0620: "Wings"

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jomala
Posts: 2
Joined: Fri Aug 07, 2009 9:40 am UTC

Re: "Wings" Discussion

This is a bricklayer's accident report, which I think has relevance (even if its apparently apocryphal and dates back over 100 years).

"Dear Sir

in Block 3 of the accident report form. I put "poor planning" as the
cause of my accident. You asked for a fuller explanation and I trust
the following details will be sufficient.

I am a bricklayer by trade. On the day of the accident, I was working
alone on the roof of a new six story building. When I completed my
work, I found that I had some bricks left over which, when weighed
later were found to be slightly in excess of 500 lbs. Rather than
carry the bricks down by hand, I decided to lower them in a barrel by
using a pulley, which was attached to the side of the building on the sixth floor.
Securing the rope at ground level, I went up to the roof, swung the
barrel out and loaded the bricks into it. Then I went down and untied
the rope, holding it tightly to ensure a slow descent of the bricks.
You will note in Block 11 of the accident report form that I weigh 135
lbs.
Due to my surprise at being jerked off the ground so suddenly, I lost
my presence of mind and forgot to let go of the rope. Needless to say,
I proceeded at a rapid rate up the side of the building.
In the vicinity of the 3rd floor, I met the barrel which was now
proceeding downward at an equal, impressive speed.
This explained the fractured skull, minor abrasions and the broken
collar bone, as listed in section 3 of the accident report form.
Slowed only slightly, I continued my rapid ascent, not stopping until
the fingers of my right hand were two knuckles deep into the pulley.
Fortunately by this time I had regained my presence of mind and was
able to hold tightly to the rope, in spite of beginning to experience
a great deal of pain.
At approximately the same time, however, the barrel of bricks hit the
ground and the bottom fell out of the barrel. Now devoid of the weight
of the bricks, that barrel weighed approximately 50 lbs.
I refer you again to my weight. As you can imagine, I began a rapid
descent, down the side of the building.
In the vicinity of the third floor, I met the barrel coming up. This
accounts for the two fractured ankles, broken tooth and several
lacerations of my legs and lower body.
Here my luck began to change slightly. The encounter with the barrel
seemed to slow me enough to lessen my injuries when I fell into the
pile of bricks and sorry to report, however, as I lay there on the
pile of bricks, in pain, unable to move, I again lost my composure and
presence of mind and let go of the rope and I lay there watching the
empty barrel begin its journey back down onto me. This explains the two broken legs.

jamesrl
Posts: 3
Joined: Mon Aug 10, 2009 9:29 am UTC

Re: "Wings" Discussion

Randall the guy in your comic got this one somewhat wrong. He needs to remember pressure != density when looking things up on wikipedia (especially when the [imath]3 T_{\text{Titan}} \approx T_{\text{Earth}})[/imath].

Using ideal gas law PV = NkT (k is boltzmann's constant, N is number of molecules) or
$density \equiv \rho = \frac{m}{V} = \frac{(m/N)*P}{k T}$
where for molecular nitrogen (which Titan's atmosphere is 98.4%, but for simplicity we'll assume 100%): [imath]m/N \cong 2*14*1.67 x10^{-27} \text{kg/molecule}[/imath].

So for Titan at using P = 146.7 kPa and T = 93.7 K (and for Earth at P = 101.3 kPa and T = 287 K):
[imath]\rho_{\text{Titan}} = 5.3 \text{kg}/\text{m}^3[/imath] and [imath]\rho_{\text{Earth}} = 1.2 \text{kg}/\text{m}^3[/imath] (and the earth's surface atmospheric density is verified as 1.2 kg/m^3 http://en.wikipedia.org/wiki/Earth%27s_atmosphere despite the somewhat poor assumption of being pure N2).
Hence Titan's atmosphere is 4.4 times larger or roughly 350% larger.

Now the lift force is proportional to density: http://en.wikipedia.org/wiki/Lift_(force). So if on earth you have a lift force of [imath]L_0[/imath] and a weight of [imath]W_0[/imath], on Titan you'd have a lift force of [imath]4.4 L_0[/imath] and a weight of [imath]0.14 W_0[/imath]. Therefore the criterion to fly on Titan is [imath]4.4 L_0 \geq 0.14 W_0[/imath] or you need $L_0 \geq \frac{0.14}{4.4} W_0 = 0.03 W_0$
that is you need to be able to lift 3% of your weight with wings.

We did make several assumptions that we probably have to investigate in more detail. For example, will the lift coefficient be the same on Earth and Titan? (The Reynolds number is density dependent and lift coefficient is Reynolds number dependent). Or will you be able to flap your wings as quickly on Titan (in the thicker medium). E.g., just as its more difficult to flap your wings in water than in air, it will be more difficult to move your wings on Titan than in earth. From an energy standpoint, it is easy to see it will be more difficult to flap your wings at the same speed in the denser medium.

PS: I first did the calculations over at reddit: http://www.reddit.com/r/science/comment ... te/c0bveln as I wasn't a member of these forums. [Though do read xkcd frequently.]

EDIT 2013: Also submitted to explainxkcd.com as user jimbob
Last edited by jamesrl on Sat Jun 08, 2013 5:36 am UTC, edited 2 times in total.

Homncruse
Posts: 11
Joined: Fri May 08, 2009 4:06 pm UTC

Re: "Wings" Discussion

neurodan wrote:I just want to know how to parse the alt-text...

Where do I put the parentheses??

I prefer:

Code: Select all

// Please do {     not (           try {               ( any of this ) and ( ( die ) or ( get arrested ) )          }     )}

Okay, so it's not perfect, but nobody else turned it into the obvious programming reference.

:edit: Adding missing ) on line 6
Last edited by Homncruse on Mon Aug 10, 2009 9:57 pm UTC, edited 1 time in total.

phillipsjk
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Re: "Wings" Discussion

Error: expected ')' on line 6.

Edit: the missing bracket was on line 5, but I assumed a C-style ambiguous error message when the closing "}" was encountered.
Last edited by phillipsjk on Mon Aug 10, 2009 11:34 pm UTC, edited 1 time in total.
Did you get the number on that truck?

You, sir, name?
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Re: "Wings" Discussion

Homncruse wrote:I prefer:

Code: Select all

// Please

Nothing instills trust like code with comments pleading for the program to work.
I edit my posts a lot and sometimes the words wrong order words appear in sentences get messed up.

Homncruse
Posts: 11
Joined: Fri May 08, 2009 4:06 pm UTC

Re: "Wings" Discussion

phillipsjk wrote:Error: expected ')' on line 6.

Oops. That's what I get for not compiling my code before I post it!

You, sir, name? wrote:Nothing instills trust like code with comments pleading for the program to work.

The comments are just vocalizing what every developer thinks in his/her head the first time code runs anyway. You can't honestly tell me you've never clicked the 'Run' button with at least crossed fingers (or, if you're writing firmware and loading it onto hardware, actually RUNNING just in case you flipped a bit somewhere...).

phillipsjk
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Re: "Wings" Discussion

Homncruse wrote:The comments are just vocalizing what every developer thinks in his/her head the first time code runs anyway. You can't honestly tell me you've never clicked the 'Run' button with at least crossed fingers (or, if you're writing firmware and loading it onto hardware, actually RUNNING just in case you flipped a bit somewhere...).

Well, there was that one time I was doing my final electronics project. Since I did not know for sure how well the custom hardware would work, I could not use ad-hoc debugging. I went through my assembly code line-by-line to make sure it was correct (It was a month over-due). After I was done, there was still some unexplained behavior (a variable was not behaving as expected), but it worked. My hardware had a few minor design flaws I haven't gotten around to fixing yet.

Edit: In case that sounded harsh, it is about the only time I have been confident a non-trivial piece of code I wrote was correct. Not sure how it relates back to the comic though.
Did you get the number on that truck?

uggie
Posts: 2
Joined: Fri Aug 07, 2009 11:27 pm UTC

Re: "Wings" Discussion

jamesrl wrote:Randall the guy in your comic got this one somewhat wrong. He needs to remember pressure != density when looking things up on wikipedia (especially when the [imath]3 T_{\text{Titan}} \approx T_{\text{Earth}})[/imath].

Using ideal gas law PV = NkT (k is boltzmann's constant, N is number of molecules) or
$density \equiv \rho = \frac{m}{V} = \frac{(m/N)*P}{k T}$
where for molecular nitrogen (which Titan's atmosphere is 98.4%, but for simplicity we'll assume 100%): [imath]m/N \cong 2*14*1.67 x10^{-27} \text{kg/molecule}[/imath].

So for Titan at using P = 146.7 kPa and T = 93.7 K (and for Earth at P = 101.3 kPa and T = 287 K):
[imath]\rho_{\text{Titan}} = 5.3 \text{kg}/\text{m}^3[/imath] and [imath]\rho_{\text{Earth}} = 1.2 \text{kg}/\text{m}^3[/imath] (and the earth's surface atmospheric density is verified as 1.2 kg/m^3 http://en.wikipedia.org/wiki/Earth%27s_atmosphere despite the somewhat poor assumption of being pure N2).
Hence Titan's atmosphere is 4.4 times larger or roughly 350% larger.

Now the lift force is proportional to density: http://en.wikipedia.org/wiki/Lift_(force). So if on earth you have a lift force of [imath]L_0[/imath] and a weight of [imath]W_0[/imath], on Titan you'd have a lift force of [imath]4.4 L_0[/imath] and a weight of [imath]0.14 W_0[/imath]. Therefore the criterion to fly on Titan is [imath]4.4 L_0 \geq 0.14 W_0[/imath] or you need $L_0 \geq \frac{0.14}{4.4} W_0 = 0.03 W_0$
that is you need to be able to lift 3% of your weight with wings.

I agree with the density corrections and didn't bother checking it when I did my initial calculations, but I don't agree with the 3% number. Lift has to equal weight or you fall. That means that the minimum lift you can generate and sustain level flight on titan is 14% of your earth weight. What changes is the velocity and or wing area required to produce sufficient lift to keep you from going splat.

On second thought I think I misread your initial post. You're saying that the same conditions that produce 3% of your earth weight in lift on earth will produce 14% of your earth weight in lift on titan. Got it now.

phillipsjk
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Re: "Wings" Discussion

jamesrl wrote:
Now the lift force is proportional to density: http://en.wikipedia.org/wiki/Lift_(force). So if on earth you have a lift force of [imath]L_0[/imath] and a weight of [imath]W_0[/imath], on Titan you'd have a lift force of [imath]4.4 L_0[/imath] and a weight of [imath]0.14 W_0[/imath]. Therefore the criterion to fly on Titan is [imath]4.4 L_0 \geq 0.14 W_0[/imath] or you need $L_0 \geq \frac{0.14}{4.4} W_0 = 0.03 W_0$
that is you need to be able to lift 3% of your weight with wings.

We did make several assumptions that we probably have to investigate in more detail. For example, will the lift coefficient be the same on Earth and Titan? (The Reynolds number is density dependent and lift coefficient is Reynolds number dependent). Or will you be able to flap your wings as quickly on Titan (in the thicker medium). E.g., just as its more difficult to flap your wings in water than in air, it will be more difficult to move your wings on Titan than in earth. From an energy standpoint, it is easy to see it will be more difficult to flap your wings at the same speed in the denser medium.

Well, after uggie brought it up, I noticed an error in your assumptions: The linked wikipedia page for "lift force" describes models for fixed-wing aircraft.

When the character in the comic lifts his wings, the dense air on titan would cause more down-force. I am fairly certain that birds actually deform their wings during flapping, otherwise, there would be no net lift during hovering (gliding would be OK). Since we don't have a detailed design of the wings, except for a reference to a science-fiction work in a Wikipedia article (Credit: Mavrisa), we many never be able to model it properly.
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jamesrl
Posts: 3
Joined: Mon Aug 10, 2009 9:29 am UTC

Re: "Wings" Discussion

I'll admit I was cavalier in citing the Lift Force equation as a justification that force from "flapping" should be propto density; however I standby the relationship that as density (assuming other things are constant such as Reynolds number, viscosity, etc) increases it is easier to create a force upwards via flapping or motion through the fluid.

Think of Newton's third law, if your flapping is essentially pushing against a gas, causing the gas to move; the equal and opposite reaction is for you to move up. As the speed [imath]v[/imath] the air moves down is to first approximation the speed [imath]v[/imath] you flapped your wing in the down direction. So picture a wing of area [imath]A[/imath] moving downwards, which in a time [imath]\delta t[/imath] causes a volume of air [imath]\delta V = A v \delta t[/imath] to move down at a velocity [imath]v[/imath]. Then the force (while the wing is moving downward) is
$F = \frac{dp}{dt} \approx \frac{m}{\delta t} v = \frac{\rho \delta V}{\delta t} v = \frac{\rho A v\delta t}{\delta t} v = \rho A v^2$
So the force of a downward flapping wing should be roughly proportional to density, wing area, velocity squared, from pretty general arguments. Now fluid flow is very complex field and I oversimplified a lot (didn't start with Navier-Stokes, etc), but you can see the back-of-an-envelope calculation.

You could also argue this from a purely dimensional analysis point of view with physical intuition that rho, v, and A are the significant quantities that intuitively will contribute to the force.
This would be $F = [N] = \frac{[kg] [m]}{[s]^2} \propto \rho ^{x} A^{y} v^{z} = \frac{[kg]^x [m]^{2y} [m]^z}{[m]^{3x} [s]^{z}} = [kg]^x [m]^{2y + z - 3x} [s]^{-z}$
Solving this we immediately see x = 1 (to match up units of mass [kg] on both sides), z = 2 (to match up units of time [s] on both sides), and then get 1 = 2y + z -3 x = 2y + 2 - 3 = 2y -1 or y=1 to match up units of length [m]. Thus we recover the same expression that [imath]F \propto \rho A v^2[/imath]

But then again, I thought we were trying to fly like birds, which do incorporate both lift and thrust. [The flapping is better used not to cause lift directly, but to cause thrust, with an angle of attack.] I talked about this a little bit over in the reddit forums the other day and posted a link on physics of flight:
http://www.ucmp.berkeley.edu/vertebrate ... ysics.html (also good info searching about physics of orinthopters).

And you are quite mistaken when you think "the dense air on titan would cause more down-force". Assume the air on titan to first approximation for this purpose isn't moving [e.g., no wind], and lets also assume the object is significantly denser than N2 so we don't have to factor in buoyancy.
For any object on Titan (or any atmosphere), the net downward force of all the air above it, is precisely balanced by the net force of all the air below it. Hence there's no force on an object from the air pressure. Or to think of it on Earth if you take a table with 1m^2 of surface area and put it outside (where atmospheric pressure is 101.3 kPa), the table will not feel a force of 101.3 kPa*1 m^2 = 10^5 N = 22000 lbs, because "the down-force" exactly balanced out by the air pressure on the other side of it giving an "up-force".

Or think of how much easier it is too provide up force in water (very dense compared with air). [Yes buoyancy also plays a helping role in water as we are about the same density (or slightly less) as water]

EDIT: On second thought I think by more down force you might be assuming that the wings are just simply moving up and down at equal speed, which obviously would drive not provide lift half the time. But as [imath]F \propto v^2[/imath] as long as you flap quickly downward (at large v), and slowly upward (small v) the time-averaged net force will still be upwards. For example say the wings move 0.5 m and flapping downward takes .25 s, but flapping upward takes 1 second.
Then the time averaged force upwards is
$\langle F \rangle = \rho A \left[ \left(\frac{0.5 \text{m}}{0.25 \text{s}}\right)^2 \left\{ \frac{0.25 \text{s}}{1.25 \text{s}} \right\} - \left(\frac{0.5\text{m}}{1\text{s}}\right)^2 \left\{\frac{1 \text{s}}{1.25 \text{s}} \right\} \right] = \rho A (0.60 \text{m}^2 \text{s}^{-2})$
However, obviously you can make significant improvements over this by more complicated wing strokes.

SECOND-EDIT: Forgot to actually time average in the equation for [imath]\langle F \rangle[/imath] (despite saying I needed to time average). Oh well lack of sleep. Corrected the equation by adding the two parts in braces.
Last edited by jamesrl on Fri Aug 14, 2009 4:35 pm UTC, edited 3 times in total.

phillipsjk
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Re: "Wings" Discussion

I think you may have missed my point: Before the character in question can do the down-stroke (creating lift), he needs to lift the wings into the air (creating down-force proportional to the top area of the wings).

Of course, this morning I thought of an exception where that assumption doesn't work: Insects have relatively fixed wings. I suppose they can still produce net lift by moving the air towards the tips of the wings (like a paper fan). However, insects beat their wings much faster than birds do.

To be honest, I am over my head with fluid dynamics calculations.
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Technical Ben
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Re: "Wings" Discussion

phillipsjk wrote:I think you may have missed my point: Before the character in question can do the down-stroke (creating lift), he needs to lift the wings into the air (creating down-force proportional to the top area of the wings).

Of course, this morning I thought of an exception where that assumption doesn't work: Insects have relatively fixed wings. I suppose they can still produce net lift by moving the air towards the tips of the wings (like a paper fan). However, insects beat their wings much faster than birds do.

To be honest, I am over my head with fluid dynamics calculations.

As far as I know, you tilt the wings during the up stroke, cancelling out some of the forces (as your moving edge on). At the very least this moves the direction of push to sideways on the up stroke, down on the down stroke. Look at a bumble bee/flys wings flap. They are a lot more complicated than just up|down.
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Faranya
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Re: "Wings" Discussion

phillipsjk wrote:I think you may have missed my point: Before the character in question can do the down-stroke (creating lift), he needs to lift the wings into the air (creating down-force proportional to the top area of the wings).

But, if that propostition were true, nothing would be capable of flight...

I would assume he took the same time Daedalus did to examine the flight patterns of birds and figured out the most effective way to work his wings.

phillipsjk
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Re: "Wings" Discussion

Technical Ben wrote:As far as I know, you tilt the wings during the up stroke, cancelling out some of the forces (as your moving edge on). At the very least this moves the direction of push to sideways on the up stroke, down on the down stroke. Look at a bumble bee/flys wings flap. They are a lot more complicated than just up|down.

Yes, so derate your lift by 30-50%.

As for insect flight: In think I would need a technological aid for that. I can't see things moving that fast (I've heard Bees are at about 300Hz)
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jamesrl
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Re: "Wings" Discussion

First, this is not really fluid dynamics at the level I'm treating it; just high school level physics where we have [imath]F \propto \rho A v^2[/imath] (which I derived above in two different ways).

If you recall the EDIT part of my previous comment (editted on same day as originally posted), you'll see I understood your point eventually. But even ignoring angles and complicated wing motions through space, as long as the down-strokes are faster than the up-strokes then you will create an overall thrust upwards. This is because the velocity is squared.

Imagine wing flaps move 0.5 m up and down with an area A, in a fluid of density [imath]\rho[/imath] and the down stroke takes 0.25 s (for a speed of [imath]v = .5 m /(.25 s) = 2 m/s[/imath]) and the upstroke takes 1 second (for a speed of [imath]v = 0.5 m/s[/imath]). Now for the 0.25 s when the wing strokes down there is a force upwards, and when the wing strokes up for 1 s there is a downward force. Since the upward force is 16 times stronger than the downward force (which lasts 4 times longer), so in time-integrated upward force is only 4 times stronger than the time integrated downward force. Anyhow, the net effect is that you get a net upward force.

Now complicated flight patterns will definitely help optimize things. My roommate interviewed here has videos of insect flight. [His advisor (who is still working to get tenure and does good research) hired media professionals to put the youtube up (and also make podcasts) to make the science "interesting" as a form of outreach. The group had little say in the finished product; they recorded an interview of him and spliced his words in; e.g., him saying "yes" is not to their specific questions (some of which are incorrect; e.g., the part about Mars, which neglects that the Martian atmosphere is ~100 less dense than earth's so much more difficult to get mechanical flies moving on.)].

frog42
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Re: "Wings" Discussion

This comic woke me up in my sleep one night, so now that I've made an account, I feel obliged to post.

The effect would be similar to being underwater. Imagine flapping wings underwater. Even with precise control, you'd be pushing yourself down pretty hard on each back-swing. Gliding would be perfectly feasible, but works just fine on Earth as well (see hang gliders).

Mokele
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Re: "Wings" Discussion

Imagine flapping wings underwater. Even with precise control, you'd be pushing yourself down pretty hard on each back-swing.

Sea turtles and dolphins manage it (both have lift-based swimming). In the case of sea turtles, propulsive force is produced on both upstroke and downstroke (along with a little lift) due to being flexible enough to turn their flipper around during the upstroke. They're actually quite fast as a result. You wouldn't think a 400lb turtle can swim that fast, but I've seen them really haul shell when spooked.
"With malleus aforethought, mammals got an earful of their ancestor's jaw" - J. Burns, Biograffiti

frog42
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Re: "Wings" Discussion

Mokele wrote:
Sea turtles and dolphins manage it (both have lift-based swimming). In the case of sea turtles, propulsive force is produced on both upstroke and downstroke (along with a little lift) due to being flexible enough to turn their flipper around during the upstroke. They're actually quite fast as a result. You wouldn't think a 400lb turtle can swim that fast, but I've seen them really haul shell when spooked.

I didn't rule it out, I just wanted to point out that it would hardly be like flying through air (as that's what woke me up). Also, the fall to the ground wouldn't really hurt.

But you have bested me, for you found a way to say "haul shell" in a perfectly natural sentence. TMNT4LIFE

Mokele
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Re: "Wings" Discussion

I just couldn't resist. Turtle power!
"With malleus aforethought, mammals got an earful of their ancestor's jaw" - J. Burns, Biograffiti

dragonfiremalus
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Re: 0620: "Wings"

Notice that he does not tell you not to try it. He simply advises against trying it and dying or getting arrested. If we do not die or get arrested, then we are free to try away. I need a good bridge.

Freonpsandoz
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Re: 0620: "Wings"

I'm surprised that nobody has mentioned that the amount of thrust (force) that the flier must produce in flight is not the same as the lift; the required thrust is normally much less. As someone else noted, the lift must equal the weight of the system (mass of aircraft & pilot times gravitational acceleration). The thrust required to remain airborne is equal to the _drag_ produced by the system. If the system has a lift-to-drag ratio (L/D) of 10:1, then the pilot must produce a thrust slightly greater than 10% of the system weight in order to climb. (BTW, I'm certain that a flying bicycle using a human-powered propeller would be _much_ more efficient than flapping.)

david.windsor
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Re: 0620: "Wings"

Hey, once hat guy melts his wings off he won't fall, the counter weight will lift him up.
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