I think the error in these calculations was pointed out before, but just to
recap......
Snow is 10% density of rain (agreed - this is an average figure, but is used,
for example, in National Building Code of Canada short form calculations).
Therefore 80 cm of snow - 8 cm of rain = 80 mm of rain - NOT 8 mm of rain.
Therefore all calculations using 8 mm of rain are out by a factor of 10.
Still an awful lot easier than any other measurement system.
John F-L
----- Original Message -----
From: Pat Naughtin
To: U.S. Metric Association
Sent: Friday, February 04, 2011 8:39 PM
Subject: [USMA:49758] Re: Super Bowl: NFL, stop with the Roman numerals
On 2011/02/05, at 06:56 , John M. Steele wrote:
Well, then, they should change the scoreboard and the clock.
--- On Fri, 2/4/11, [email protected]
<[email protected]> wrote:
Probably because using Roman numerals gives a
Gladiator-like aura to the game.
Carleton
Dear All,
To get this back on to a metrication track, let me resend an email from last
October:
On 2010/10/22, at 05:10 , Michael Payne wrote:
The collapsed Hanger at Dulles Airport, Length 100 x 50 m. Snow
weighs 10% of rain. If we have 80 cm of snow, it's 8 mm of rain. 1 millimeter
of rain is 1 liter per square meter or 1 kg per square meter. Therefore the
roof area of that hanger is 100x50=5000 m2 times 8 kg/m2 = 40,000 kg or 40
metric Tons (Tonnes) additional load. I did not need a calculator.
Dear Michael,
I couldn't help myself.
Here is your calculation done in Roman numerals:
The Dallas hangar is approximately CD Roman feet by CC Roman feet. The
snow that fell was •••• Roman feet so the volume of snow is obviously CD x CC x
•••• and, given that the density of water in Rome is about XLV libra per cubic
Roman foot, the mass of this snow is clearly CD x CC x •••• x XLX and the
overall mass of the snow on the roof is (CD x CC x •••• x XLX) ÷ MM Roman tons
(if they existed). I used Wikipedia as the reference for these figures and I
will leave the final calculation for your amusement; I would appreciate it if
you kept track of the time it took you to complete this calculation and let me
know.
Here is your calculation done in customary measures:
The Dallas hangar is approximately 325 feet by 160 feet. The snow that
fell was 1/3 of an inch or 1/36 of a foot so the volume of snow is obviously
325 x 160 x 1/36 and, given that the customary density of water is about 62 1/2
pounds per cubic foot, the mass of this snow is clearly 325 x 160 x 1/36 x 62
1/2 and the overall mass of the snow on the roof is (325 x 160 x 1/36 x 62 1/2)
÷ 2000 short tons. I used Wikipedia as the reference for these figures and I
will leave the final calculation for your amusement; I would appreciate it if
you kept track of the time it took you to complete this calculation and let me
know.
And here is my statement of the problem using metric system units:
The Dallas hangar is approximately 100 metres by 50 metres. The snow
that fell was 80 centimetres, which is the equivalent of 8 millimetres of rain
so the volume of water is obviously 100 x 50 x 8 litres and, given that the
density of water is exactly 1 kilogram per litre, the mass of this snow is
clearly 100 x 50 x 8 x 1 and the overall mass of the snow on the roof is (100 x
50 x 8 x 1) ÷ 1000 = 40 tonnes. I didn't need to use Wikipedia or any other
reference for these figures and, like you, I did the calculations in my head
(but then checked them with a calculator). Time taken for calculation: less
than 5 seconds for mental arithmetic and 10 seconds for calculator checking.
Cheers,
Pat Naughtin LCAMS
Author of the ebook, Metrication Leaders Guide, see
http://metricationmatters.com/MetricationLeadersGuideInfo.html
Hear Pat speak at: http://www.youtube.com/watch?v=_lshRAPvPZY
PO Box 305 Belmont 3216,
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