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Air in tires adding to weight?
Our robot is on the edge of the weight limit [119-121lbs].
We do have pneumatic tires, 6 of them 10inches, would taking the air out of them help make a difference in weight? Also would that be legal? |
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Re: Air in tires adding to weight?
I doubt it will make any difference. Air is fairly light (hence the term light as air). Also if you do this and then re-inflate your wheels and that gets you over weight you are illegal, you may have passed inspection with your under inflated wheels but its still breaking the rules even if you don't get caught. You are better off speed holing your bot in many places
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Re: Air in tires adding to weight?
Air is a low density form of matter. Unless you compressed the air to an extreme level I would find a solid piece of matter a lot better to take off for weight reduction.
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Thanks of the replies, I guess I will get that hole drill warmed up...
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Re: Air in tires adding to weight?
1 word . . . helium. :D
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Is there a rule you can reference that verifies this claim? |
Re: Air in tires adding to weight?
FYI, at 35psi and room temperature, air weighs about 1/4 lbs per cubic foot (source).
So... ...yeah you better plan to save weight someplace else. Joe J. |
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This is why I think letting the air out of the tires is legal. 1/4 LBS for 1 sq foot. 6 Tires that are 10x3inches Hm that might help. |
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The reason pneumatics are supposed to be empty when weighing (and all through inspection other than the power-on check) is for safety, so that something doesn't go totally haywire. Pneumatic tires don't have that problem. Therefore, I can conclude that pneumatic tires, if deflated for inspection, must compete as inspected; i.e., deflated. Oh, and Dr. Joe is right about air's lack of weight savings. Just trust me on this (other than: I was dealing with about 3 IWC air last summer). |
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Is helium a legal robot building substance?
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Re: Air in tires adding to weight?
There are plently of ways to lose weight.
I whould try cutting holes or trinagles in your bot if possibale. If we can see a pic of if we might be able to give some ideas. |
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I figured this thread had to be trolling...but then it was serious???
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*Assuming the wheel's inner tube is a torus of R=3.5, r=1.5 in. Volume is V=2(pi^2)(R)(r^2). |
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