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# Re: SQL Fun Challenge #3

From: Mark C. Stock <mcstockX_at_Xenquery>
Date: Fri, 12 Mar 2004 07:40:56 -0500
Message-ID: <rImdnWwcO_t5MszdRVn-jQ@comcast.com>

"Noel" <tbal_at_go2.pl> wrote in message news:c2salh\$sbu\$1_at_inews.gazeta.pl...
|
| Uzytkownik "FlameDance" <flamedance_at_gmx.de> napisal w wiadomosci
| news:c2s3qo\$kcc\$03\$1_at_news.t-online.com...
| > Daniel Morgan wrote:
| >
| > As the bird (constant 20mph) flies slower than the trains (constant
| > 50mph) it won't fly back and forth at all, instead it will lag behind
| > "Train 1". It will fly one-way until the trains collide, at 2/5th the
| > trains speed. The trains collide after 114/2 miles, so the bird flies
| > (114 miles *2)/(2*5) = 114/5 = 22.8 miles in the same time span.
| >
| > Or did I misunderstand the setup? It appears to me that to make sense
| > the question requires that the bird flies faster than the trains.
|
| The trains colide in (114/(50+50)) hours, so if bird fly with 20mph, then
| he flies (114/100) * 20 = 22.8 miles.
| If bird speed would be 70 mph then it flies 79,8 miles.
|
| Ah, SQL
| SELECT (114/(50+50)) * 20 FROM DUAL;
| ;-)
| --
| Noel
|
|
|

i really don't think we know enough details about the operating environment

what about crosswinds? tunnels? what type of power are the trains running under -- could smoke/noise/electromagnetic fields have a bearing? how does the bird connect to the database...

;-{ mcs Received on Fri Mar 12 2004 - 06:40:56 CST

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