Re: Metric Time (was Re: Why not 13 months? (Was La Systeme Metrique))
Robert A. Uhl (ruhl@phoebe.cair.du.edu)
12 Oct 1995 16:51:07 GMT
In article <ACA20A99966861F326@ehrice.his.com>,
Edward Rice <ehrice@his.com> wrote:
>
>
>In article <4578mk$6jd@sol.sun.csd.unb.ca>,
>t08o@europa.sun.csd.unb.ca (Keith Morrison) wrote:
>
> > If you have an engine that uses 3 fluid ounces of gas for every foot
> > it travels, how many gallons does it burn travelling 30.5 miles?
Well, a fl. oz. == a jigger == 1/8 cup == 1/16 pt. == 1/32 qt. ==
1/64 pottle == 1/128 gallon. So it burns 3/128 gal. for every ft.
Now, there are 5,280 feet in a mile, so 30.5 mi. == 158400 + 2,640
ft. == 160,000 ft. Yes, I worked this out in my head.
Three by 160,000 is 480,000. 480,000 divided by 128 is roughly
3,000. Of course, I am very bad at mental division, so I had to
resort to a piece of paper for the division, which comes out to 3,750.
Any car burning this much gas would die a painful death at the hand of
the owner.
Total time for mental approximation: 5 minutes.
>Zero, because it would be towed -- on the face of it, it burns too much gas
>to run the engine while travelling that distance. No arithmetic is
>necessary.
Yep
> > If you have an engine that uses 15 cubic centimeters of gas for every
> > meter that it travels, how many liters does it burn travelling 30.5
> > kilometers?
>
>If you have to do this with precise numbers, you're in the wrong business.
Methinks that any who advocates metric numbers is in the wrong
business.
--
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| Bob Uhl | Spectre | `En touto nika' + |
| U of D | PGG FR No. 42 | http://mercury.cair.du.edu/~ruhl/ |
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