57
$\begingroup$

I just ran into this meta-answer which includes this image:

enter image description here

My initial guess is that the wrecking ball is far too heavy for the helicopter to lift, but I'd be interested in the numbers to see just by how much.

  1. From this silhouette, can the make/model of helicopter be determined? (I'm hoping this image is based of some real-life craft)

  2. If so, presumably the vehicle length is known, and the diameter of the wrecking ball may be calculated (spikes may be ignored) according to the scale of this image

  3. Assume the wrecking ball is solid mild steel with a density of 7.85 g/cm³. The mass of the wrecking ball can then be calculated

  4. If the make/model is known, then presumably the max payload weight is also known, and we can see if staying aloft is a possibility or not.

Close-voters: I asked this question on Aviation.SE because the key piece of information I was missing was the identification of the craft in the cartoon image. I think is on-topic here.

Improve this question
$\endgroup$
15
  • 8
    $\begingroup$ Even if the ball can be lifted I can be more worried about it's swing nasty effects on the fly not to say the catastrophic (for the helicopter) effect of it hitting the ground $\endgroup$
    – jean
    Commented May 22, 2018 at 12:09
  • 9
    $\begingroup$ Wrecking balls are not usually solid steel - it's much cheaper to fill them with concrete instead. $\endgroup$
    – Toby Speight
    Commented May 22, 2018 at 12:51
  • 30
    $\begingroup$ You know that's photoshopped, right? Quite apart from anything else, wrecking balls don't have spikes. $\endgroup$
    – David Richerby
    Commented May 22, 2018 at 13:55
  • 26
    $\begingroup$ Existing answers give analysis on "helicopter lifting a metal sphere of this apparent size", but if you're also interested in whether a helicopter with a dangling wrecking ball is possible/useful at all then it reportedly actually happens. $\endgroup$
    – Kamil Drakari
    Commented May 22, 2018 at 14:59
  • 7
    $\begingroup$ Less of a "wrecking ball" and more of a morningstar flail, I'd say. $\endgroup$
    – Beanluc
    Commented May 22, 2018 at 23:43

9 Answers 9

Reset to default
119
$\begingroup$

EDIT

Huge apologies to one and all, in my first version of this answer I mistyped 184 as 84, and then proceeded to do all my calculations based on that. This is the corrected version.

Also, huge thanks for all the up-votes and comments. I'm so glad it's not just me that is enjoying this question immensely.

The radius of the ball in the image is 21 pixels. The length of the fuselage is 184 pixels, which is 50' 7.5" (15.43m) according to widely available statistics. Hence, a pixel represents approximately 84mm

So, the radius of the ball is 21 x 84 = 1764mm = 1.764m

And the volume of the ball (if solid) is 23.0 m³.

For it to be liftable by the helicopter, using Pilothead's quote of 9000lb (4082kg), the density would need to be 177.5 kg/m³. Mild steel, at 7,850kg/m³, is way too heavy. Expanded polystyrene has a density of 28-45kg/m³, so that would be feasible.

However, if the ball were not solid, and were made of mild steel, the total volume of steel permissible would be 0.52m³. At an outer radius of 1.764m the inner radius must be at least 1.750m to keep the volume of metal down to 0.52m³, so the ball would only be 14mm thick (9/16") and would only cost about $3,300.

So, let's try Titanium (darling of the aircraft industry). With a density of 4,506kg/m³, we could have a volume of 0.90m³. That would allow a thickness of 24mm (15/16"). However, the price of that much titanium would be $229,500 (today's prices - 6 years ago, it would have been over a million US dollars).

Aluminium, however, has a density of 2,700kg/m³. So now we could have a volume of 1.51m³, allowing a thickness of 40mm (1 9/16"), and would only cost $9,360

Pumice, at 250kg/m³, would allow a volume of 16.33m³, allowing a thickness of nearly 0.6m (23½"). That would be very sturdy, I reckon. I can't find a price for a solid lump of pumice that size though.

Incidentally, the people in the picture are also about 21 pixels tall, making them around 1.76m (5'9") tall.

This has now taken up about 3 hours of my time, which like all other hours up to now, are hours I'm never going to get back. However, it has been so much fun.

FURTHER EDIT

Following a suggestion from Jules in the comments, I've discovered Lithopore 75-150, which is a ridiculously low density foamed concrete. This could give us the magical solid sphere that we want, with sufficient strength to squish multiple personnel without significant damage to itself. At the 75 kg/m³ end of the range, we could possibly even have some spikes, and a non-negligible cable mass too.

More maths and engineering research required.....

Improve this answer
$\endgroup$
18
  • 18
    $\begingroup$ Or the photo is completely real, the ball is made of marshmallow, and the people are just twice as close to the camera as the helicopter! $\endgroup$
    – jumps4fun
    Commented May 22, 2018 at 9:16
  • 15
    $\begingroup$ Marshmallow, alas, is still way too dense at about 500kg/m³. I do like the idea of the people being closer though, and being completely unrelated to the helicopter situation. $\endgroup$
    – Phil M Jones
    Commented May 22, 2018 at 9:19
  • 16
    $\begingroup$ I tend to doubt that an 8mm thick sheet of aluminum would be "flimsy" according to most people's definition of the word. Certainly it would be adequate for squishing soft-ish things like a couple of fleeing giants. $\endgroup$
    – aroth
    Commented May 22, 2018 at 9:54
  • 8
    $\begingroup$ The ball is not made of steel or marshmallow. The caption makes it clear: it is rock. Don’t say this to gravity. $\endgroup$
    – Emil Jeřábek
    Commented May 22, 2018 at 12:45
  • 7
    $\begingroup$ Hmmm, Emil, you may have a point. I'll assume it's Pumice (a volcanic rock) at a density of 250kg/m³ (half that of marshmallow, and a mere 9.26% of the density of Aluminium). As it's impossible to cast pumice, you could find a lump of it, machine it down to a sphere of 3.86m radius, cut it in half, then machine out the inside of each to leave two hemispheres of 9cm (3½") thickness, then somehow join them back together. You would then have a 9000lb ball of rock. $\endgroup$
    – Phil M Jones
    Commented May 22, 2018 at 13:00
51
$\begingroup$

The helicopter would be a UH60 Blackhawk. Max payload on a hook is 9000lb. At your mild steel density this is half a cubic meter, so the picture as you describe is impossible.

UH60

Improve this answer
$\endgroup$
5
  • 5
    $\begingroup$ Eyeballing it, I'd say the ball diameter is about the same as the height - 5.13m. Thus ball's volume is ~70m³ and mass is ~555 metric tonnes! Did I get that right? Definitely no fly. $\endgroup$
    – Digital Trauma
    Commented May 22, 2018 at 0:58
  • 13
    $\begingroup$ Maybe they are cheating and the ball is actually hollow :) $\endgroup$
    – Jeremy Friesner
    Commented May 22, 2018 at 4:50
  • 4
    $\begingroup$ Look at the slack on the rope though! $\endgroup$
    – transistor09
    Commented May 22, 2018 at 6:49
  • 4
    $\begingroup$ The helicopter isn't a solid hunk of metal. Why would the ball be? $\endgroup$
    – aroth
    Commented May 22, 2018 at 9:42
  • 6
    $\begingroup$ @aroth: because that's what the question says it is: "3) Assume the wrecking ball is solid mild steel" If we can say its a helium filled mylar balloon then the question becomes pointless. Or even more pointless :) $\endgroup$
    – RedGrittyBrick
    Commented May 22, 2018 at 10:38
48
$\begingroup$

Of course the photo is absolutely real. I know a guy who knows a guy who says he met the guy in front!

3) Assume the wrecking ball is solid mild steel with a density of 7.85 g/cm3.

That's where you went wrong. The angle of the line and the drag equation allow us to determine the ball's mass, given a known velocity. The ball presents a frontal area of 11.6 m2, which at 80 m/s, a high estimate for the helicopter's speed with an external load, would create ~18 kN of drag. Call it 20 with the spikes.

The cable is angled 35 degrees off vertical, which means the horizontal force on the line is approximately 0.7 times the vertical force. That indicates the vertical force to be approximately 28.5 kN, and thus the ball to weigh ~2,800 kg.

This results in force on the hook of about 42 kN. So, the load is well within the UH-60's capabilities, and the ball's density at 240 m^3 volume is a mere 11.65 kg/m^3. Still quite an impact even with the kinds of foam available at these densities, though!

For a ball of solid steel (Fv~=20 MN) to deflect that much, the "helicopter" would have to be doing something in the range of 3,000 m/s, depending on the ball's exact Reynolds number.

Improve this answer
$\endgroup$
9
  • $\begingroup$ the total force in the angled case is greater, not less. The vertical force being reduced doesn't lighten the load on the helicopter. $\endgroup$
    – Erin Anne
    Commented May 22, 2018 at 19:25
  • 1
    $\begingroup$ Yes. The actual load on the hook is (Fh^2+Fv^2)^0.5, or about 42 kN. But a force of more than 1 g is already accounted for in its design. The angle of the line allows us to determine the ball's mass, given a known velocity. At a lower velocity, the ball would be lighter. For a ball of solid steel (Fv=20 MN) to deflect that much, the "helicopter" would have to be well over Mach 10. [edit: Added to the answer] $\endgroup$
    – Therac
    Commented May 22, 2018 at 20:09
  • $\begingroup$ Ohhhh I completely misunderstood the argument you were making. The edit helps. Thanks! $\endgroup$
    – Erin Anne
    Commented May 22, 2018 at 20:46
  • 4
    $\begingroup$ This is assuming stable flight. The line might be angled because the helicopter maneuvered to catch it's target. $\endgroup$
    – Loren Pechtel
    Commented May 22, 2018 at 21:57
  • 1
    $\begingroup$ That would take a longer calculation. The ball's got gravity to do some catching up of its own - helicopters can't accelerate so fast as for the image to work with arbitrary mass. $\endgroup$
    – Therac
    Commented May 23, 2018 at 4:36
21
$\begingroup$

Indeed helicopter/wrecking ball combo exists and do some useful job, check this video.

The film shows how the Norwegian Public Road Administration cleared a potential rockfall debris prior to establishing an additional rockfall barrier fence.

First they smash the cliff with a wrecking ball, then drop water over the same area using a water bucket.

Improve this answer
$\endgroup$
3
  • $\begingroup$ I'll note they use a significantly smaller ball. Looks to be about 1m to 1.5m in diameter. $\endgroup$
    – Schwern
    Commented May 24, 2018 at 18:22
  • 2
    $\begingroup$ @Schwern Yep, something like that. Anyway the task still very dangerous, especially when the rock must be smashed somewhere int the middle of the height and the rope is not long enough for helicopter to be above the cliff. $\endgroup$
    – Alex Zherebtsov
    Commented May 25, 2018 at 11:00
  • 2
    $\begingroup$ @Schwern - Yeah, small but heavy for sure. I think they said the ball was 1,800 kg (about 4,000 pounds), so almost half the (9,000 pound) max load of the helicopter depicted in the question. $\endgroup$
    – Kevin Fegan
    Commented May 28, 2018 at 8:24
6
$\begingroup$

If the ball is empty inside, it can probably be made even larger than the helicopter. ZMC-2 was a balloon made of metal, light enough to fly on its own. It even has some rudders that resemble the spikes of the pictured ball. It is probably possible to adjust the lift of the balloon slightly below its weight, so that a helicopter could "lift" it up.

ZMC-2 was made of Alclad, the aluminium alloy.

Improve this answer
$\endgroup$
4
$\begingroup$

Even if the wrecking ball is within the payload capacity of the helicopter, a problem will remain.

The big issue with any sling loads is oscillation. A load that swings through an arc exerts a load on the helicopter that pulls the helicopter off the horizontal. If the swing angle is big enough, you'll lose enough lift to lose control and crash. So pilots are careful to keep the load in a stable position.

A wrecking ball is the antithesis of this: it only works when it's swinging. So instead of the 9000 lb quoted above for the maximum load, the ball has to be much lighter to make sure you can keep flying when the ball swings around.

Improve this answer
$\endgroup$
2
$\begingroup$

As Therac discusses, the ball has a manageable mass. Now as to concern about speed, those calculations assume static forward flight, and it is likely this ball is being swung all around. That could account for the slack in the lines.

As far as stabilization, we can take a cue from the other people who control swinging loads from high above them: those who load/unload container ships, dropping the container right on the pins of the lower container or the haulaway truck or train. Their rule is simple: Always stay above the load. If you are at position X and it is at position Y, X - Y becomes new impulse to make it swing. You don't want to do that except to cancel out existing swing, and even then you have to be "in phase" (or to be more precise, not).

Of course in the wrecking ball business, you'd want to swing it. But it'd take careful management (especially post-strike!) to make sure the cable didn't jerk the helicopter in unexpected directions and cause an areodynamic upset, or throw debris at the helicopter. Even dust repeatedly getting into the turbine could be a problem.

Improve this answer
$\endgroup$
1
$\begingroup$

Of course it's photoshopped, but I think this is meant to depict a mine, not a wrecking ball.

Per wikipedia, "Floating mines typically have a mass of around 200 kg, including 80 kg of explosives" - see https://en.wikipedia.org/wiki/Naval_mine. That's well within the carrying capacity of a Blackhawk.

Improve this answer
$\endgroup$
0
$\begingroup$

The image is fake. There are no known slingloads for the UH-60 which correspond to the image shown.

Improve this answer
$\endgroup$
2
  • 10
    $\begingroup$ Fake? Are you sure??? Maybe it's a concept demo by the 160th - they do wild stuff like this, don't they? :-) $\endgroup$
    – Ralph J ♦
    Commented May 22, 2018 at 3:54
  • 2
    $\begingroup$ Not to mention the stability and CG havoc being played if that thing was to swing as depicted. This answer is a reasonable frame challenge, since the question is based on utter nonsense. As an answer, though, it could use a bit more meat to spell out why the situation is absurd. $\endgroup$
    – KorvinStarmast
    Commented May 22, 2018 at 21:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .