the watch works, or is the the answer independent of that too ?

 

[jdavel]

Gokul,

How he gets the watch on is not important.

How the watch works IS important.

 

[hemmul]

one healthy astronaut's body-temperature is around 36.6° C. when he puts the watch on - there is certain heat transfer between the two. As the responce to the heating is different for different parts of the mechanism inside the watch - the relative size of the elements changes - so the watch becomes disbalanced :)
well, it is crazy idea, because as far as i remember the clocks produced in 1905 in Zurich were highly accurate ;)
The other possiblity is that the astronaut waves his hands, and the constant random accellerations can lead to some errors in the mechanism, but the average effect of those is zero... so getting back to the change in the conditions for the two clocks, the only parameter that is changed, AFAIK is temperature (thermal expansion)

 

[jdavel]

I'm beginning to think that my attempt to wrap a story around this problem has left the physics too obscure and wide open to interpretations other than the one I had in mind. So I'm going to give a hint (actually answer part of the question for you) to redirect and focus your thinking where I intended.

The watches don't keep the same time. The cause (at least the one I'm looking for) is purely mechanical, having to do with the way that watches made in 1905 (or in any year before about 1960) worked. Also, discovering the cause leads unambiguously to whether the watch the astronaut wears runs faster or slower.

So now the question is: Does putting the watch on cause it to speed up or to slow down, and why?

 

[hemmul]

The watches don't keep the same time. The cause (at least the one I'm looking for) is purely mechanical, having to do with the way that watches made in 1905 (or in any year before about 1960) worked. Also, discovering the cause leads unambiguously to whether the watch the astronaut wears runs faster or slower.

So now the question is: Does putting the watch on cause it to speed up or to slow down, and why?

Well, well, well :) the question turns to be more and more interesting :)

Actually, by the end of IXX's century they understood that the dust was dangerous for the watch. There were several attempts to prevent it from entering the mechanism - like making additional covers etc, but unfortunately they didn't lead to full isolation and impermeability of the watch...

in 1926 the first licenses for water-resistant cases began to appear, while in 1946 the first impermeable hand-watch - Rolex - was born!

Later, in 1969, the fully water-resistant watch and their "brothers" were very popular. and very useful! So popular, that Neil Armstrong and Edwin Eldrine had the highly-accurate Omega Speedmaster on their hands while walking on the moon :)

Well, now back in 1905 - the inner mechanism of the watch was not protected from the outer world.
While there was no danger for the watch floating in space, the one placed on the hand of an astronaut will definitely live in the environment where there is air, water dust etc... i hope it's clear why that should be the reason for different functioning ;)