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	<title>Comments on: Rigging, Leverage Formula and The &#8220;M&#8221;</title>
	<atom:link href="http://www.maxrigging.com/rigging-leverage-formula-and-the-m/feed" rel="self" type="application/rss+xml" />
	<link>http://www.maxrigging.com/rigging-leverage-formula-and-the-m</link>
	<description>how to get more out of your rowing, and rowing equipment</description>
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		<title>By: Steve</title>
		<link>http://www.maxrigging.com/rigging-leverage-formula-and-the-m/comment-page-1#comment-1286</link>
		<dc:creator>Steve</dc:creator>
		<pubDate>Tue, 06 Jul 2010 14:17:06 +0000</pubDate>
		<guid isPermaLink="false">http://www.maxrigging.com/?p=778#comment-1286</guid>
		<description>Okay so I have used the same leverage formula ([(L-I)-B/M]/S) for years....but I have been nagged by this problem. 

With adjustable oars (this formula came out prior to adjustable oars), this formula does not take into account the inboard, or more accurately the inboard/outboard ratio, of the rigging. This formula only really uses the outboard (L-I)...so as long as the outboards are the same, it never considers how long the inboard is, and what sort of impact that has on the rigging. 

For example, and using extremes to prove my point....assuming you had two shells with the same spread, and the same blades, this formula would say that a 374 length oar with 115 inboard would have the same leverage as a 370 oar and a 111 inboard.

As a rigging nut, I have played around with this and they most definitely are not the same...

Any thoughts on this?  Am I missing something?  Is there a way to take into account the inboard when using this formaula???</description>
		<content:encoded><![CDATA[<p>Okay so I have used the same leverage formula ([(L-I)-B/M]/S) for years&#8230;.but I have been nagged by this problem. </p>
<p>With adjustable oars (this formula came out prior to adjustable oars), this formula does not take into account the inboard, or more accurately the inboard/outboard ratio, of the rigging. This formula only really uses the outboard (L-I)&#8230;so as long as the outboards are the same, it never considers how long the inboard is, and what sort of impact that has on the rigging. </p>
<p>For example, and using extremes to prove my point&#8230;.assuming you had two shells with the same spread, and the same blades, this formula would say that a 374 length oar with 115 inboard would have the same leverage as a 370 oar and a 111 inboard.</p>
<p>As a rigging nut, I have played around with this and they most definitely are not the same&#8230;</p>
<p>Any thoughts on this?  Am I missing something?  Is there a way to take into account the inboard when using this formaula???</p>
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		<title>By: Mike</title>
		<link>http://www.maxrigging.com/rigging-leverage-formula-and-the-m/comment-page-1#comment-902</link>
		<dc:creator>Mike</dc:creator>
		<pubDate>Wed, 28 Oct 2009 02:14:37 +0000</pubDate>
		<guid isPermaLink="false">http://www.maxrigging.com/?p=778#comment-902</guid>
		<description>Hi Ian:

Darn good question, and something I&#039;ve been looking for myself. I&#039;ve tried with several physicists to develop one,
but no luck. I&#039;m not saying it is not out there, I just cannot find it . . . 

Hmm . . . anyone have a lead on this???</description>
		<content:encoded><![CDATA[<p>Hi Ian:</p>
<p>Darn good question, and something I&#8217;ve been looking for myself. I&#8217;ve tried with several physicists to develop one,<br />
but no luck. I&#8217;m not saying it is not out there, I just cannot find it . . . </p>
<p>Hmm . . . anyone have a lead on this???</p>
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	<item>
		<title>By: Ian</title>
		<link>http://www.maxrigging.com/rigging-leverage-formula-and-the-m/comment-page-1#comment-901</link>
		<dc:creator>Ian</dc:creator>
		<pubDate>Wed, 28 Oct 2009 02:01:30 +0000</pubDate>
		<guid isPermaLink="false">http://www.maxrigging.com/?p=778#comment-901</guid>
		<description>Mike,
Is there a formula available that WILL compare between different blade shapes?  I&#039;ve looked everywhere and can&#039;t find one.  Is there a way that we can derive blade surface area into the length value?</description>
		<content:encoded><![CDATA[<p>Mike,<br />
Is there a formula available that WILL compare between different blade shapes?  I&#8217;ve looked everywhere and can&#8217;t find one.  Is there a way that we can derive blade surface area into the length value?</p>
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	<item>
		<title>By: Mike</title>
		<link>http://www.maxrigging.com/rigging-leverage-formula-and-the-m/comment-page-1#comment-863</link>
		<dc:creator>Mike</dc:creator>
		<pubDate>Mon, 17 Aug 2009 02:37:40 +0000</pubDate>
		<guid isPermaLink="false">http://www.maxrigging.com/?p=778#comment-863</guid>
		<description>@Ali:

Two thoughts here. First, it would depend on the length of the oar blade, as I use approx 1/2 the length to find the &quot;M&quot;. Second, you cannot use this formula to compare the two types of blade design. It will not give you a valid comparison.</description>
		<content:encoded><![CDATA[<p>@Ali:</p>
<p>Two thoughts here. First, it would depend on the length of the oar blade, as I use approx 1/2 the length to find the &#8220;M&#8221;. Second, you cannot use this formula to compare the two types of blade design. It will not give you a valid comparison.</p>
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	<item>
		<title>By: Ali</title>
		<link>http://www.maxrigging.com/rigging-leverage-formula-and-the-m/comment-page-1#comment-862</link>
		<dc:creator>Ali</dc:creator>
		<pubDate>Tue, 11 Aug 2009 21:04:03 +0000</pubDate>
		<guid isPermaLink="false">http://www.maxrigging.com/?p=778#comment-862</guid>
		<description>How does the M value of the C2 Fat Smoothie sweep oar compare to that of the Smoothie2 with the vortex edge?</description>
		<content:encoded><![CDATA[<p>How does the M value of the C2 Fat Smoothie sweep oar compare to that of the Smoothie2 with the vortex edge?</p>
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