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Bricks falling over


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Flapping Flight Dec. 6, 2017, 8:47 p.m. PST
@Ken Burner
Yes:) if we imagin the bricks as a bars of an grid ,as many bars in the grid as less empty spaces area so higher resistance for the escaping air- higher force on every single brick.

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Flapping Flight Dec. 6, 2017, 8:41 p.m. PST
@Ken Burner
Yah this truth (and because in US including NASA work many people originaly from all around the word , the president Carter who is a phisicist by proffesion wanted in his mandate to move the mesurment system to metric but I understand resistance of the people it's difficult to change the people habits .The good side is that in the US you can find almost equally easy components of both systems when in Europe non metric one is difficult to find .

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Ken Burner Dec. 6, 2017, 8:34 p.m. PST
@Flapping Flight
I hope that you interpretation is correct. The less the number of bricks the better!

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Ken Burner Dec. 6, 2017, 8:28 p.m. PST
"NASA lost its $125-million Mars Climate Orbiter because spacecraft engineers failed to convert from English to metric measurements when exchanging vital data before the craft was launched." News from 1999 and yet we still use the old Imperial system a lot!

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Flapping Flight Dec. 6, 2017, 8:24 p.m. PST
@ExpAir
Sorry about mishmashing (') and(") still wait North Americans to move to metric system :)

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Flapping Flight Dec. 6, 2017, 8:18 p.m. PST
@Ken Burner
Ok , understand where difference is coming from I leaning all the time to most easiest for us :)
Will wait untill tommorow to see which one is correct interpretation.
In one of my earlier posts I wrote that we will spent more efforts on clearifing the rules then to do our actual work of designing building
Something crazy:)

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Ken Burner Dec. 6, 2017, 7:49 p.m. PST
Ah, Team work! OK so rules say "The bricks will be arranged approximately every 24 inches." To many some, but not all that means center to center distance, while others may interpret as inside to inside dimensions. Perhaps Paul can help.

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ExpAir Dec. 6, 2017, 7:45 p.m. PST
Ken its just that you are calculating points and then placing bricks on points, but flapping flight is allowing for a 2foot space between bricks

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Ken Burner Dec. 6, 2017, 7:45 p.m. PST
If the bricks were longer than 2 foot each, rather than 8", there would be less than 47 bricks I think.

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Ken Burner Dec. 6, 2017, 7:42 p.m. PST
@Flapping Flight
94 feet divided by 2 feet = 47

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Ken Burner Dec. 6, 2017, 7:40 p.m. PST
@Flapping Flight
Yes you stated that the number of bricks is equal to the number of spaces. Therefore if there are 47 spaces there are 47 bricks if I'm thinking correctly.

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ExpAir Dec. 6, 2017, 7:34 p.m. PST
Flapping flight - slight correction on your message
inches is represented "
and feet '

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Flapping Flight Dec. 6, 2017, 7:29 p.m. PST
@Ken Burner
Ken, your result is wrong !
Perimeter is 30"x3.14=94,2"
Converted in inches 94.2"x12=1130.4'
The brick length is 8' and the space betwin every two bricks is 2"=24'
The number of brics is equal to number of spaces because the circle is closed curve so we have to divide the perimeter of the circle on to the length of the brick + the length of the space and will receive the number of bricks( which is equal to number of spaces .
So 1130.4/32=35.325 bricks and exactly the same number of spaces
My previous result was 35.316309 is slightly different ( not significantly)because I'm from Europe and every time convert to metric when calculate something :)

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ExpAir Dec. 6, 2017, 7:28 p.m. PST
the only reason I was bothering about bricks at all is because i thought they were being stood upright. you can try an experiment with your garden hose - just take any brick and stand it upright and then using your finger find a nozzle jet setting so to speak that just allows you to knock over an upright bright. now lay the brick down on the second side and my guess is you will struggle to blow it over at all - no matter what you do. your air blower will never blow it over I would hesitate to guess.

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Ken Burner Dec. 6, 2017, 7:22 p.m. PST
The smallest two sides are 2 x 4", the second smallest two sides are 2 X 8", the largest two sides are 4 x 8". The rules seem to say to put the 2 X 8" side on the ground and aim the 4 x 8" towards the 30 ft. circle center.

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ExpAir Dec. 6, 2017, 7:16 p.m. PST
really - are you sure. because in that case you really dont need to worry - you will never blow a brick over lying on its 2nd smallest side unless you direct turbine thrust right at it (within the power realities of this competition)

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Ken Burner Dec. 6, 2017, 7:12 p.m. PST
@ExpAir
Actually the bricks will be lying on their second to smallest side . The smallest would be the 2 X 4" side :)

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ExpAir Dec. 6, 2017, 7:08 p.m. PST
@Paul Musille
Sorry Paul, Ive just realised what you said, "standing on the 2 x 8 inch side."
since the 8 inch side is the long side - what you meant to say was it will be resting on the smallest side with the largest face facing the machine

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Ken Burner Dec. 6, 2017, 6:59 p.m. PST
Now if I can just come up with the bricks. No Lowe's sells the recommended bricks within 100 miles nor do they deliver around here. So, now I need find a local source for "full common red clay pavement and walkway ASTM C-902 standard bricks." And here I thought that bearings were going to be my main problem!

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Ken Burner Dec. 6, 2017, 6:45 p.m. PST
I come up with 47 bricks. 30 foot diameter times 3.14 / 2 (for one brick every 2 feet) = 47.