Calcuation of Moments of Inertia for Simple Fractional Rig

Hi,

I am trying to cacluate the longitudinal and transverse moments of inertia for a small boat sailing rig. I have gone for a simple (no spreader) fractional rig at this time.
Skenes (Kinney) does not cover this kind of rig, so I am next trying Eulers forumulas, in the form of:

I-Longitudinal = (Mast Load x 3) / (2 x (3.14 / panel length)^2 x (10 x 10^6, for aluminum))

I-Transverse = (Mast Load x 3) / ((3.14 / panel length)^2 x (10 x 10^6, for aluminum))

The problem is, because there are no spreaders and thus only one panel, that the only difference between the two is the factor of 2 multipler in the denominator is left out for T-Transverse. This results in a transverse moment of inertia that is twice the longitudinal, which is the reverse of what most masts are designed to support.

My question is, am I doing something fundamentally wrong, or does Eulers just not work for simple, no spreader, rigs?

I have seen that Larsson does include an approach for simple rigs, and am trying to figure that out from the rather disjointed descriptions and forumlas, but because I have been taught to use Eulers, I first want to know if I can use that instead.

Thanks,

Grant

 

OK, I keep looking at my reference texts and finally understood that the '2' difference was actually the 'end fixity constant' - the degree of freedom of movement at the ends of the column. In the reference I use, the mast is consider to be held better in place at the top of the mast, than it is at the spreaders by a factor of two.

THis leads me to think that for a simple no spreader, fractional rig, I should use the same constant.

Anyone have any idea if I am thinking straight?

If that is correct, then I need to decide what that constant should be... something between 1 and 2 I guess, probably closer to 1 than 2..

The version of Eulers I was using was:
Longitudinal: W = 2(Pi/L)^2 *E * I
Transverse: W = (Pi/L)^2 *E * I
or, for I:
Longitudinal: I = W / (2(Pi/L)^2 *E)
Transverse: I = W / ((Pi/L)^2 *E)

Cheers,

Grant

 

Further research...

Larsson, in Principles of Yacht Design covers simple rigs. Its interesting to see that using their forumulas, the transverse moment of inertial (Ix) also comes out higher than the longitudinal - 1.2 times - but less than by the Euler´s formula I mention above (2 times).

The Australian standard, I am told, also covers simple rigs, but with just the opposite result of larsson, Iy, is 1.2 times greater.

Now, I wonder, how is it possible that three well respected methods of calculating mast size can vary by more than 100% for Iy relative to Ix?

Grant

 

Just read your postings and would ask are all models assuming no spreaders or intermediate shrouds? Also, do they suggest a different calculation if there are running back stays or staysail stays? My hunch is that the formula with the difference of "2" is actually assuming a single set of lower shrouds and no running backstays, thus the 2X larger load longitudinally.

Hope that helps.

BV

 

Hi BV,

Indeed the particular Eulers I used was given in connection for masts with spreaders, and also a backstay as far as I can tell. Thus the formula for transverse I is for panels that have no for and aft support, other than angled shrouds. Thus the factor of 2 is added as these have less 'fixity', which menas more strenght is needed, but at the same time the panels (distance between spreaders) are shorter than the total height of the mast which is used for for the longitudinal I, and so the total value still comes out less for Ix than Iy. But since I was trying to use it for a mast with no spreaders, where the x and y panels are the same length, I get an Ix that is 2x larger than for Iy.

So, at least for Eulers, I would be asking our esteemed forum membership to remark if there is version of Eulers that can be used for simple, no spreader (fractional) rigs. Mainly its a question of the contant to apply in regard to the fixity of the ends of the mast (or where the shrouds and stays meet the mast).

THanks for taking a look and making your suggestions. I was giving up hope... ;-)

Grant

 

Grant, I'm not NA, but I'd hazard a guess that you could use the formula for the fore-n-aft calculation for both directions, with the understanding that you'll need a bit more metal to deal with the compression load caused by the side shrouds being a great deal closer to the mast step (horizontally) than the bow and back stays. Also, without any intermediate stays of either kind, you might want ton consider a round section, rather than oval. My understanding is that designers went to an oval section with flat sides on the beam sides of the mast due to the desire to get away from running back stays and staysail or inner forestays.

Good luck,

BV