# Milano Bridge allides with gantry crane in Busan today

I don’t know how to read that table. Presumably T is thrust and Q is torque. I don’t know what the n is.

If this table is applicable we could assume the Milano Bridge prop is about 1/4 out of the water what results does that give?

Google tells me that n is efficiency which is: Efficiency = 100% - Slip.

In this case it looks like its the third column divided by the forth column. ( 8th row: 0.259/0.314 = 0.824)

I think it is safe to interpolate this value which is half of the half blade distance, thus in between 1 and 0.789 which amounts to about 0.89 of thrust meaning that 89% of the full thrust is available at that moment.

The efficiency, generally indicated with the nabla sign η, is the quotient of thrust and torque.

1 Like

I was thinking that on the Milano the Bridge the blade was likely more than 1/4 out of the water so 1/4 would be h/R = 0.5.

From h/R = 1.0 at 2 rps thrust drops off from 0.944 to 0.789 which is -20.6%.

You mean that the interpolation is between .944, not 1, and 0.789. The remaining thrust is then a little bit less, about 87%.

I see now that the rpm of the ship is 76 and thus the rps 1.27. The slower the rpm or rps the more thrust so that is not a bad thing. Hard to tell the difference.

I’m saying that from looking at this:

That h/R = 0.5 or maybe less. Assume 0.5.

Don’t know what it’s called. The part fwd of the prop that holds the shaft can be seen:

MB closer to h/R = 0 than 0.5 possibly.

I don’t get it. Is the last photo of the Milano Bridge? Just dipping with the propeller blades in the water…

The top picture is the MB, the top of the shaft housing can be seen just above water level. It’s from the video taken by the crew member.

The photo of the ship with the blue paint is a photo of the same area but of some random ship.

A different somewhat crude approach. I measured on the photo the width of the ship. In my case it was 17.5 cm but that might be different on larger screens. Then I measured the top of the rudder to the waterline which was 1.2 cm. A bit difficult because of the wake. So the rudder and the prop as seen are 1.2/17.5 * 51 m ≈ 3.5 m above the waterline.

Normally the propeller’s tip is in line with the top of the rudder so then the part of the blade that is out of the water is also 3.5 m. For a ship of this size I estimated the propeller’s diameter at 9.6 m, could not find this value on the internet. Interpolating this as done earlier the efficiency is 84%, still not too bad considering.

2 Likes

The 84 % assumes that efficient is max at h/R = 0. But the tip of the blade at the surface is not considered efficient. At least not in my experience.

We measured propeller immersion by the radius of the prop, not the diameter. The minimum required immersion was 0.7, that number being the distance the tip was submerged. For us it worked out to an aft draft of about 7.2 meters or thereabouts.

Using the h/R method minimum allowable submersion would be h/R = 1.35

I posted this before but here at the min drafts for the Panama Canal.

Here’s the min drafts for transit through the Panama Canal for a 200 meter ship.

Over 625′ (190.50 m) – 24′ (7.32 m) forward, 26′ (7.93 m) aft, TSW

We had a min draft fwd as well for bow thruster effectiveness.

1 Like

Company report that came with the chartlet noted 60% (looks like) immersion:

Anyway what matters here is not prop thrust directly but reduction in turning power. So rudder immersion matters as well.

The immersion of the prop was 60% according to this paper. With the somewhat crude calculation which I made above the rudder and prop immersion is 61%, so that is not too far off!

What matters is the Y-axis velocity (breaking down the chartlet into a Cartesian system). A toll in advance distance was going to be exacted past a certain point with the velocity she had approaching the dock. If the chartlet is scale, you could calculate the speed and accel vectors of the port quarter. And of course the propeller efficiency table doesn’t really tell anything about the astern efficiency.

But of course there is the matter of tugs and environmental conditions… while the math might amuse, the recommendation seems obvious without calculation.

With a normally always-immersed propeller:
Is it known, how an accidentally emerged part would influence the strength of the propeller walk?

The flow spiral, created by the ejected water, could not be a complete one, anymore.
More or less lateral displacement of the stern?

My first thought was the transverse thrust of the RH fixed pitch propeller would be increased due the the incomplete immersion in an astern movement and I haven’t seen much to change my mind. The rudder operating in highly aerated water would have little effect.

2 Likes

I don’t have a source but as the rough rule of thumb I recall is backing power is 70% of ahead power. It has been mentioned on this thread in some cases it’s much less.

As far as advance the general rule of thumb there is 3-4 ship lengths.

Without getting bogged down in the specifics doesn’t look like the Milano Bridge had a lot of room to turn.

It seem odd that not only a lot of speed but started out with only 20 degrees of rudder. Like they didn’t expect to have any trouble making the turn. Didn’t try to tighten up till way late.

It’s from SOLAS.

Class rules will vary but match up to IACS requirement guiding compliance :

M25.1 In order to maintain sufficient manoeuvrability and secure control of the ship in all normal circumstances, the main propulsion machinery is to be capable of reversing the direction of thrust so as to bring the ship to rest from the maximum service speed. The main propulsion machinery is to be capable of maintaining in free route astern at least 70% of the ahead revolutions.

That could have been where I saw it but my recollection, fwiw, is that at the same rpm the prop astern would have roughly 70% of the thrust of the prop when ahead at the same rpm.

I agree with the power astern being 70% of ahead. I don’t think you would get much response from the engine until the ship lost some way. All you would do is run out of air.

I think 70% must be the standard they use in the shiphandling books. I seem to have mislaid mine at the moment or I would have looked it up.

In any case I don’t think it applies here as it looks like the Milano Bridge was ahead till the end.

Also, yeah, I think a full astern with a light ship and 60% immersion; she’s gonna want to take a walk.