<?xml version="1.0" encoding="utf-8"?><!-- generator="wordpress/2.2.2" -->
<rss version="2.0" 
	xmlns:content="http://purl.org/rss/1.0/modules/content/">
<channel>
	<title>Comments on: The Horizon</title>
	<link>http://www.mikedidonato.com/2007/12/05/the-horizon/</link>
	<description>It's hard being a superstar</description>
	<pubDate>Tue, 02 Dec 2008 01:08:08 +0000</pubDate>
	<generator>http://wordpress.org/?v=2.2.2</generator>

	<item>
		<title>By: Paul Scott</title>
		<link>http://www.mikedidonato.com/2007/12/05/the-horizon/#comment-69049</link>
		<author>Paul Scott</author>
		<pubDate>Mon, 24 Dec 2007 15:09:13 +0000</pubDate>
		<guid>http://www.mikedidonato.com/2007/12/05/the-horizon/#comment-69049</guid>
		<description>Hi Mike - Wikpedia deals with Horizon Distance Directly. As can be seen from their analysis, you get less than 3 miles visual horizon (your answer) if  you're a dwarf.  You get 12 miles at a 100 feet vantage point.

See Below:

Distance to the horizon

The straight line of sight distance d in kilometers to the true horizon on earth is approximately

d = \sqrt{13h},

where h is the height above ground or sea level (in meters) of the eye of the observer. Examples:

    * For an observer standing on the ground with h = 1.70 m (average eye-level height), the horizon appears at a distance of 4.7 km.
    * For an observer standing on a hill or tower of 100 m in height, the horizon appears at a distance of 36 km.

To compute the height of a tower, the mast of a ship or a hilltop visible above the horizon, add the horizon distance for that height. For example, standing on the ground with h = 1.70 m, one can see, weather permitting, the tip of a tower of 100 m height at a distance of 4.7+36 ≈ 41 km.

In the Imperial version of the formula, 13 is replaced by 1.5, h is in feet and d is in miles. Examples:

    * For observers on the ground with eye-level at h = 5 ft 7 in (5.583 ft), the horizon appears at a distance of 2.89 miles.
    * For observers standing on a hill or tower 100 ft in height, the horizon appears at a distance of 12.25 miles.

The metric formula is reasonable (and the Imperial one is actually quite precise) when h is much smaller than the radius of the Earth (6371 km). The exact formula for distance from the viewpoint to the horizon, applicable even for satellites, is

    d = \sqrt{2Rh + h^2},

where R is the radius of the Earth (note: both R and h in this equation must be given in the same units (e.g. kilometers), but any consistent units will work).

Another relationship involves the arc length distance s along the curved surface of the Earth to the bottom of object:

    \cos\frac{s}{R}=\frac{R}{R+h}.

Solving for s gives the formula

    s=R\cos^{-1}\frac{R}{R+h}.

The distances d and s are nearly the same when the height of the object is negligible compared to the radius (that is, h</description>
		<content:encoded><![CDATA[<p>Hi Mike - Wikpedia deals with Horizon Distance Directly. As can be seen from their analysis, you get less than 3 miles visual horizon (your answer) if  you&#8217;re a dwarf.  You get 12 miles at a 100 feet vantage point.</p>
<p>See Below:</p>
<p>Distance to the horizon</p>
<p>The straight line of sight distance d in kilometers to the true horizon on earth is approximately</p>
<p>d = \sqrt{13h},</p>
<p>where h is the height above ground or sea level (in meters) of the eye of the observer. Examples:</p>
<p>    * For an observer standing on the ground with h = 1.70 m (average eye-level height), the horizon appears at a distance of 4.7 km.<br />
    * For an observer standing on a hill or tower of 100 m in height, the horizon appears at a distance of 36 km.</p>
<p>To compute the height of a tower, the mast of a ship or a hilltop visible above the horizon, add the horizon distance for that height. For example, standing on the ground with h = 1.70 m, one can see, weather permitting, the tip of a tower of 100 m height at a distance of 4.7+36 ≈ 41 km.</p>
<p>In the Imperial version of the formula, 13 is replaced by 1.5, h is in feet and d is in miles. Examples:</p>
<p>    * For observers on the ground with eye-level at h = 5 ft 7 in (5.583 ft), the horizon appears at a distance of 2.89 miles.<br />
    * For observers standing on a hill or tower 100 ft in height, the horizon appears at a distance of 12.25 miles.</p>
<p>The metric formula is reasonable (and the Imperial one is actually quite precise) when h is much smaller than the radius of the Earth (6371 km). The exact formula for distance from the viewpoint to the horizon, applicable even for satellites, is</p>
<p>    d = \sqrt{2Rh + h^2},</p>
<p>where R is the radius of the Earth (note: both R and h in this equation must be given in the same units (e.g. kilometers), but any consistent units will work).</p>
<p>Another relationship involves the arc length distance s along the curved surface of the Earth to the bottom of object:</p>
<p>    \cos\frac{s}{R}=\frac{R}{R+h}.</p>
<p>Solving for s gives the formula</p>
<p>    s=R\cos^{-1}\frac{R}{R+h}.</p>
<p>The distances d and s are nearly the same when the height of the object is negligible compared to the radius (that is, h</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: mike d.</title>
		<link>http://www.mikedidonato.com/2007/12/05/the-horizon/#comment-68122</link>
		<author>mike d.</author>
		<pubDate>Thu, 06 Dec 2007 22:00:14 +0000</pubDate>
		<guid>http://www.mikedidonato.com/2007/12/05/the-horizon/#comment-68122</guid>
		<description>Those jerks.  They copied my MSPaint.</description>
		<content:encoded><![CDATA[<p>Those jerks.  They copied my MSPaint.</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Andrew</title>
		<link>http://www.mikedidonato.com/2007/12/05/the-horizon/#comment-68118</link>
		<author>Andrew</author>
		<pubDate>Thu, 06 Dec 2007 16:19:09 +0000</pubDate>
		<guid>http://www.mikedidonato.com/2007/12/05/the-horizon/#comment-68118</guid>
		<description>Yeah, it's a short three miles if you're in a flat field.  If you get up onto some higher ground -- say, 3500 feet  from the top of Mount Greylock -- you can see a lot farther.  72 miles by the above calculation, which is clear to Connecticut.</description>
		<content:encoded><![CDATA[<p>Yeah, it&#8217;s a short three miles if you&#8217;re in a flat field.  If you get up onto some higher ground &#8212; say, 3500 feet  from the top of Mount Greylock &#8212; you can see a lot farther.  72 miles by the above calculation, which is clear to Connecticut.</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Mykal</title>
		<link>http://www.mikedidonato.com/2007/12/05/the-horizon/#comment-68114</link>
		<author>Mykal</author>
		<pubDate>Thu, 06 Dec 2007 15:17:04 +0000</pubDate>
		<guid>http://www.mikedidonato.com/2007/12/05/the-horizon/#comment-68114</guid>
		<description>That sounds right, as a general rule I learned that the horizon is usually around 3 miles away if you are looking out over a flat area.</description>
		<content:encoded><![CDATA[<p>That sounds right, as a general rule I learned that the horizon is usually around 3 miles away if you are looking out over a flat area.</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Aaron</title>
		<link>http://www.mikedidonato.com/2007/12/05/the-horizon/#comment-68110</link>
		<author>Aaron</author>
		<pubDate>Thu, 06 Dec 2007 13:30:38 +0000</pubDate>
		<guid>http://www.mikedidonato.com/2007/12/05/the-horizon/#comment-68110</guid>
		<description>Having spend a large part of my life in extremely flat, treeless areas, that sounds about right to me.  

But wait!  There's a problem!  The dude's eyes aren't at the top of his head!  Make your observer 6.1 feet tall, and your math is still probably accurate though.</description>
		<content:encoded><![CDATA[<p>Having spend a large part of my life in extremely flat, treeless areas, that sounds about right to me.  </p>
<p>But wait!  There&#8217;s a problem!  The dude&#8217;s eyes aren&#8217;t at the top of his head!  Make your observer 6.1 feet tall, and your math is still probably accurate though.</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Adam</title>
		<link>http://www.mikedidonato.com/2007/12/05/the-horizon/#comment-68108</link>
		<author>Adam</author>
		<pubDate>Thu, 06 Dec 2007 13:13:27 +0000</pubDate>
		<guid>http://www.mikedidonato.com/2007/12/05/the-horizon/#comment-68108</guid>
		<description>If you were on Wikipedia already, why didn't you just head on over to "horizon"?
http://en.wikipedia.org/wiki/Horizon</description>
		<content:encoded><![CDATA[<p>If you were on Wikipedia already, why didn&#8217;t you just head on over to &#8220;horizon&#8221;?<br />
<a href="http://en.wikipedia.org/wiki/Horizon" rel="nofollow">http://en.wikipedia.org/wiki/Horizon</a></p>
]]></content:encoded>
	</item>
	<item>
		<title>By: shamus</title>
		<link>http://www.mikedidonato.com/2007/12/05/the-horizon/#comment-68099</link>
		<author>shamus</author>
		<pubDate>Thu, 06 Dec 2007 04:59:32 +0000</pubDate>
		<guid>http://www.mikedidonato.com/2007/12/05/the-horizon/#comment-68099</guid>
		<description>you know what this problem needs?

more sig figs

also this site's calculator jives with the 3 mile ball park http://www.boatsafe.com/tools/horizon.htm

Another way to do the problem is to set up that same triangle and then find acos(r/h).  If you do this in radians just multiply the radius by the angle to get the arc length rather than the other side of the triangle. which incidentally yields 2.948 miles as well.</description>
		<content:encoded><![CDATA[<p>you know what this problem needs?</p>
<p>more sig figs</p>
<p>also this site&#8217;s calculator jives with the 3 mile ball park <a href="http://www.boatsafe.com/tools/horizon.htm" rel="nofollow">http://www.boatsafe.com/tools/horizon.htm</a></p>
<p>Another way to do the problem is to set up that same triangle and then find acos(r/h).  If you do this in radians just multiply the radius by the angle to get the arc length rather than the other side of the triangle. which incidentally yields 2.948 miles as well.</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Jes Saint</title>
		<link>http://www.mikedidonato.com/2007/12/05/the-horizon/#comment-68094</link>
		<author>Jes Saint</author>
		<pubDate>Thu, 06 Dec 2007 04:04:20 +0000</pubDate>
		<guid>http://www.mikedidonato.com/2007/12/05/the-horizon/#comment-68094</guid>
		<description>I have no idea, but it sounds ok to me.  I just love that the stick figure in the diagram is holding a guitar.</description>
		<content:encoded><![CDATA[<p>I have no idea, but it sounds ok to me.  I just love that the stick figure in the diagram is holding a guitar.</p>
]]></content:encoded>
	</item>
</channel>
</rss>
