The physics relates to something called moments (force x perpendicular distance). It turns out that the passengers – whether they knew it or not – did just the right thing by pushing the wings. Because there was a large distance between the wings and the wheels it gave the people pushing greater leverage. “It’s the principle of the spanner – the longer the spanner the less force you have to apply in order to shift the nut,” Andrews says.

http://www.bbc.co.uk/news/blogs-magazine-monitor-30235844

Is it just me, or is Andrew talking rubbish? Pushing a plane horizontally, it doesn’t matter where you push.

Is it on a conveyor belt?

😀

Not if the leverage from the wing caused it to rotate around the vertical axis, thus breaking the stiction in the bearings so it could then start rolling in a linear fashion.

Maybe he means they pushed the wings alternately, pivoting round the further set of locked wheels. They wouldn’t have been able to get much force behind it with their arms over their heads though, looks dubious to me.

I thought of that, but that’d be pretty hard with the nosewheel, and the article talked about pushing it backwards.

Is it on a conveyor belt?

Quiet in the cheap seats!

johnners is about the only explanation which makes any possible sense, but then to get leverage they’d have to push on the ends of the wings, not near the root where they’re pictured. Almost certainly yet another journo who doesn’t understand physics 😉

I’ve pushed a lot of light aircraft and gliders and I’m very convinced that that is bollocks. To rotate, yes it makes sense, but move along in a straight line? No, simply not so

I’m no good at this sort of stuff, but it sounds sensible to me….. wheel = nut, undercarriage strut = spanner, with one proviso- if they managed to push the plane backwards before the bearings were freed then their energy was used to overcome the friction between the ice and wheel rather than between the wheel and bearings?

I look forward to imminently being told why I am wrong.

A spanner rotates. This airplane isn’t rotating, it’s moving in a straight line.

The basics of leverage is one end of the lever moves further than the other. A prybar with both ends the same length would be a rubbish prybar. But if the airplane is moving in a straight line,all parts of the airplane move the same distance, so there’s no mechanical advantage.

I’ve just got off a plane and that had 2 engines, both of which rotated. Nice

Pretty sure the pilot just stuck it in reverse.

lol

don’t you mean “aeroplane” ?

is it just me, or is Andrew talking rubbish?

He is indeed talking a load of utter, utter, cobblers.

Almost certainly yet another journo who doesn’t understand physics

says Dr John Andrews, visiting fellow in physics at the University of Bristol.

Is this guy on drugs? I don’t much rate the idea of going to his physics lectures.

Reckon the journo may have not been able to read his notes and googled & printed something he doesn’t quite get 😉

The mechanical advantage that they are referring to is the moment require to overcome the friction caused by the frozen grease, not the friction along the ground. The plane may not be rotating but I would hope that the wheels are.

I wouldn’t have thought (based on precisely nothing) that the grease in the bearings would freeze, but could the pads have frozen onto the disks?
Either way, maybe his theory isn’t as crackers as first appears.

The mechanical advantage that they are referring to is the moment require to overcome the friction caused by the frozen grease, not the friction along the ground. The plane may not be rotating but I would hope that the wheels are.

Yes, but the plane is not rotating around the wheels, so there’s no mechanical leverage from pushing the ends of the wings. So the moment is the size of the wheel only.

It’s like pushing a car to bump start it – doesn’t matter if you push from the middle or a corner, it’s still the same force required.

The distance he’s taling about is the vertical height from the wheels to the wing (at least that’s how I read it) not the horizontal distance along the wing. To use the car analogy it is like trying to bump start by pushing the bumper rather than roof.

Bencooper is right – this is simply pushing the vehicle along a path, not rotating it around a fixed point. It’s even worse, as the place where there is a rotation is in the length of the pushers’ bodies – to try and push something into a high shelf is harder than pushing the same mass from the centre of your body. All those outstretched arms mean I doubt the passengers really made any significant difference.

The distance he’s taling about is the vertical height from the wheels to the wing (at least that’s how I read it) not the horizontal distance along the wing.

Even so that doesn’t help. The wheel hub will move at exactly the same rate as the wing, the cockpit and the top of the tail because it’s all being pushed horizontally.

And even if somehow you removed the nosewheel and rocked the plane back and forth, it would rotate around the contact patch not the bearings, so still wouldn’t help.

It’s like pushing a car to bump start it – doesn’t matter if you push from the middle or a corner, it’s still the same force required.

Unless you’re applying the force along a vector in line with the CoG then some of the push is being wasted, pushing at the corner will result in a moment about the CoG; this moment is reacted by side forces in the tyres.

In the plane example, I think that the plane would rotate around the wheels (around the Z-axis), unless the CM and ACM moments created by people pushing on each wing were in absolute equilibrium then it has to; if you consider the 3 points of the undercarriage as being fixed with a variable release force. Then as the force is applied to the wing then one of the points will break free first (due to variations in the applied moment and the friction in the wheels) so the plane will rotate; then as the pushers compensate the other wheels will break free. It’s like when you have to ‘walk’ a washing machine into place under a worktop, it always skews off at an angle to start with.

I hope this ‘expert’ doesn’t teach. That’s a really bad understand of moments and in actual fact that would be an inefficient way to push a plane due to energy losses in the wing flex and suspension movement. Pushing on the legs of the undercarriage would be the most efficient method but it looks like bullshit to me

Unless you’re applying the force along a vector in line with the CoG then some of the push is being wasted, pushing at the corner will result in a moment about the CoG; this moment is reacted by side forces in the tyres.

Okay, yes, that was a bad example – better to say two people both pushing in the middle or at the corners.

In the plane example, I think that the plane would rotate around the wheels (around the Z-axis), unless the CM and ACM moments created by people pushing on each wing were in absolute equilibrium then it has to

The nosewheel would prevent that happening, unless you deliberately turned the nosewheel 90 degrees. Because the nosewheel is so far forwards, the force required to scrub the nosewheel sideways would be much greater than any offset force on the wings.

I’d bet that, even if everyone just pushed on one wingtip, the aircraft would still move linearly instead of rotating.

Nobody is pushing in the one place where there really is an advantage to be had, which is the tops of the tyres.

Nobody is pushing in the one place where there really is an advantage to be had, which is the tops of the tyres.

Spot on.

What utter horsepoop. Visiting fellow my hairy arse. Ben, I’m rather disappointed that you doubted your own understanding of the problem 🙂

(Anyone’s welcome to that set-up for a line about fellows visiting my hairy arse. Knock yourself out.)

I thought I completely understood it, but as usual STW made me doubt myself 😉

pdw has got it.
I’ve pushed a few dead trucks around and pushing the tops of the tyres gives you a 2:1 advantage over pushing the chassis or bodywork. Even more so if the brakes or bearings are dragging, by putting a turning force directly on the wheels.

Fag packet calculation;
Reaction force at nose wheel = 33kN, coefficient of friction =0.6 so slipping load for nose wheel = 20kN

Assume that thrust of engines wasn’t quite enough to break the friction of the bearings, and that all bearings are equally stuck, so breakout force per wheel = 14kN, giving a nose wheel breakout force =28kN and main gear breakout force = 56kN

So a linear push would need 28+56+56 = 140kN to release the stuck bearings
If you pushed on the top of the tires 14+28+28 = 70kN

Assuming pushing force on one wingtip and taking moments about one of the main gear wheels;
20*push force = (56*10)+(28*5)+(20*18)
Push force = 53kN

(Wingspan = 30m, Distance between main gear wheel = 10m, Distance to nose wheel = 18m)

The guy who’s wrong seems to be assuming that the wheels are clamped to stop them from rotating (so if things got flexy the lever arm of the undercarriage could help rather than pushing at the hub as the tow vehicle would do) when, as most above posters have alluded to, the moment arm is just determined by the radius of the wheels.

Seems like a wind up – although passengers plus all other available means of pushing would obviously be more likely to get things moving. Especially as the thickness of the leading edge skin is probably far too thin for the passengers to push hard on it without putting dents all over, regardless of any other points.

Of course if it’s on a conveyor belt then it depends which way the wind is blowing.

The guy who’s wrong seems to be assuming that the wheels are clamped to stop them from rotating

Which is a reasonable assumption given the article states;

leaving the brake pads frozen in the minus 52C temperature……there was an additional problem in that the wheels were stuck with frozen grease

If the wheel were clamped to the ground, then pushing higher would make a difference (although not a lot, as nothing’s going to happen until you push it hard enough to lift the nose wheel, which isn’t going to happen)

If it’s just frozen brakes, then the moment is the radius of the wheels, wherever you push.

Anybody who thinks pushing a plane on the wingtips would give them more power is an idiot. Yes; if you wanted to spin the plane around to face in a different direction you woud push the end of the wing so as to get the longest leverage.

Here in Nigeria I’ve seen a BAC 1-11 pushed by about 50 staff from the airline office when it got wedged between a wall and a broken-down fuel tanker. The pilot applied reverse thrust and gave it full gas but a wingtip just scraped alarmingly along the side of the fuel tanker making a dent and removing the paint so we all got out and waited in the shade under the wings until all the clerks from the airline’s office walked out and pushed on the leading edges of the wings. The ‘plane moved smoothly backwards out of the gap and we all jumped back in and shot off to Kano, taking the bend from the taxiway to the runway on one wheel as the pilot yelled: “Cabin Crew: doors to automatic! Let’s get de hell out of here!”

I think spent a day with Chris Sheperd once and he knows what he’s talking about

I think we now have the answer. If the wheels are “stuck” to the ground the pushing higher up might help

(Imagine a front wheel in a fork and the wheel clamped to the ground and the front brake on. Extra leverage helps us rotate the contact point)

If the bearing are jammed or the brakes on the extra level doesn’t help if the wheel is fixed to the ground. In this case the moment about the axle is simply the frictional force multilpied by the wheel radius

(Imagine a front wheel in a fork and the front break on. using the end of the steerer as a leaver doesn’t help if the wheel can just rotate about the contact point)

No, that’s nonsense. Wherever you push on the plane or bike (wheels excluded), the force is only transmitted to any given wheel radially from its axle. Push as high or as low as you like, but the moment at the stuck bottom of the wheel can never exceed the pushing force times the radius of the wheel, because that’s where the force on the wheel is applied.

My instinctive thought is that the best height at which to apply the pushing force is at the height of the axle, but until I can find a pen and an envelope I’m going to be cautious on that 🙂

(Edit: hmm, thinking more bout it, wheels frozen at the ground *and* at the bearing means the axle force transfer’s a little different up to a point… might need two envelopes. And jeez, it’s been a while…)

Having pushed quite a few aeroplanes about over the years, pushing on top of the wheels gets them rolling the best.