<?xml version="1.0" encoding="UTF-8"?><rss version="2.0"
	xmlns:content="http://purl.org/rss/1.0/modules/content/"
	xmlns:dc="http://purl.org/dc/elements/1.1/"
	xmlns:atom="http://www.w3.org/2005/Atom"
	xmlns:sy="http://purl.org/rss/1.0/modules/syndication/"
	xmlns:itunes="http://www.itunes.com/dtds/podcast-1.0.dtd"
		>
<channel>
	<title>Comments on: How Much Torque (NM) Does a Golf Swing Put on the Average Person/Tiger&#8217;s Knee? What Can I Compare this To?</title>
	<atom:link href="http://golfguideonline.net/114/torque-nm-golf-swing-put-average-persontigers-knee-compare/feed/" rel="self" type="application/rss+xml" />
	<link>http://golfguideonline.net/114/torque-nm-golf-swing-put-average-persontigers-knee-compare/</link>
	<description>Golfing Tips</description>
	<lastBuildDate>Sat, 10 Oct 2009 20:52:31 +0000</lastBuildDate>
	<sy:updatePeriod>hourly</sy:updatePeriod>
	<sy:updateFrequency>1</sy:updateFrequency>
	<xhtml:meta xmlns:xhtml="http://www.w3.org/1999/xhtml" name="robots" content="noindex" />
	<item>
		<title>By: Joe M</title>
		<link>http://golfguideonline.net/114/torque-nm-golf-swing-put-average-persontigers-knee-compare/#comment-394</link>
		<dc:creator>Joe M</dc:creator>
		<pubDate>Thu, 03 Sep 2009 18:57:36 +0000</pubDate>
		<guid isPermaLink="false">http://golfguideonline.net/114/how-much-torque-n-m-does-a-golf-swing-put-on-the-average-persontigers-knee-what-can-i-compare-this-to/#comment-394</guid>
		<description>&lt;a href=&quot;&quot;&gt;Lolita&lt;/a&gt;


Here it is in simple terms.
Torque = Force X lever. 
The lever is the distance from the center of curvature of the swing. 
Force= mass X acceleration

Mass is the wt of the club, his arms and the ball itself when he hits it.
Acceleration if from the equation 
accel=w^2(r)
Here is how you find the r. Find the point where half of the mass of the swing is further out, and half is closer to the center. That is the r part of the acceleration.
Now let me get in to the w. Just at there are 360 degrees in a complete circle. There is 2*pi radians in a complete circle. Approx 6.28 radians in a complete circle.
w=radians/second
Try to find out how much time it takes to complete the swing and how much of a complete circle it takes. Take the percentage of a complete circle, multiply it y 2 * pi and then divide it by the amount of time it takes to complete that swing. That would be w. Take the w to the second poser, this is w^2.</description>
		<content:encoded><![CDATA[<p><a href="">Lolita</a></p>
<p>Here it is in simple terms.<br />
Torque = Force X lever.<br />
The lever is the distance from the center of curvature of the swing.<br />
Force= mass X acceleration</p>
<p>Mass is the wt of the club, his arms and the ball itself when he hits it.<br />
Acceleration if from the equation<br />
accel=w^2(r)<br />
Here is how you find the r. Find the point where half of the mass of the swing is further out, and half is closer to the center. That is the r part of the acceleration.<br />
Now let me get in to the w. Just at there are 360 degrees in a complete circle. There is 2*pi radians in a complete circle. Approx 6.28 radians in a complete circle.<br />
w=radians/second<br />
Try to find out how much time it takes to complete the swing and how much of a complete circle it takes. Take the percentage of a complete circle, multiply it y 2 * pi and then divide it by the amount of time it takes to complete that swing. That would be w. Take the w to the second poser, this is w^2.</p>
]]></content:encoded>
	</item>
</channel>
</rss>

