## What-If 0133: "Flagpole"

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sevenperforce
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### Re: What-If 0133: "Flagpole"

So I figured out (in an edit I subsequently lost) that an orangutan could very well manage to pull off a 50-story drop if we take the lower bound for orangutan weight, the upper bound for orangutan arm length, and the "seven times stronger than a human" value.

Flumble wrote:
Neil_Boekend wrote:quantum7
How you decelerate doesn't matter. To decelerate in x seconds by y m/s you need to apply certain force.
Unless you're being caught by Superman apparently.

The "super" is for "(supermassive) black holes at my command". (either classical black holes or something with quantum gravity) He doesn't so much "fly" as he bends spacetime to make him fall in the direction he points his hands. It's an elegant explanation for why people survive being caught by him. This is now my headcannon. (same goes for The Matrix, although "it's a computer program" covers it, too)

They're still going to be ripped to shreds by the deceleration. a = v2/2x, so the faster you're falling, the greater distance you will need in order to decelerate safely, and it's a squared-term relationship.

sevenperforce wrote:Imagining that you were on the right trajectory that you could get your arms and legs around it and basically hug it tightly all the way down, what would the limiting factor be? Hug strength? Frictional heating?

If the person can feel pain, the frictional heating wins, I believe.

Not if their shirt and pants are able to be the primary contact with the flagpole.

Supposing someone has dropped from 25 stories -- five times what would be feasible with a horizontal flagpole. That's going to be a falling speed of 38 meters/second, which is still under terminal velocity. If we go with 65 kg or so for mass, then the person's kinetic energy is a nasty 95.6 kJ. That's what we need to burn off on a slide down the flagpole.

Suppose that Rex is (again) in incredibly good physical shape and can exert a lot of force on the flagpole. Weightlifters can do 400+ Newton dumbbell flies using their chest muscles and 450+ Newton bicep curls, so if we imagine that Rex is in great physical shape and can exert this kind of force with both chest and arms, and pretend his legs can exert the same kind of grip, that's 3400 Newtons total.

With a generous sliding frictional coefficient of 0.5, that's 1700 Newtons of upward force retarding his motion, for a deceleration of 26 m/s2. But gravity is still tugging him down at 9.8 m/s2, reducing his effective deceleration to only 16.3 m/s2. He's going to need 44 meters of flagpole to slow to a stop. Not gonna work out too well for Rex.

What about the temperature? Let's say Rex can manage to wrap their body completely around the flagpole, so that they're able to spread out the friction over about 1 meter of it vertically. I'm going to guesstimate the diameter of the flagpole at a good solid 15 cm (this is going to be a big flagpole), making the surface area about 0.0056 m2. Not much to work with, really. That deceleration is going to take place over only 2.33 seconds, which means we need to somehow transfer 95.6 kJ to that surface area in that time period. Not enough time for heat to dissipate at all. The specific heat of water is, of course, 4.184 J/cc/degree, so if we use this value as a ballpark and say that the heating is limited to the top 1 cm layer of clothing, then that's 56 square centimeters to work with. He can manage 234.3 Joules per degree.

Which will heat his clothing by over 400 degrees Celsius. Guess you were right.

Tyndmyr
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### Re: What-If 0133: "Flagpole"

Oneknown wrote:So... I was curious how far you could fall and pull off this stunt given the variables used here.

First, I converted mi/h to km/h, because metric system! Then I reduced the velocity until the force was less than 10 kN.[1]

Then using this freefall calculator[2] I found that a freefall time of 1.840 seconds yields a velocity of 64.96 km/h, which is a fall from 16.60 meters.

So you can jump off a skyscraper and pull this trick off. You just need to make sure the flagpole is no further than 16 meters down.

[1] Wolfram alpha dot com/input/?i=100+lbs+*+%2865+kmph%29%5E2+%2F+%281.5+meters%29
[2] keisan dot casio dot com/exec/system/1224835316

If the flagpole has any flex in it, that would probably increase the distance. Now, granted, I am not sufficiently musclebound as to be towards the top end of this spectrum in any case, but if you've got flex in the pole, that eases the whole arm-ripping-off thing, and gives you some additional upward velocity as it snaps back. Seems like a win all round.

In fact, if you've got enough distance between you and the building, and between the flagpole and the ground, there should be an optimal solution of flex for essentially any height.

Provided, of course, that you're able to actually grab the flagpole at that speed, which seems eventually problematic.

rmsgrey
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### Re: What-If 0133: "Flagpole"

sevenperforce wrote:
Flumble wrote:
Neil_Boekend wrote:quantum7
How you decelerate doesn't matter. To decelerate in x seconds by y m/s you need to apply certain force.
Unless you're being caught by Superman apparently.

The "super" is for "(supermassive) black holes at my command". (either classical black holes or something with quantum gravity) He doesn't so much "fly" as he bends spacetime to make him fall in the direction he points his hands. It's an elegant explanation for why people survive being caught by him. This is now my headcannon. (same goes for The Matrix, although "it's a computer program" covers it, too)

They're still going to be ripped to shreds by the deceleration. a = v2/2x, so the faster you're falling, the greater distance you will need in order to decelerate safely, and it's a squared-term relationship.

Nope - the point of using gravity to decelerate them is that the effect of the gravity and the effect of the acceleration cancel perfectly, so it's like they don't change speed at all (except for the bit where the rate of change of their position relative to other objects changes)

speising
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### Re: What-If 0133: "Flagpole"

rmsgrey wrote:
sevenperforce wrote:
Flumble wrote:
Neil_Boekend wrote:quantum7
How you decelerate doesn't matter. To decelerate in x seconds by y m/s you need to apply certain force.
Unless you're being caught by Superman apparently.

The "super" is for "(supermassive) black holes at my command". (either classical black holes or something with quantum gravity) He doesn't so much "fly" as he bends spacetime to make him fall in the direction he points his hands. It's an elegant explanation for why people survive being caught by him. This is now my headcannon. (same goes for The Matrix, although "it's a computer program" covers it, too)

They're still going to be ripped to shreds by the deceleration. a = v2/2x, so the faster you're falling, the greater distance you will need in order to decelerate safely, and it's a squared-term relationship.

Nope - the point of using gravity to decelerate them is that the effect of the gravity and the effect of the acceleration cancel perfectly, so it's like they don't change speed at all (except for the bit where the rate of change of their position relative to other objects changes)

aka "free fall"

sevenperforce
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### Re: What-If 0133: "Flagpole"

speising wrote:
rmsgrey wrote:
sevenperforce wrote:
Flumble wrote:
Neil_Boekend wrote:quantum7
How you decelerate doesn't matter. To decelerate in x seconds by y m/s you need to apply certain force.
Unless you're being caught by Superman apparently.

The "super" is for "(supermassive) black holes at my command". (either classical black holes or something with quantum gravity) He doesn't so much "fly" as he bends spacetime to make him fall in the direction he points his hands. It's an elegant explanation for why people survive being caught by him. This is now my headcannon. (same goes for The Matrix, although "it's a computer program" covers it, too)

They're still going to be ripped to shreds by the deceleration. a = v2/2x, so the faster you're falling, the greater distance you will need in order to decelerate safely, and it's a squared-term relationship.

Nope - the point of using gravity to decelerate them is that the effect of the gravity and the effect of the acceleration cancel perfectly, so it's like they don't change speed at all (except for the bit where the rate of change of their position relative to other objects changes)

aka "free fall"

Well, unless the g-potential gradient is sufficiently smooth then you'll still end up with some nasty effects from tidal forces, though perhaps at an order lower than the alternative. Tidal forces still act the same on a body in free fall.

For an object in free fall, the only thing you feel is the difference between the gravitational acceleration at your feet and the gravitational acceleration at your head. This feeling can be approximated as if you were hanging upside down from your feet -- at this point, your body is experiencing the equivalent of a 9.81 m/s2 difference. Definitely uncomfortable, but not dangerous. If you were to add about 10 kg of weight to your upper torso, it would be more uncomfortable. 20 kg would probably begin to be painful, and 50 kg would be unbearable (your body is not nearly as good at stretching as it is at compressing). A difference of twice your weight would almost definitely do permanent damage unless you were in fantastic physical condition.

Let's say that around 1.25 g of difference is the point of significant discomfort. Assuming 2m for the height of a normal adult human, that means the maximum allowable gradient of Superman's gravitational deceleration field is about 6.1 s-2 or 6.1 billion eotvos (which is a weird unit if there ever was one). That's the distance derivative of acceleration...I think? Anyway, it's the same as 6.1 m/s2 per meter.

Stopping someone moving at terminal velocity (120 mph or 54 m/s) requires 5.5 seconds at 1 g, 2.8 seconds at 2 g, 1.8 seconds at 3 g, and so forth. The problem, though, is that the acceleration won't be a constant -- can't, in fact, be a constant. If we use a constant 6.1e9 eotvos gradient, though, then the slope of acceleration is a constant, greatly simplifying integration. The way it works out, Superman is going to need no less than 31 meters to make the gravitational gradient low enough to avoid causing significant discomfort.

rmsgrey
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### Re: What-If 0133: "Flagpole"

If you're lying across the gravity field (as though cradled in someone's arms, say) then the field gradient could be about six times as steep...

And a suitable arrangement of point-masses can produce any smooth gravitational field you care to specify - so, in principle, Superman could create a locally near-uniform field producing the required acceleration which rapidly dropped off beyond the target volume...

koreiryuu
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### Re: What-If 0133: "Flagpole"

Pfhorrest wrote:One big caveat that strikes me: you're falling from "a height" above the tallest building in your town, and "partway" down is the flagpole.

I signed in to post the same mistake. Randall uses "half-way," when the question gives the non-specific distance "partway."
I understood the question as, "At what distance would you have to fall in order for this to be possible, if it's even possible at all?"

(Rhyme unintended.)
Wilson: "How'd you get here?"
House: "Osmosis."

Flumble
Yes Man
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### Re: What-If 0133: "Flagpole"

Are there any good relativity physics simulators? I would really like to play around with some micro black holes and supermen.

rmsgrey wrote:And a suitable arrangement of point-masses can produce any smooth gravitational field you care to specify - so, in principle, Superman could create a locally near-uniform field producing the required acceleration which rapidly dropped off beyond the target volume...

Though, doesn't that require a lot of point-masses inside the body? Isn't that dangerous?

mathmannix
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### Re: What-If 0133: "Flagpole"

koreiryuu wrote:
Pfhorrest wrote:One big caveat that strikes me: you're falling from "a height" above the tallest building in your town, and "partway" down is the flagpole.

I signed in to post the same mistake. Randall uses "half-way," when the question gives the non-specific distance "partway."
I understood the question as, "At what distance would you have to fall in order for this to be possible, if it's even possible at all?"

(Rhyme unintended.)

do "have to fall" and "possible" rhyme for you?
I hear velociraptor tastes like chicken.

Quercus
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### Re: What-If 0133: "Flagpole"

mathmannix wrote:
koreiryuu wrote:
Pfhorrest wrote:One big caveat that strikes me: you're falling from "a height" above the tallest building in your town, and "partway" down is the flagpole.

I signed in to post the same mistake. Randall uses "half-way," when the question gives the non-specific distance "partway."
I understood the question as, "At what distance would you have to fall in order for this to be possible, if it's even possible at all?"

(Rhyme unintended.)

do "have to fall" and "possible" rhyme for you?

"fall" + "at all" maybe?

Pfhorrest
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### Re: What-If 0133: "Flagpole"

koreiryuu wrote:At what distance
would you have to fall
in order for this to be possible,
if it's even possible at all?

Meter would be nicer if you struck "in order" and "even".
Forrest Cameranesi, Geek of All Trades
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mathmannix
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### Re: What-If 0133: "Flagpole"

Yeah, I definitely read* it as three lines:
koreiryuu wrote:At what distance / would you have to fall
in order for this / to be possible,
if it's even / possible at all?

* - both present and past tenses
I hear velociraptor tastes like chicken.

sevenperforce
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### Re: What-If 0133: "Flagpole"

rmsgrey wrote:If you're lying across the gravity field (as though cradled in someone's arms, say) then the field gradient could be about six times as steep...

True. That's 12.6 meters, which is better, but still pretty significant.

And a suitable arrangement of point-masses can produce any smooth gravitational field you care to specify - so, in principle, Superman could create a locally near-uniform field producing the required acceleration which rapidly dropped off beyond the target volume...

The problem is the "rapidly dropped off" part. It's the dropping-off that causes the tidal distortion and stretching. You want it to drop off less rapidly, not more rapidly.

koreiryuu wrote:
Pfhorrest wrote:One big caveat that strikes me: you're falling from "a height" above the tallest building in your town, and "partway" down is the flagpole.

I signed in to post the same mistake. Randall uses "half-way," when the question gives the non-specific distance "partway."
I understood the question as, "At what distance would you have to fall in order for this to be possible, if it's even possible at all?"

(Rhyme unintended.)

Likewise.

rmsgrey
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### Re: What-If 0133: "Flagpole"

sevenperforce wrote:And a suitable arrangement of point-masses can produce any smooth gravitational field you care to specify - so, in principle, Superman could create a locally near-uniform field producing the required acceleration which rapidly dropped off beyond the target volume...

The problem is the "rapidly dropped off" part. It's the dropping-off that causes the tidal distortion and stretching. You want it to drop off less rapidly, not more rapidly.[/quote]

Except you really don't want the field to stay anywhere close to uniform beyond the target volume - it's not worth catching someone successfully if everyone nearby gets accelerated up into the air at the same time. You want a near-uniform field around the falling person and as near normal gravitational conditions away from them as you can manage.

sevenperforce
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### Re: What-If 0133: "Flagpole"

rmsgrey wrote:You want a near-uniform field around the falling person and as near normal gravitational conditions away from them as you can manage.

The problem arises during the person's transition from Earth's near-uniform field to a completely different near-uniform feed.

Neil_Boekend
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### Re: What-If 0133: "Flagpole"

If you match the change in gravitational field perfectly with the change in acceleration then there is no problem.
Mikeski wrote:A "What If" update is never late. Nor is it early. It is posted precisely when it should be.

patzer's signature wrote:
flicky1991 wrote:I'm being quoted too much!

he/him/his

ps.02
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### Re: What-If 0133: "Flagpole"

Izawwlgood wrote:As someone who is healing from surgery to repair a torn bicep,

Uh, biceps, like series and kudos, is already singular. No need to drop the s.

ps.02
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### Re: What-If 0133: "Flagpole"

sevenperforce wrote:What about the temperature? Let's say Rex can manage to wrap their body completely around the flagpole, so that they're able to spread out the friction over about 1 meter of it vertically. I'm going to guesstimate the diameter of the flagpole at a good solid 15 cm (this is going to be a big flagpole), making the surface area about 0.0056 m2. Not much to work with, really.

Wait, what? You're saying 1m × 0.15m × 3.14 = 0.0056m2? Because I get 0.47m2, which is pretty different. In fact I can account for about 56cm2 just on the grip side of the fingers of one hand.
That deceleration is going to take place over only 2.33 seconds, which means we need to somehow transfer 95.6 kJ to that surface area in that time period. Not enough time for heat to dissipate at all.

The metal pole won't carry any of that heat off? Seems like at least half the energy should go into the metal. Maybe even more than half - I suspect the pole would help to cool Rex's skin/clothing. It has 44 times as much surface area to heat (if the stopping distance is 44m), and at least initially, the heat is being carried away at 38m/s.

All of which is to say I think your calculation might be off by a factor of 160 or so.

sevenperforce
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### Re: What-If 0133: "Flagpole"

ps.02 wrote:
sevenperforce wrote:What about the temperature? Let's say Rex can manage to wrap their body completely around the flagpole, so that they're able to spread out the friction over about 1 meter of it vertically. I'm going to guesstimate the diameter of the flagpole at a good solid 15 cm (this is going to be a big flagpole), making the surface area about 0.0056 m2. Not much to work with, really.

Wait, what? You're saying 1m × 0.15m × 3.14 = 0.0056m2? Because I get 0.47m2, which is pretty different. In fact I can account for about 56cm2 just on the grip side of the fingers of one hand.

Oops! I did make a mistake, though not quite THAT large of a mistake. It should be 1m * 0.15m/2*3.14=0.2355m2, because 0.15 is the diameter, not the radius. I must have exponentiated when I should have divided.

So the actual area is 0.2355 m2, which I admit will definitely produce a different result.

That deceleration is going to take place over only 2.33 seconds, which means we need to somehow transfer 95.6 kJ to that surface area in that time period. Not enough time for heat to dissipate at all.

The metal pole won't carry any of that heat off? Seems like at least half the energy should go into the metal. Maybe even more than half - I suspect the pole would help to cool Rex's skin/clothing. It has 44 times as much surface area to heat (if the stopping distance is 44m), and at least initially, the heat is being carried away at 38m/s.

All of which is to say I think your calculation might be off by a factor of 160 or so.

You're right, I didn't consider the heat going into the pole. But the distance won't have much of an effect; what will have an effect is the heat capacity of the metal vs the heat capacity of Rex's skin/clothing. However, I was giving Rex's skin/clothing the same heat capacity as water for the purposes of the estimate, so that shouldn't be much of a difference. I suppose that Rex's skin/clothing WILL heat up significantly as it goes down, thus reducing its capacity for absorbing energy and transferring more energy to the pole proportionally.

But if we consider this difference to be negligible, then we should assume that half the energy goes into heating the pole and half the energy goes into heating Rex. So Rex, with a surface area of 2355 cm2 and a corresponding surface volume of 2355 cm3 will be able to manage 9853 Joules/degree. So he'll heat up by only 4.85 degrees Celsius...noticeably uncomfortable, but quite survivable. As long as the flagpole is at least 44 meters long, that is.

KittenKaboodle
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### Re: What-If 0133: "Flagpole"

sevenperforce wrote:Oops! I did make a mistake, though not quite THAT large of a mistake. It should be 1m * 0.15m/2*3.14=0.2355m2, because 0.15 is the diameter, not the radius. I must have exponentiated when I should have divided.

err, um, well, you did reduce your error by an order of magnitude, which is good. But...http://en.wikipedia.org/wiki/Circle#Length_of_circumference

In the spirit of "What if" (https://what-if.xkcd.com/84/). 0.15 * 3 is about 1/2, or, maybe 1, or 0.1. That last one looks good 0.15 rounds to 0.1, pi rounds to 1, and 1 rounds to 1: 0.1*1*1 = 0.1m2. So, looks like your answer is kind of ok after all (well, except for that significant digits thing)

what if wrote:"all that matters is getting in the right ballpark; that is, the answer should have about the right number of digits"

ps.02
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### Re: What-If 0133: "Flagpole"

sevenperforce wrote:It should be 1m * 0.15m/2*3.14=0.2355m2, because 0.15 is the diameter, not the radius.

Eh, if you're going to calculate using the radius, you need to use tau (6.28), not pi (3.14).
You're right, I didn't consider the heat going into the pole. But the distance won't have much of an effect; what will have an effect is the heat capacity of the metal vs the heat capacity of Rex's skin/clothing. However, I was giving Rex's skin/clothing the same heat capacity as water for the purposes of the estimate, so that shouldn't be much of a difference. I suppose that Rex's skin/clothing WILL heat up significantly as it goes down, thus reducing its capacity for absorbing energy and transferring more energy to the pole proportionally.

I still think the pole will help to cool Rex's skin as he slides down. At any given point the section of pole he is touching is close to ambient air temperature, and will feel cool to the touch because of how well steel conducts heat.

Your initial assumption about Rex making contact with the whole circumference of 1m of pole is pretty generous, though, as is the assumption of his clothing/skin being evenly heated to a depth of 1cm.

Finally, Rex doesn't have to come to a dead stop. A final velocity of, say, 8m/s (equivalent to a fall from 10 feet, if my mental estimate is close) seems comfortable enough. Thus the flagpole could be a bit shorter, or his grip a bit lighter, or his contact surface area a bit smaller, and he'll still be OK.

Wiz
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### Re: What-If 0133: "Flagpole"

ps.02 wrote:Finally, Rex doesn't have to come to a dead stop. A final velocity of, say, 8m/s (equivalent to a fall from 10 feet, if my mental estimate is close) seems comfortable enough. Thus the flagpole could be a bit shorter, or his grip a bit lighter, or his contact surface area a bit smaller, and he'll still be OK.

Exactly what I was going to say. (Well, except for "comfortable enough", maybe.) Per a quick Google search, your terminal velocity would be around 8 m/s using a parachute, so given some practice, you should be able to hit the ground at that speed and walk away without any injuries.

sevenperforce
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### Re: What-If 0133: "Flagpole"

ps.02 wrote:
sevenperforce wrote:It should be 1m * 0.15m/2*3.14=0.2355m2, because 0.15 is the diameter, not the radius.

Eh, if you're going to calculate using the radius, you need to use tau (6.28), not pi (3.14).

Holy crap, that's a horrific error. Remind me not to drink and math again.

You're right, I didn't consider the heat going into the pole. But the distance won't have much of an effect; what will have an effect is the heat capacity of the metal vs the heat capacity of Rex's skin/clothing. However, I was giving Rex's skin/clothing the same heat capacity as water for the purposes of the estimate, so that shouldn't be much of a difference. I suppose that Rex's skin/clothing WILL heat up significantly as it goes down, thus reducing its capacity for absorbing energy and transferring more energy to the pole proportionally.

I still think the pole will help to cool Rex's skin as he slides down. At any given point the section of pole he is touching is close to ambient air temperature, and will feel cool to the touch because of how well steel conducts heat.

When transferring heat, the rate of heat transfer is proportional to the difference in temperature between the two objects. This particular example is somewhat different, as energy is being converted into heat by friction, but I think the same principle holds: the hotter something is, the more slowly it absorbs heat. So that, I think, is where the cool metal pole will act as a proportionally better heat sink as Rex slides down it. But Rex will definitely never stop heating up until he comes to a stop.

The generosity of other estimates probably outweighs any advantage from the metal/skin comparison.

Finally, Rex doesn't have to come to a dead stop. A final velocity of, say, 8m/s (equivalent to a fall from 10 feet, if my mental estimate is close) seems comfortable enough. Thus the flagpole could be a bit shorter, or his grip a bit lighter, or his contact surface area a bit smaller, and he'll still be OK.

If he only needs to lose 30 m/s rather than 38 m/s, the flagpole will only have to be 28 meters high, which is quite the improvement.

Neil_Boekend wrote:If you match the change in gravitational field perfectly with the change in acceleration then there is no problem.

lesto
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### Re: What-If 0133: "Flagpole"

i would concentrate a bit more with the "indestructible" pole flag: after the arm rips off, what is the speed? and what is the high someone in a good shape CAN do the trick?

then we can add the "dynamic pole" (i thing you are missing its maximal structural integrity, at some point it will break) and see how much (if) improve. (bonus: how much heat will the pole generate during bending?)

ps.02
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### Re: What-If 0133: "Flagpole"

sevenperforce wrote:
ps.02 wrote:Eh, if you're going to calculate using the radius, you need to use tau (6.28), not pi (3.14).

Holy crap, that's a horrific error. Remind me not to drink and math again.

Or, a great example of why pi is confusing and should be replaced everywhere by tau.

The only cogent pro-pi argument I've heard is that, rather than think of it as "half the ratio of circumference to radius", you can think of it as "area of a unit circle". But I've never heard anyone teach it that way, and I am highly skeptical that a ratio based on area makes more sense as a fundamental constant than a ratio based on distance.

Other pro-pi arguments ("if you're using calipers, you measure diameter, not radius", "lowercase tau is used for other stuff in physics", "πr2", Euler's Identity, etc.) don't really hold water.

Wait ... I guess there was another pro-pi argument that carried weight with me. Or, more precisely, whose weight I am probably still carrying. And that is the pie that somebody ordered last week that I ate three or four pieces of. A delicious and satisfying argument, but ultimately still not convincing.

sevenperforce
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### Re: What-If 0133: "Flagpole"

Eh, I only mixed it up because I calculated the volume of a cylinder rather than its area, then corrected my mistake but used radius again instead of diameter.

The simplest reason, I think, that pi is taught rather than tau is a historical one. I remember learning about Aristotle using polygons to find the ratio of a circle's diameter to its circumference back before I knew three digits of pi.

Jackpot777
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### Re: What-If 0133: "Flagpole"

For our non-baseball-loving readers (or just those in other countries), the baseball player's last name "Pujols" is pronounced "poo holes".

Maybe your name is something funny in another language too.

mathmannix
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### Re: What-If 0133: "Flagpole"

Jackpot777 wrote:Maybe your name is something funny in another language too.

Nope, mine means nothing.
I hear velociraptor tastes like chicken.

x7eggert
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### Re: What-If 0133: "Flagpole"

mccdyl001 wrote:Came to say the same as SuperSteve.

Some quick estimating has 3kN = 700lb or 350kg in proper units. So 10kN = 1 ton. 100kN = 10 tones. Of nearly instant downward force. On a flagpole sticking out the side of a building? That thing is either bending or pulling out the wall long before the arms are torn off.

Those are imperial pounds (about 450 g) as in "we in the US are proud to have been a colony of ancient GB". 3 kN =^= 600 metric pounds (500 g) in Europe, but sane people don't use metric pounds for counting (One pound is OK, two pounds is "insane").

OTOH, pounds and kg are never proper units of force.

x7eggert
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### Re: What-If 0133: "Flagpole"

Whizbang wrote:
Neil_Boekend wrote:quantum7
How you decelerate doesn't matter. To decelerate in x seconds by y m/s you need to apply certain force.
Unless you're being caught by Superman apparently.

Evidently Superman has psychokenetic powers that he uses unconsciously to blunt the effects of his strength (I dunno, ask Second Talon), so I can only assume this also is used when he catches people to, somehow, catch them without breaking them.

Gwen Stacy(Spider-Man) was not that lucky.

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### Re: What-If 0133: "Flagpole"

Here is a quote from the article that isn't quite correct:
A gymnast on the uneven parallel bars pushes the human body to its limits while performing maneuvers very similar to Rex's flagpole stunt.[3] A 2009 study used 3D motion capture to measure the forces involved in an elite gymnast's routine. They found that the athlete's hands exerted a force of over 3 kN on the bar at the bottom of a swing. In other words, the gymnast was briefly supporting almost 700 lbs of weight.
(emphasis mine)

Here is a quote from the linked study's abstract (I cannot access the paywalled paper at this time)
The displacement of a calibrated horizontal bar is used as a measure for forces acting on the bar itself during dynamic performances in artistic gymnastics. The high bar is loaded with known forces...
(emphasis mine)

As a Gymnast myself, this part was of particular interest to me.

There are a few things that should be cleared up.

The first is that the uneven parallel bars are an event in Women's gymnastics. They are made from wood or a Composite (depending on the age of the equipment). The Parallel bars are made from a similar material and are an event in Men's gymnastics.

The Horizontal bar or "high bar" is a steel bar and is an event in Men's Gymastics.

The linked study clearly says in it's abstract that they studied the High bar. This makes sense because you routinely see larger displacements in this event's bar than any of the others (though you might see more forces in the old "swinging rings" event, I don't know much about this, and it hasn't been a thing for at least 50 years so it's safe to leave it out I think)

You can see the sort of deflection you get in a highbar routine in this video: (NOTE 1)
(three release moves in a row and especially in the dismount). The specific skill is a "tap swing" or "chinese tap", which in intended to put a large amount of energy into the bar prior to a release move or a dismount.

So, even if it weren't clear from the paper's abstract, they are obviously studying the deflection of the metal horizontal bar (or "high bar" as it is called by everyone other than FIG officials). The wood or composite parallel bars and uneven bars have less forces on them.

If you watch videos of parallel bars and uneven bars you may see more deflection than on the men's highbar, but this is because of the different material used and different set-up. Why Randall used the "uneven parallel bars" term instead of the "high bar" term in the paper could be for a number of reasons:

1. The people writing the paper only studied the uneven bars and erroneously called it the "high bar" in their abstract. I can't imagine why, as the tallest gymnasts are male and therefore they can put a lot more energy into the horizontal bar than a shorter female gymnast can put into their uneven bars. People writing papers aren't usually Gymnastics expects so I guess I can't really say if they studied the right event or not. I suspect this is not the case as "horizontal bar" is listed as a keyword.

2. A simple mistake on Randall's part, confusing the two events is common, and the official names don't help.
---

Anyway it was an excellent article, and I enjoyed it despite the factual innacuracy.

NOTE 1: I cannot link the youtube video because of the spam filter, but the video is ?v=UPJ1YXI6D54 "Athens 2004 mens all around part 8" and the specific set of moves starts at 0:31