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RE: TR3A Ignition timing

To: "'KTRIUMPH@aol.com'" <KTRIUMPH@aol.com>, lermanis@netspace.net.au
Subject: RE: TR3A Ignition timing
From: "Gambony, Jim" <jim.gambony@eds.com>
Date: Mon, 26 Oct 1998 16:19:53 -0600
Cc: triumphs@Autox.Team.Net
Or if you don't have any pies in the house... measure the circumference of
the pulley and divide by 36 (360 degrees in a circle/10degrees) to figure
out how far the 10 degree mark would be.

Jim

> -----Original Message-----
> From: KTRIUMPH@aol.com [SMTP:KTRIUMPH@aol.com]
> Sent: Monday, October 26, 1998 4:06 PM
> To:   lermanis@netspace.net.au
> Cc:   triumphs@autox.team.net
> Subject:      Re: TR3A Ignition timing
> 
> 
> In a message dated 98-10-26 06:31:12 EST, you write:
> 
> << It is at this point that I seek your assistance:
>  
>  If the hole in the crank shaft pulley is TDC for Number1 cylinder,how far
>  along the circumference of the pulley would the 10 degree position be? I
>  ask this with the hope that someone has had to do this calculation in the
>  past. Also I presume the 10 degree BTDC mark would go to the right(ie
>  Clockwise)of the TDC hole in the pulley.
>  
>  All suggestions appreciated.
>  
>  Chris Lermanis.
>  TR3A 1958 >>
> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
> Elementary, my dear Watson!
> 
> Measure the diameter of the pulley, multiply by pi (3.1414 I believe), and
> divide by 36.
> Dividing by 360 will yield 1 degree. 
> Shoulda paid more attention in geometry class!
> 
> Ken Nuelle
> 58 TR3A
> 62 TR3B
> 64 TR4
> By the way if it isn't handy to measure the diameter for you I think I
> could
> dig up an old pulley and measure it for you - let me know.

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