DAR measurement

[vyv_cox]

Yesterday we discovered that the hub of Cecilia's propeller is riddled with cracks. Two repairers tell me that they would not attempt to fix it, which is what I expected. Researching a new one I came across DAR, Disc Area Ratio, a new expression for me. I need to choose between 43% and 52%, either of which it seems are possibles for the boat.

To measure it, I took a photo of the old prop, transferred it into Photoshop, added small circles over the blade and the space between, then counted them. Pleased to discover that the method gave a surprisingly accurate estimate of the ratio.

The prop is marked up 08305 18 x 14 RH BE. I assume the number is a maker's derail number or ID, the rest is standard apart from BE. Anyone know what that means?

 

[rob2]

Can't help with the BE, I'm afraid - could it be the manufacturer? As I'm sure you're aware, but maybe for the benefit of others, Sailing boats tend to use a significantly lower DAR blade pattern than a motor boat. It is a design factor which is unfortunately not directly used in most prop calculations, although as it has a major effect on efficiency and slip, most calculators adjust for it by adjusting the slip. It is assumed that the lower DAR causes less drag when sailing, but I wonder how it moves the goalpost in the perennial argument as to whether a prop drags more when spinning or locked. For myself I looked at the "proof" a series of test on a prop without defining DAR characteristics and considered it a perfect example of a failure in design of experiment. Intuition suggests that props with varying characteristics fitted to boats with differing characteristics will give different results leaving us with the most frequent result in boating "Well, it depends..." You really need to identify every variable and prove their interactions before you can hope to calculate with any accuracy. That explains the spin/lock confusion and why some props supplied just don't perform as expected.

Rob.

 

[rob2]

Maybe the BE is indicating where you are supposed to hit it "Beat 'Ere"

Sorry not very helpful I know....

Who needs help when you can have a giggle?

 

[nigelmercier]

An easier way:
Make a selection of the propeller, or a blade.
Use Histogram window in Photoshop (click on "extended view"). In the lower left corner you'll find the number of pixels of your selection.
Repeat for other area.

 

[PuffTheMagicDragon]

Perhaps the bronze contains Beryllium in the alloy?

 

[vyv_cox]

An easier way:
Make a selection of the propeller, or a blade.
Use Histogram window in Photoshop (click on "extended view"). In the lower left corner you'll find the number of pixels of your selection.
Repeat for other area.

Tried it but found it far more difficult in practice. Not sure my version of Photoshop is the same as yours.

 

[Seajet]

Vyv Cox,

pardon my forgetfulness, but is this a saildrive hub ?

The reason I ask is that the Volvo saildrive job ( hub / boss ) holding the 2 blade folding prop' cracked on my Carter.

I took it to BAe Kingston which at the time had about 6 departments varying from medium to high tech welding, and despite them being chums all pronounced it " **** high zinc alloy' can't do anything with it ! "

Cost me £480 for a new Volvo saildrive prop boss in 1988...

Hope you have better luck.

 

[VicS]

To measure it, I took a photo of the old prop, transferred it into Photoshop, added small circles over the blade and the space between, then counted them. Pleased to discover that the method gave a surprisingly accurate estimate of the ratio.

I'd have drawn round it, cut the sections out and weighed them. You do need a sensitive weighing machine, of course, but I have an old fashioned 2 pan chemical balance.

 

[nigelmercier]

I'd have drawn round it, cut the sections out and weighed them...

"Old School" Rocks!

 

[vyv_cox]

Seajet, No it's a conventional prop on a conventional shaft drive. I have no problem with their statements that they cannot repair it, it would be a very tricky job, not because of the alloy but the position of the cracks. Now I have looked at it with magnification I can see it is a network of corrosion cracks, not a fatigue or mechanical single one.

 

[Seajet]

VyV_Cox,

oh well it means a new shaft then ? I have no experience in the Med', whether a supplier here would be preferable to a local, though my instincts from what I hear say 'get a supplier here from UK and sort out cheapest transport, ideally a chum / forumite heading that way ' !

A new shaft should fit in a hatchback car ?

Good luck.

 

[AntarcticPilot]

I'd have drawn round the blade, and computed the area of the resulting polygon.

The area of the total disc is simple; πR².

If you don't have software that will compute the area of an arbitrary polygon, it is actually pretty easy to do manually from the raw coordinates - for each pair of adjacent points, compute the area between the line joining them and a base-line outside the polygon (either the minimum Y coordinate or the Y-axis will usually work; the latter makes the sums simpler). Add the area to the sum if the second polygon has a greater X coordinate than the first, and subtract it is the second x-coordinate is less than the first.

It should be straightforward to prepare a spreadsheet to do the calculation. Given that the curves of the blade outline are smooth, you wouldn't need too many points to define the polygon close enough to give a good result.

 

[Jegs]

Hi Vyv,

An ill-informed stab in the dark - Right Hand Blade Edge?

John G

 

[vyv_cox]

VyV_Cox,

oh well it means a new shaft then ? I have no experience in the Med', whether a supplier here would be preferable to a local, though my instincts from what I hear say 'get a supplier here from UK and sort out cheapest transport, ideally a chum / forumite heading that way ' !

A new shaft should fit in a hatchback car ?

Good luck.

It's our 'winter' boat, Colvic Northerner now in Port Dinorwic (North Wales, not France as someone seems to think). Just needs a new prop but Castle Marine can supply at lower cost than most others and they are about 5 miles away.

 

[vyv_cox]

I'd have drawn round the blade, and computed the area of the resulting polygon.

The area of the total disc is simple; πR².

If you don't have software that will compute the area of an arbitrary polygon, it is actually pretty easy to do manually from the raw coordinates - for each pair of adjacent points, compute the area between the line joining them and a base-line outside the polygon (either the minimum Y coordinate or the Y-axis will usually work; the latter makes the sums simpler). Add the area to the sum if the second polygon has a greater X coordinate than the first, and subtract it is the second x-coordinate is less than the first.

It should be straightforward to prepare a spreadsheet to do the calculation. Given that the curves of the blade outline are smooth, you wouldn't need too many points to define the polygon close enough to give a good result.

Doing it the way I did using Powerpoint took me about ten minutes. I cannot see any calculation method being quicker, although a spreadsheet might well be more fun.

 

[AntarcticPilot]

Doing it the way I did using Powerpoint took me about ten minutes. I cannot see any calculation method being quicker, although a spreadsheet might well be more fun.

I'm used to using software that automagically computes areas on the fly using the algorithm I outlined, or a similar one.

You would only need to get the coordinates of about ten points and shove them into a spreadsheet. The spread-sheet equation would look something like:

Area = (Y2+Y1) * (X1-X2)/2 for each pair of coordinates. You then simply sum the results of this formula.

I would attach a spreadsheet but it won't let me!