Absurd, I suppose is in the eye of the beholder. [Although they may be right in this case.] So I tried to figure out an objective analysis of the problem.

At 10 m (meters, right), that is a lot of leverage to contend with if the

flat starts to tip forward. You say it is on wheels, so I assume that its on some kind of trolley/carriage, and for the sake of the discussion, I’ll assume that the trolley is 1.5

meter wide by 3

meter deep (but of course it could be deeper than that) (I’ll be honest at this

point, I am having trouble with how stable this structure may be). But forging on, I’m assuming that the problem is a balance of torques: one for the tipping vertical

flat and the other for the trolley and

counterweight.

For the vertical

flat, I am assuming that weight of the vertical

flat is “concentrated” at the midpoint vertical distance (5 m). For the trolley, I am assuming that weights must be placed about three quarters of the way from the

flat, 0.75 x 3 m = 2.25 m.

The real question in this problem is how far forward will you let it tip before it falls over? (My gut feeling is that the answer could be “not at all”, but I don’t know if any flats are held to that standard.) But, for the sake of discussion, assume that the

flat can tip forward 10 degrees (that’s about 2 m forward at the top.). Any assumption of acceptable tipping assumes that the

flat will not get pushed accidentally and will be carefully maneuvered. Deciding the “acceptable tipping angle” will require serious thought. For example if a “normal” 2.5 m to 4 m high

flat were to fall forward, it stays on the

stage and only the actors are in the way. But a 10 m

flat may reach the audience. (Not to trivialize the actors’

safety, but they may see it coming and get out of the way, or turn/react to protect themselves,

etc.)

Okay, back on

track:

Torque (T) = F x L x cosine a

Where F is the weight in kg (I’m fudging units here a little, I usually use US lb, but the force conversion units will cancel down in the calculation anyway)

L is the distance from the pivot

point to the weight

a is the angle between horizontal and the arm L

If the

flat has tipped forward 10 degrees, the angle a for the

flat is 80. Meanwhile, the trolley has lifted up 10 degrees and angle a for the trolley is 10 degrees.

So the equation to balance:

T of

flat = T of trolley

Now, assume that the front

flat weighs 82 kg (about 180 lb – I’ve assumed using 2x4s, something substantial to support 10 m), then you have:

82 kg x 5m x cos 80 = Weight on trolley x 2.25m x cos 10

82 x 5 x 0.17 = Weight x 2.25 x 0.98

Weight = 32 kg (70 lb)! With the weight of the

flat (82 kg), that’s a lot of weight to be wheeling around.

I suspect that I’m being conservative here (for example, I have ignored the weight of the trolley and weight of any back bracing and I’ve made some conservative assumptions on the design materials and weight). (On the other

hand, if a more conservative tipping angle of 15 degrees is used, the weight is 50 kg. [110 lb]) [When I apply the same math to a more typically sized flat (2.5 m to 4 m high), I get a reasonable counterweight value (3 to 5 kg).]

Although not related to the original question, erecting the

flat from horizontal to vertical may be a structural problem in and of itself, depending on the materials of construction and design. That is, 10 m (30 feet) of a span supported at only one end; it could buckle under its own weight as it is being pushed up.

Further, if the

flat isn’t that rigid, the

flat might wobble back and forth if it is moved quickly or hit by something – if that wobble moves beyond the angle of the design tipping

point, it will fall over.

Also, this will have to moved smoothly. A sudden start or sudden stop will get it tipping forward.

Any chance you can fly this in as fabric drops?

Joe