Albury High School Popsicle Bridge record breaker!

We run a competition here at Albury High in Australia for our Yr 12 Engineering Studies students involving max. 50 popsicle sticks, PVA wood glue and 2m of extra strong thread.  Structures have to span 400mm and are centrally loaded.  Here’s a photo of our 2010 winner which weighed in at 69 grams and held 58kgs giving an efficiency using the formula on your website if I used it correctly of around 12000!

10 Responses to “Albury High School Popsicle Bridge record breaker!”

1. ricky gonzalo kariA - April 26, 2012 at 2:58 pm

   WE WOULD LOVE TO DO THIS PROJECT BECAUSE WE NEED TO BUILD ONE, AND WE WOULD ENJOY THIS EXACLY HOW DO U DO THIS BRIDGE

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2. Ziggy_boarder - April 23, 2011 at 2:36 am

   A 12,000 ratio would require it to hold 828kg, or 1,821lbs.

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3. Ziggy_boarder - April 23, 2011 at 2:30 am

   I’ve seen 5 level Baily bridges from WW-II that supported Sherman tanks using the same general design principals. Heavy cables where supported at the center by a built-up underside structure, as this level of arching was not possible. Still, the cables added hugely to the strength of the structure.

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4. Eileen Sher - March 13, 2011 at 9:57 am

   Hi, would you be able to show me some ropes as to how I can build a bridge like yours? For instance, what do I have to take into consideration in order to be able to build a strong bridge? Would appreciate it very much for your help. Thanks.

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5. Physics Kid - December 9, 2010 at 4:05 pm

   Can we see a short how to on this? This would be an awesome home project

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6. Tyler - November 1, 2010 at 12:56 pm

   this bridge is friggen awesome. I was surprised to see how many popsicle sticks you used and how much weight you put on there.

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7. Britni - March 31, 2010 at 3:56 pm

   That is so cool but do you know if you can make one of those out of 200 pop sicle sticks.

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8. Garrett Boon - March 1, 2010 at 2:12 pm

   Thanks for sharing the photo.
   The efficiency equation works like this:
   Take the mass held: 58kg and covert it to grams, 58,000g. Then divide that by the mass of the bridge: 69g. That equals 841.

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   • Britni - March 31, 2010 at 3:57 pm

     Yeah Garrett i dont really understand what you just said can you explain that?

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     • AndrewL - April 1, 2010 at 3:30 pm

       The bridge efficiency is the ratio of the mass supported to the mass of the bridge.
       mass_supported / mass_of_bridge = efficiency

       In this case:
       58000 / 69 = 840.58 efficiency

       This is different to the original posters claim of 12,000 efficiency.

       The use of string/thread in this design is a major structural element.

       Reply