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    Re: Angular inclination of an interstate highway
    From: Gary LaPook
    Date: 2013 Dec 31, 11:25 -0800
    I know 10% is about 6° from the flight navigation margin of uncertainty so 7% must be a little bit less than 6°. So I just multiplied your magic number of 3438 by .07 and came up with (in my head) 240.66 parts per 3438. Dividing 240.66, which is the number of minutes, by 60 produces 4.11°. As confirmation I did the same with my known quantity of 10% which then produces 5.73° a little bit less than 6°, as I expected. 

    Using the same method with my approximation of 1/3600 produces 4.2° but it is much easier to do in your head because you combine the "multiply by 3600" with the "divide by 60" and end up just having to multiply the percent by 60. So the easier to do method is to simply multiply the percent by 60 to find degrees.

    gl



    From: Frank Reed <FrankReed@HistoricalAtlas.com>
    To: garylapook@pacbell.net
    Sent: Tuesday, December 31, 2013 10:27 AM
    Subject: [NavList] Angular inclination of an interstate highway


    Here's a road sign from Google Street View imagery in southern Texas. According to someone who posted a similar image on flickr, this is the steepest incline on I-10, which runs across the southern US from Florida to California.
    What is the angular inclination of this downhill slope in degrees (to the nearest tenth)? No calculator and no trig allowed. You can do it in your head with that magic number, 3438. Again: angles ARE ratios. We only quote them in degrees and minutes for historical continuity..
    After you've done the calculation in your head, consider the possible trig solutions. Are they any better?
    -FER
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