Maths Headscratcher

[Stu]

'lose' the 'scope?

Put it outside the dock chained up? The thicker the chain the quicker it will go.

Trust me I'd find a way, I hate the bloody things. ;) I'd quite happily risk my job by breaking in one night with an axle grinder, and take the bloody thing to bits! Either that or push it into Poole Harbour. Mind you, I don't know how I'd get it anywhere near the harbour without looking a tad silly...

 

Answers on a postcard.

 

Stu

[Piers Shepperd]

I think I just found Massive Jim.

 

Picture of Massive Jim

 

Does this help?

[Thirdtap]

Monkey suits?

We have been working in teenage muntant ninja turtle outfits from the panto ten years ago.

Should I ask management for funding?

There was talk of euro funding for Captain Pugwash gear but nothing ever come of it.

[robloxley]

I was thinking about this the other day, though in a slightly different situation - how exactly vertical does the scope need to be when you're focussing on a rake? It's not easy with just that bubble spirit level thingy.

 

Anyway, I've not yet dusted my mechanics notes off so haven't done the calculations. However, I think the assumptions that were made by a previous poster are so great as to make the result sufficiently inaccurate as to be misleading. For a start, the scope has a width, so regardless of how the person at the top moves his weight should always stay within the base. Surely the major cause (if it really does happen) of scopes tipping is people trying to pull themselves sideways or along a bar from the top - that's the only way I can see you generating enough lateral force.

Also, an 83kg Dave at the bottom will have a weight of about 830N, and should be able to exert more than a 2.5kg horizontal force!

[Bryson]

Also, an 83kg Dave at the bottom will have a weight of about 830N, and should be able to exert more than a 2.5kg horizontal force!

 

You're right of course...10N per kilgram , not 1N...

 

If it make you feel any better, I miscalculated the other side of the equation by the same factor of 10...so the calculation is still (nominally) correct.

 

As for width...again, you're right. I have a mechanical engineer friend of mine looking at the calcs and may supply us with a stupendously complex calculation that takes width, weight of tallescope (uneven) and factors like flex into account...if I understand a word of it, I'll post it...

 

Anyone got any better ways to calculate how much of your own body mass you can supply in horizontal force? That's the real problem with the calcs - 30% was a "top of my head" kind of figure.

 

Or perhaps someone would like to just try it out for real? ;)

[James]

You could always go round the loading doors of theatres up and down the country, with a set of bolt cutters, and start up a second hand tallescope business?

 

So,

 

All the tallescopes from around the country mysteriously go missing, at which point there is a whole bunch of finance cases made for genies. Then you anounce your second hand tallescope business, but who are you aiming it at as all these theatres now hopefuly have nice new genies? ;)

 

James

[James C]

I think these posts go some way to answering the perennial question about the value of a formal education in this business...or at least the need to know a mechanical engineer if you haven't had the education ;)

[Ben...]

If The weight on the top is 83kg, the tallescope is 7m with it's mass evenly distributed along its length, the width of the base is 2m, and the weight on the top is 1m out from the centre, there is one sideways force (from Wee Johnny) of 27.6 kg at 1.5m from the base, then the 'scope can get to 13degrees from vertical before it tips ;)

 

For the mathematically inclined, I got this by taking the moments through the pivot point;

The scope is h metres tall, with the platform and base both W metres from centre to edge (makes it simpler)

If F1 is the force on top (Doomed Dave)

F2 is the lateral force (Wee Johhny) parallel to the ground, x metres from the base

F3 is the mass of the 'scope, acting h/2 metres from the ground

The reaction forces on the pivot can be ignored because that's where we're taking the moments

The angle from the vertical is T

 

(and force = mass * 9.81)

 

so sum of moments =

F1*h*sin(T)-F2(W*sin(T) + x * cos(T)) + F3( (h/2) * sin(T) - W * cos(T) )

 

We want the point of equilibrium, beyond which the thiing is moving (either back to stability, or onwards to certain death)

so set that lot = 0.

Rearrange and solve for T, gives the value above.

 

The full answer being;

T = arccos(2*(F1*h-F2*W+1/2*F3*h)/(4*F1^2*h^2-8*F1*h*F2*W+4*F1*h^2*F3+4*F2^2*W^2-4*F2*W*F3*h+F3^2*h^2+4*F2^2*x^2+8*F2*x*F3*W+4*F3^2*W^2)^(1/2))

 

so there

 

Ben

[Ben]

Ben that would work except you have assumed that the mass evenly distributed along its length, this is not really the case as it depends on how much of the scope is extended, and it is designed to not have the weight distributed evenly as when it is vertical you woant the weight to be at the bottom as this lowers the center of gravity making it less likly to tip in the first place, hence my previous post about weighing down the bottom more and it will take a lot more to tip it. a scope is designed so that the centre of gravity is low and therefore is is incorrect to assume the mass is evenly distributed along its length. you also have to take into acount the mass of the base which adds to the overall mass, and puts the equation off even more. come to that, how is the weight at the top 1m out from the centre? thats quite a distance and I think you will find it means that Doomed Dave is outside of the basket compleatly and if thats the case he deserves to fall, well unless hes a cartoon I doubt he can stand in mid air.

also have you taken into acount the fact that only downwards forces are mass*9.81? as the horizontal force that Wee Johhny is applying is totaly independant of the force of gravity (IE *9.81) so the force is not that same as any horizontal force. unless I am wrong you have not acounted for this and therefore once again the equation would be invalid.

 

The maths in your equation is very good but there are far too many assumptions that are way off reality for it to be a fair model for real life, sorry. ;)

[Ben...]

The maths in your equation is very good but there are far too many assumptions that are way off reality for it to be a fair model for real life, sorry.   ;)

It is far from perfect which is why I stated the assumptions.

But if anyone knows the mass distribution of a tallscope, or at least how high it's center of mass is, then it'd be easy to correct the calculation - just subsititute the height of the center of mass where I've put h/2. If someone knows the mass of the base and the mass of the platform that'd probably do for working it out.

 

And regarding the 1m away from the top, that was me trying to keep things simpler. If the base is W1 from centre to edge (where it's pivoting) and the weight at the top is W2 from the centre, then if the force from the top mass is F1, the moment will be

+ F1 * { W1*cos(T) + h * sin(T) - W2*cos(T) }

 

with the other two moments the same (except that read W2 for W) the final equation gets quite evil;

arccos(2*(F1*h-F2*W2+1/2*F3*h) / (4*F1^2*h^2 - 8*F1*h*F2*W2 + 4*F1*h^2*F3 + 4*F2^2*W2^2 - 4*F2*W2*F3*h + F3^2*h^2 + 4*F1^2*W1^2 - 8*F1^2*W1*W2 - 8*F1*W1*F2*x - 8*F1*W1*F3*W2 + 4*F1^2*W2^2 + 8*F1*W2*F2*x + 8*F1*W2^2*F3 + 4*F2^2*x^2 + 8*F2*x*F3*W2 + 4*F3^2*W2^2)^(1/2))

(at this point I got a headache and got a computerised algebra programme to do the rearranging)

 

So if the values are all the same as above, except that W1 (the distance from the centre of the platform) is 0.4m, and the base is still 1m from centre to edge, the angle becomes 14.6 degrees.

 

And if the centre of mass of the 'scope is a fifth of it's height from the ground, the moment of F3 becomes;

F3 * ( h/5 * sin(T) - W2 * cos(T) )

and the angle becomes a rather comical 27 degrees.

 

If young Johnny ambles off for a Tea break, (ie F2=0) the angle becomes 25 degrees, so draw your own conclusion about the usefulness of him as an outrigger. Much like all such things, the lower the centre of mass the better.

[Dave]

Hang on a moment guys!

 

I don't mind big Brian stealing my name, but when did I become doomed??? ;) This is unfair to the world's population of Daves, most of whom (myself included) treat scopes with respect and don't lean out of them.

 

I will make a formal complaint to the Royal Society for the Prevention of Cruelty to Daves.

 

Seriously, I must confess to having done A level maths myself, fifteen years ago. I'm getting old. As an aside, my maths teacher (who was very big) once gave me a saturday morning detention for missing a lesson to rig lights for the school play. But that story's for another time.

 

The tallescope, like any other rigid object, can be modelled as a single mass which acts through its centre of gravity, and the big unknown in solving this problem is where this is located. Given that a scope is approximately symmetrical in two axes, the CoG is likely to lie somewhere along the centre of the ladder, but how far up is anyone's guess. Who dares ask the makers? The other forces involved are the reaction from the ground through the wheels, the weight of the idiot leaning out on top, and of course wee Johnny's desperate efforts.

 

The other thing we need to know is the the location of the centre of gravity of my namesake on top. I'd expect a normal person's CoG is going to be somewhere in their middle. The scope will always be stable if this CoG is between its wheels, and, as Ben just said, it seems rather unlikely that it will be anywhere else.

 

So, to make this this exercise more realistic, we need to tilt the scope to start with. That's easy - just introduce a raked stage!!!

 

To make things easier, assume that the guy up top is centred in the basket. Then, calculate how far the scope can be tilted sideways before it begins to topple.

 

Answers on a postcard...

 

Dave (not big and not doomed!)

[robloxley]

It might be useful for calculation purposes if someone could inform us of the exact documented circumstances in which a tallescope has 'fallen over' - was it due to the technician simply leaning out, was it due to the technician trying to move the tallescope from the basket by pulling on some bar or other, was it due to use on a raked stage, was it the people at the bottom pulling it over when trying to move it sideways etc.?

[Ben...]

of course, when I said mass evenly distributed along its length, I did of course mean evenly distributed vertically

 

Which in the first case just meant that the CoG was half way up. In the second pile of numbers I put it a fifth of the way up, which is why the danger angle got a lot bigger (25 degrees.) And I was using a 'scope mass of 800kg, which is a bit of a guess.

 

Anyway, I reckon you'd have to be trying damn hard to knock one over.

 

Incidentally, when I let the computer do the algebra, a slightly snappier equation for the tipping angle is

T = arctan((-F1*W1+F1*W2+F2*x+F3*W2)/(F1*h-F2*W2+F3*h*CoG))

 

where CoG is how high the centre of mass is, as a fraction of h.

 

 

 

I think that's quite enough of that for one day ;)

[Ben]

It might be useful for calculation purposes if someone could inform us of the exact documented circumstances in which a tallescope has 'fallen over' - was it due to the technician simply leaning out, was it due to the technician trying to move the tallescope from the basket by pulling on some bar or other, was it due to use on a raked stage, was it the people at the bottom pulling it over when trying to move it sideways etc.?

 

you are missing the point, it doesnt matter how its tipping the fact is that the scope is tipping, how far can it tip before the poor Doomed Dave at the top is written off???

[Dick The Cabin Boy]

Circumstances in which a tallescope has 'fallen over' - was it due to the technician simply leaning out, was it due to the technician trying to move the tallescope from the basket by pulling on some bar or other, was it due to use on a raked stage, was it the people at the bottom pulling it over when trying to move it sideways etc.?

 

A couple of years ago I actually managed to tip one of the old fellas...we were taking moving lights down (Mac 600 from what I remember) I had the out riggers extended although not fully and the top of the basket was about 3 foot away from the bar. We slung a line round the buggers, I lifted them off the bar then the guys on the ground lowered them down before moving me (moving with the scope with someone in the basket!! ;)

 

This was all fine until at nearly the end of our evenings work when I lifted the head off the bar the dimwits below just let in run down without watching. One of the clamps from the Head got snagged on the bottom of the basket and as I reached down to release it the whole thing started to topple forwards.

 

I reached up to grab the bar (now above my head) as the scope went from under my feet. The out riggers stopped it from going over too fast and so the guys caught it and pushed back under me so I could climb in... and I live to tell the tale