Date: Thu, 18 May 2000 00:04:25 -0700 To: balloondeco@balloonhq.com From: Mark Balzer <mbalzer@balloonhq.com> Subject: Re: Dance Floor Canopy Calculation Needed >I need a bit of help figuring out a calculation... >The lift capacity for an 11" balloon is .35 oz. >so using 1000 balloons as an example am I correct in the following: >1000 x .35 = 350 ounces >350 ounces divided by 16 oz. = 21.88 lbs. so far so good. >Is this 21.88 lbs., pounds per base pole or total? It's the total lift (vertical) force. >Or am I completely nuts.... How about I answer that with a question... What is heavier, a pound of feathers, or a pound of gold? :-) :-) >and not have the calculation even close. Well, you got off to a good start but then you forgot about the arch lines. The balloons are tied to the arch lines, not the base poles. The lines change the magnitude and direction of the lift force, then apply that new force vector to the poles. This problem needs to be solved using analytic geometry and vector mechanics, but you can do a quick experiment to get a feel for it. Take a rope and tie one end to a doorknob. Have a helper pull the rope taught horizontally (lock the door first). grab the rope in the middle and lift upwards with a few pounds of force. Because of the force multiplication (also called mechanical advantage) it doesn't take much lift force to completely overpower your helper! > What I mean by leverage is that the canopy pulls against your poles. > And the poles are 6 to 8 feet tall and as it is with a lever it > takes less effort to move something with leverage. The balloons > are pulling against the poles, not lifting the bases. Actually, the balloons are pulling on the lines, and the lines are are always both "pulling against the poles" (horizontally) _and_ lifting the bases (vertically). To get a feel for it, think of the extremes. A very tall arch with a given number of balloons would exert very little "pulling against the poles" (tension in the line drawing the poles together); the force on the poles would be predominately that from the balloon lift. As you flatten the arches in the canopy (same number of balloons), the lift will be the same, but the "pulling against the poles" tension will grow larger. > I'm sure a physicist could come up with a calculation to determine > the amount of weight needed to keep the poles from pulling in and > tipping the bases. Well... it's hard to get anything practical out of a physicist :-) :-) I'm sure an engineer could come up with a calculation, but it might take the backs of _two_ envelopes. > The other thing to remember about the leverage (torque arm effect) > is that the poles in the columns themselves are going to bend, > also. I would use 1" electrical conduit for the poles. I always > use 3/4", but with that much pull they bend. You can even slip a > smaller pole inside the main pole to add stiffness. Excessive deflection may be a sign of an unsafe structure. The above advice is excellent for minimizing deflection. But let's assume you did your homework and the structure is safe, the stresses are acceptable, etc. Let's further assume your only worry is objectionable visible deflection, but that for whatever reason you can't use the aforementioned larger diameter poles. In that case you might want to consider counteracting the bending of the poles as follows: a) Prop your baseplates to angle each pole away from the direction it wants to bend. Then your pole will look bowed overall by only a fraction of the deflection you'd see at the top of an initially vertical pole. b) The more elegant solution is to pre-bend the poles by just the right amount in just the right places, so that they become straight when the lines exert their pull. I'm sure an engineer could come up with a calculation to determine the amount of pre-bend needed to make the poles straighten when loaded - they do it when they "camber" bridge beams, flatbed truck beds, etc. I hope this helps, Mark o Balloon HQ <M> The most complete collection of balloon info on the web _/ \_ http://www.BalloonHQ.com