I know for a fact that together, my partner and I can inflate and tie 100 balloons in 12 minutes.
Some of the following comments include amounts of money in the imaginary unit called "C-shells." These units are used to avoid any hint of illegal price fixing in the balloon industry.
Don't let the pole stop you. Use it to your advantage. Why not wrap the umbrella pole with balloons? Or twirl #9 or #40 ribbon around pole? Or twirl 260's and round balloons? What about using the 4 Square balloons that are used in SDS. Make a box shape, place oasis down around the pole and add some foam wire, some small balloons and some foil tuffs and you're done.
Take a defective, separated 18" mylar, silver side down and place one cup of sand into the center. Tie with a piece of balloon ribbon leaving both ends of the ribbon approximately 12" long. Trim top to 2". Air inflate the clusters - colors can be mixed, banded, or grouped - and size them down to about 4 1/4". You can make your own sizer from a glass, bowl or whatever. Tie duplets and twist into clusters of 4. Arrange the colors and place the first cluster onto the weight. Bring one of the ribbons around two opposite balloons (figure 8) and then around the third. Place the second cluster atop the first, move into position and use that same ribbon wrapping the same as the first cluster. Pull the wrapping ribbon up through the center and bring up the second piece of ribbon. Knot together twice and trim. Use two pieces of onion grass - cut off the ends with a wire cutter to approximately 2" and wrap with floral tape to cover all sharp ends. Dab with either cool glue or floral adhesive and position into the center of the top cluster. Use your thumb nail on the underside of 4-6 strands of the onion grass to curl downward. Bag them up in long bags - 12 per bag.
They are indestructible - can be done easily a week ahead. Cost: 8 x 5" latex = .32, 2 sprays of onion grass = .30, Misc. sand, mylar, ribbon, glue, tape = .20. Add about 5 minutes labor. I charge a base fee for basic colors in this design and upgrade components, colors and floating balloons with collars, mylar, tulle, etc.
If you want to use floating helium balloons - I suggest just using a single 16" so it won't be tangling and blowing into guests' faces. You will be able to inflate all the clusters and have everything ready. Tie the floating balloon to the weight and use ] that ribbon to wrap the clusters - this keeps everything pulled together - continue as above and use the loose ribbons from the weight to retie everything before putting in the accents.
They may cost slightly more (a 1,000 count bag runs me about $16.00) but I find that with my LePouf machine, I can use them for almost any type of occasion. I can make pretty "poufs" using tulle, star shaped mylar paper, scalloped circle mylar paper, etc. and I put approx. 26 marbles in each weight. I find this is enough to hold down a centerpiece consisting of (6) 11" latex balloons and (1) 18" mylar balloon.
Also, if there should be anyone who wants to "investigate" what is inside the weight, there is no mess on the tables, just pretty colored marbles!
We use the appropriate amount of weight needed to hold what ever number of balloons - i.e. - a bouquet of twelve 11" requires approximately 7-8 oz or a 1/2 cup. If it is going outdoors you may want to add a little extra. Our cheapo postal scale that we purchased for $10.00 worked great to give us the proportions we needed - i.e. 1/2 c for twelve 11", 3/4 - 1 c. for nine 16" etc.
When you know you are going to be using this method, you might want to use a little heavier monofilament line than you would normally use to avoid any chance of "slicing" the balloons while you are condensing or uncondensing them.
This condensing technique will work better with a short arch than it will with a long arch because our plastic bags will roughly hold a 30 balloon arch.
H = height W = width L = length L = W/2*SQRT(1+((4*H/W)^2))+(W^2/(8*H))*LN((4*H/W)+SQRT(1+(4*H/W)^2))
W+H(1.5) = LUsing the above formula an arch length of L would give an arch slightly higher than it is wide which is probably the most common arch look around. If you used the formula:
W + H = LYou would get a flattened arch and the formula:
W + H(2) = LWould give you a high arch (St. Louis Gateway Arch???)
There's also an easy way to get a rough approximation by using the pythagorean theorem. If you draw a line from the top of the arch, which is the midpoint, straight down, you will get the height (here, 9 feet). This is one leg of a right triangle. The other leg runs from the midpoint of the base line of the arch to the end of the arch, (so it's half of your 15 feet, or 7.5 feet) The hypotenuse of that right triangle is a straight line from the midpoint of the arch to the end point of the arch, which is a little shorter than the curve of that half of the arch, but close enough to get your estimate. According to the theorem, A(squared) + B(squared)=C(squared) you can calculate this hypotenuse 9(squared) + 7.5(squared) = "that half of the arch"(squared). Now that you know half the arch you can figure out the whole arch by doubling it.
This all revolves around understanding simple geometry. If you don't know what a right triangle is, or a hypotenuse, or you don't know how to square a number or find a square root (yes, you needed to take the square root of C(squared) back there, even though I didn't come right out and say it) then the best way to figure out the length of an arch is to put balloons on a line until it's the right length, and that's your answer.
As for the number of balloons needed, You will need to convert the length of your line into inches, and then if you don't know how to figure out how many 11" balloons it will take to do, say 220" of line... uh, you'd better call your accountant.
There's also a common sense answer to your question. If it's 15' wide, then it's gonna be somewhat longer than that in overall length. Is it double? Well, picture it in your mind. Take a string, run it the width of the arch and back. Is that going to be too much string? I think so. So, we're between 15 and 30 feet. Why do you want to know the exact length. Surely it's not so you can bring the right length of string, you're going to bring a roll of string, if you've only got 30' of string left, go buy another roll before you go to the job. No, you just want to know how many balloons it's going to take. Well, if it's just 15' (which it's not - it has to be a little longer than that) then, using 11" balloons, (if you figure in a handy 1" gap between balloons, then you'll use exactly 1 balloon foot) it'll take 15 balloons. If it's 30' (which it's not, we're figuring it's somewhat shorter than that) then it'll take 30 balloons. If we figure it'll be, say, 5' longer than 15' or 5' shorter than 30', then we're estimating a length of 20-25 feet, or 20-25 balloons. If you're using 9" balloons then they're about 3/4 of a foot long, so you'll need 4 of them per every 3', so use an estimate of 21 - 24 feet (because those numbers are divisible by 3') to find out that you'll need 28 to 32 balloons.
I can't imagine doing a job where a difference of 5 -10 balloons in an arch is going to affect your supplies or your cost. You should have enough extra balloons on hand to accommodate this. The actual material cost is at most $2-$3. The time involved in putting 10 more balloons on an arch is about 2 minutes.
So, in conclusion... If you want the formula, I've provided it, but if you're smart enough to use the formula you probably don't need it, you should be able to rough it out in your head on the spot.
The columns that support the arch are made from a single RMS Builder. It takes a little less than one half a Builder to make a supporting column. We usually use a total of 100 balloons sized to 8" to make the two columns. We recommend under-inflated 11" balloons tied in doublets.
The RMS Queen's Arch and the RMS Builder are sold separately.
The instructions which come with the Queen's Arch give detailed information on sizing, tying, and installing the balloons in both the arch and columns. You will get the best results if you follow the tips in the instructions. The triangular cross section of the columns and the arch give a very sophisticated look.
For the best results, it is important to coat both sides of the Matrix used in the Queen's Arch with spray adhesive before you stretch it open. While the arch section is self-forming and self-supporting, it is necessary to use base plates and pipes inside the two columns to stabilize the structure.
All balloons are helium filled. I think that they used pearl balloons in the demo, and it came out very dramatic. The wind caused a lot of "fluttering".
We sat and figured out afterward that the cost on the product and helium alone was about $27,000.00.
Yes it was a HUGE project, but we had a great time and learned ALOT... If you thought the pictures looked gigantic, to be there in person would have blown your mind... the pictures did not do it justice!
When I can't get to the ceiling of a hall to attach my tulle. I use 3 foot balloons. The bigger the dance floor the more tulle will be needed. Therefore more 3 foot balloons will be needed. Weigh the length of tulle you plan on using. Check the QBN book that comes with the first video of their series. It has the amount of weight 3 foot balloons can lift. Just use the correct amount and you should have no problems.
Make one string of pearls corner to diagonal corner. Make sure you have an odd number of balloons. Make the second string of pearls identical to the first. Same length of line same number of balloons then pop the middle one. The middle balloon of the first arch doubles as the middle balloon of the second arch. If you practice with one color you will find it easy. Then you need to balance you color scheme to do the arches you want. Practice with two arches - 5 balloons long on the first, 4 on the second and you'll get the picture immediately.