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Re: [Shop-talk] math wizards ??? come here

To: "john niolon" <jniolon@bham.rr.com>, "Shop-Talk List"
Subject: Re: [Shop-talk] math wizards ??? come here
From: "Karl Vacek" <kvacek@ameritech.net>
Date: Thu, 9 Apr 2009 20:34:01 -0500
What's so complicated here ?  No trig needed - simple geometry no matter 
what the angle (the angle must theoretically be less than 90 degrees, 
practically it has to be less than about 75 degrees because of physical 
constraints of most engine cranes)

As drawn, 1" of lift at the jack point (1' from the pivot) results in 
exactly 8" of lift at the lift point (8' from the pivot).  As someone said - 
similar triangles.

Karl



> use a sine function, not a tangent, since you have the hypotenuse and
> the side opposite the angle given.
> side note:  tan and  sine of relatively small angles are just about
> the same number since tan = sin/cos and cos of small angles is close
> to 1.0
>
> -Roland
>
> On Thu, 9 Apr 2009 19:32:24 EDT, you wrote:
>
> ::In a message dated 4/9/09 12:49:19 P.M. Pacific Daylight Time,
> ::cak@dimebank.com writes:
> ::
> ::> You  need a few more pieces of information in order to do the trig.
> ::
> ::Not really ... as Randall pointed out, 'similar triangles' gets
> ::it  done with what he's got, since all the ratios remain the  same.
> ::
> ::
> ::
> ::Only if the boom has a slide on it, allowing the boom to lengthen as it
> ::rises;  Given that he has the length from the hinge to the hook and  the
> ::length from the hinge to the jacking point, a trig. table or a 
> scientific
> ::calculator should give  the angle of which the tangent is  1"/12"; then 
> that
> ::angle's tangent, multiplied by 96" should give the height  of the hook 
> above
> ::level.
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