measurment of starting block of swimming pool above the water surface.?
146239
From the usaswimming.org rules: \nA: In pools with water depth less than 4' (1.22 meter) at the start end, the swimmer must start from the deck or from within the water\nB: In pools with water depth 4' (1.22 meter) or more at the starting end, starting platforms shall meet the height requirements of 103.12.1."\n\n103.12.1 can be found here http://web.usaswimming.org/USASWeb/_Rainbow/Documents/53de9128-16c3-4a9c-91ba-bd95f1330038/(019-056)%20PART%20ONE.pdf, but says essentially somewhere between 1 foot 8 inches and 2 foot 6 inches above the water.
500588
A regulation size pool table is 4 1/2 x 9 feet.
1027487
Not enough info to answer correctly.\n\nDo you mean and 18" diameter circular pool\n\nOr an eighteen inch deep pool of undertermined dimensions\n\nfurther what is the weight and mass of the sand.\n\nWhat I can tell you is that will will take 2 cubic inches for every square inch of pool\n\nan 18" square pool has an area of 324" therefore 648 cubic inches  for example
344788
hey man, dont worry ill help. first of all our given:\nh=280m\nsg=1.03\ngc=9.7m/s^2\n\nreq'd Pressure(absolute)\n----------------------------------------------\nsol'n:\nfor DENSITY of subs:\nformula\nSpecific Gravity=density of subs/density of water\ndensity of subs=1.03 x 1000(density for water)\ndensity of subs=1030 kg/m^3\n\nFOR PRESSURE GAGE:\nP(gage)=Vh\nP(gage)=density x gravitational constant x height\nP(gage)=1030 x 9.7 x 280\nP(gage)=2,797,480 Pa\n\nFOR PRESSURE ABSOLUTE:\nP(abs)=P(gage) - P(atm)\n\nwhy minus? bec. the submarine is pressurized, meaning its vacuumed\n\nP(abs)=2,797,480 Pa - 1atm(101.325x10^3 Pa)\nP(abs)=2,696,155 Pa\n\nconvert Pa to Kpa\n=2,696,480 /1000\n=2,696.155 Kpa\n\nconvert to bar\n=2696.155kPa x (1.01325 bar / 101.325 kPa)\n=26.96155 bar\n\ndont mind the others, they dont even know how to answer this simple question. They just want to earn points!
255850
I assume the shape to be rectangle.\n\nMeasure the longer side in feets and name it L\nand the shorter side as B\n\nThen multiply L*B you get measure in square feets.\n\nIf it is irregular in shape, you have to cut it into parts and draw lines and measure each area separately.
1036141
56" or 56'?? What's the depth of the pool?
754418
You can have different sizes of the pool, for example: \n20 ft. wide pool 1800:20=90ft. long \nor 25ft. wide pool 1800:25=72 ft. long\nor 30 ft. wide pool 1800: 30=60 ft. long\nI would clarify with the owner what pool width he or she prefers.\nFor the side decks we add 2x5 ft.=10ft to the chosen width of the pool \nFor the end decks we add 2x10ft.=20 ft.to the length of the pool \n\nthus the pool site should be over\n\n(20 +10) x (90+10)=30'x100'\nor\n(25+10) x (72 + 10)=35' x 82'\nor\n(30 + 10)x (60+10)= 40'x 70'\n\nPlease remember to add the pool clearances from the property line as per local municipal codes. Also, the machines used for digging have to have access from this property, not from the neighbours yards.  The smaller they are the more difficult and expensive the job is. \nIf we add about 10 ft. for access (and clearance) on each side we end up with a poll site: \n(20+20)x(90+20)=40'x110'\n(25+20)x(72+20)=45'x92'\n(30+20) x(60+20)=50'x80'\n \nAn addition of the area needed for the house will give you the minimum lot size required.\nGood luck!
1111557
I not sure about this. Water has a lot of Weight. Ever see those "Slinky" toys. You not want you Pool to get the same idea as a "Slinky". Maybe you could Build Up the ground to get it level? Maybe a few Landscaping Timbers or even rocks laid nice and fill in to get it more level. I have wanted to do the same as you, but, with the slope I not think it will work, hate to spend money to watch it Collapse and be Ruined.\nQ: Where (and on what surfaces) can I safely and successfully set up my Simple-Set pool?\nA: For proper pool functionality and your safety the pool MUST be set up on an extremely flat, level, and stable surface. Suitable surfaces include grass, compacted earth, concrete, or pavement. Use of a ground cloth, if provided, is always recommended and is REQUIRED if the pool is to be set up on concrete or pavement. Under NO CIRCUMSTANCES should the pool be set-up on wooden decks or other wooden surfaces, or on ground that is muddy or subject to shifting. DO NOT USE SAND to level the ground, as it is also subject to shifting under the weight of the pool. Do not attempt to 'jack up' the pool on any type of frame, support, or riser(s), as this will cause the pool to collapse!
350827
Lemonade bottle (fins will help) and a water/air ratio of 1/3 : 2/3 makes a good starting point. The rest is good old trial and error!
137596
Approximately 20 laps, for a standard high school basketball court measuring 84 by 50 feet.\n\n84 + 84 + 50 + 50 = 268\n5280 / 268 = 19.7015 laps\n\n18.33 laps for an NBA/NCAA court\n22.76 laps for a standard jr. high school court
220777
use the following formula .. \n\ntotatal surface area = \n[(H/D)*4(pie)(0.5D*0.5D)] + [ (H-t)/(D-2t)*4(pie)*(0.5D-t)(0.5D-t)] + [(pie)(0.5D*0.5D)-(pie)(0.5D-t)(0.5D-t)]\n\nthis is considering D = tank diametere is from outer surface and not the cavity diameter \n\n\nouter surface area of cover = (H/D)*4*(pie)(0.5D*0.5D)\ninner surface area of cover = [(H-t)/(D-2t)*4*(pie)*(0.5D-t)(0.5D-t)] \n\nthe base line are = [(pie)(0.5D*0.5D)-(pie)(0.5D-t)(0.5D-t)]
840248
Vertical components\n\nv = u + at\n0 = u + (-32)(2.1s)\nu = 67.2ft/sec\n\nMaximum height the ball reached\n\nv^2 = u^2 +2gs\n0 = (67.2)^2 + 2(-32)(s)\n64s = 4515.84\ns = 70.56\n\ndownward velocity 0.8 seconds after the ball reached the max.height.\n\nv = u +at\nv = o +(32)(0.8)\nv = 25.6 ft/sec\n\ndistance from the maximum height\n\nv^2 = u^2 +2gs\n(25.6)^2 = 0 + 2(32)s\ns = 10.24 ft\n\nHeight of the fence = 70.56 ft - 10.24 ft\n\n\nHeight of the fence is 60.32 ft...\n\nthere ya go...
1034094
if its not my swimming pool
454640
try this site\n\nhttp://www.shopzilla.com/8B--Swimming_Pools_Jacuzzis_-_cat_id--12180104__keyword--inground%20pool__kw--inground%20pool%20prices__mkt_id--35580167__qcid--419278446397
471462
Multiply meters by 3.28084 to get feet.
908785
very close to 2 meters.
325473
Measure the weight of 1 cup of sand and figure it from there.\n\n :-)