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Date: Wed, 28 Dec 2005 10:21:21 -0500
From: Ernest Christley <echristley@nc.rr.com>
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To: Rotary motors in aircraft <flyrotary@lancaironline.net>
Subject: Re: [FlyRotary] Re: Alternate alternators
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William wrote:

> I don't understand something
>
> 0.577HP = 19041 #-ft/min,
>
> 19041 #-ft/min / 400 R/min = 47 ft-#
> Where did the 63.024 factor come from?
> Bill Schertz


Good catch, Bill.  The factor comes from the conversion from linear to 
rotary motion. 400 RPM equates to 2512 ft/min (each revolution covers 
6.28ft at a radius of 1ft).  I took the conversion from Bob Knuckoll's 
website, but I transcribed it wrong.  (It should be 63,024.) 

 Using the linear to rotary conversion, your equation becomes:

19041 #-ft/min / (400 R/min * 6.28ft/R) = 7.58 ft-#

Substituting the CORRECT constant into my equation below gives:

(.577HP x 63024) / 400RPM = 90.12 in-#
90.12 in-# x 1ft/12in = 7.576 ft-#

I think the 90.12in-# figure is the most important.  I don't have an 
exact measurement, but as I remember it the distributor drive gear is 
about an inch across, give or take a bit.  There will be something on 
the order of 90# of force trying to break the gear teeth off of the 
distributor shaft.  Someone else may want to actually run an experiment 
to see if the distributor drive gears can handle the load of a 
generator, but I'm going to move on and say that Richard was right.

Jesse, I think the temperature is just a little to high to bother with 
checking this lake for its skating potential. 8*)


>> One of the things Richard and I discussed yesterday was going with 
>> the RX-8 style of pickup and using the distributor shaft to drive an 
>> alternate power source.  He rightly noted that the gears probably 
>> aren't rated to be very strongs, since all they do is drive the 
>> distributor and oil pumps.  The questions are 1)How much load would a 
>> generator add?  and 2) How much load can the gears take?
>>
>> I don't have anything for the second question, but I got some numbers 
>> for the first part using: 
>> http://www.aeroelectric.com/articles/alternators/UA/Alternators_3.html
>>
>> Assuming a 20A load at an 800RPM idle as the highest torque a 65% 
>> efficient generator would be applying:
>>
>> (14V x 20A)/.65 = 431W needed to drive the generator.  431/746 yields 
>> .577HP.
>>
>> (.577HP x 63.024) / 400RPM = .091 in-lb of torque on the gear.
>>
>> .091in-lb?  I keep thinking that I missed something or 
>> miscalculated.  Couldn't even cheap nylon gears could handle this 
>> sort of load?  Is there any reason not to trust the distributor drive 
>> gears to be able to drive an alternate power source?
>>
>> -- 
>>         ,|"|"|,                                    |
>> ----===<{{(oQo)}}>===----        Dyke Delta         |
>>        o|  d  |o          www.ernest.isa-geek.org  |
>>
>> -- 
>> Homepage:  http://www.flyrotary.com/
>> Archive and UnSub:   http://mail.lancaironline.net/lists/flyrotary/
>
>
> -- 
> Homepage:  http://www.flyrotary.com/
> Archive and UnSub:   http://mail.lancaironline.net/lists/flyrotary/
>


-- 
         ,|"|"|,                                    |
----===<{{(oQo)}}>===----        Dyke Delta         |
        o|  d  |o          www.ernest.isa-geek.org  |