Hydrodynamics of the Nautilus - some thoughts and observations

[bob the builder]

So, according to my calculations,

So, according to my calculations, a 1/32nd scale Nautilus will need to cut through the wet stuff at about 18.5 mph to emulate the wave formation from the movie.

That's a fair old clip.

That works out to about 16 knots or 8.27 meters per second!

I'm thinking that without rocket engines, there ain't no way my model will ever be getting up to those kinds of speeds. It would probably tear off the salon windows, speed indicator, atomizer, and anything else sticking out of the hull even if it ever did...

You know.. that sounds awefully fast.. I get a decent bow wave even at 5mph (as fast as I've gotten her). Triple that speed sounds kind of excessive. Perhaps the scale movement/reaction of the water means that we wouldn't have to get quite that fast?

[Guest]

Well the formula I used

Well the formula I used is supposed to take that into account (I believe it is based on the principles of Froudes law).

However, it's all assuming the figures quoted for the Nautilus 'sled' filming miniature are in fact accurate.

The speed decreases as the scale gets smaller. So a little 31" nautilus wouldn't need the same amount of push to accurately render collision speed.

Andy

[carcharadon]

Well as best as I

Well as best as I can tell from my limited searchings of Froudes law, nowhere do they consider or mention how a model LOOKS. It seems to me that these equations are more for extrapolation of a hull design from a model to a full-scale in terms of flow resistance. How a fast-moving model LOOKS in comparison to a full-scale does not seem to even be a consideration.

But even if I'm wrong in my vague recollection, that the 22 ft. sled Nautilus was pulled at a scale speed of 300 mph it is not unreasonable to assume that a truck pulling the sled could obtain a speed of 37 mph instead of 190 mph. These are simple linear relationships. In other words a full-scale Nautilus traveling at 300 mi. an hour would take X seconds for the ship to pass a point from the tip of its nose to the furthest most point of its tail. Likewise in a linear relationship (which this is) it would take a 22 ft. Nautilus X seconds at 37 mph to pass the same point. Where X. equals X. So it is not unreasonable to believe that a truck pulling a 22 ft. sled could achieve a speed of 37 mph instead of 190 mph. For a 7 ft. sub this would be 11 mph. Again as I said these are simple linear relationships where X would-be the constant for both (for all three) subs. Now, there are two questions; 1; if it was at 37 mph or 300 mph (maybe it was 60 mph) would the equivalent speed of 11 mph for a 7 ft. model look the same? 2; I'm not sure what scale or speed they said it was can't find it on the DVD.

I seriously doubt that when they were filming this they took into consideration Froudes law.

[Guest]

Well as best as I

Well as best as I can tell from my limited searchings of Froudes law, nowhere do they consider or mention how a model LOOKS. It seems to me that these equations are more for extrapolation of a hull design from a model to a full-scale in terms of flow resistance. How a fast-moving model LOOKS in comparison to a full-scale does not seem to even be a consideration.

But even if I'm wrong in my vague recollection, that the 22 ft. sled Nautilus was pulled at a scale speed of 300 mph it is not unreasonable to assume that a truck pulling the sled could obtain a speed of 37 mph instead of 190 mph. These are simple linear relationships. In other words a full-scale Nautilus traveling at 300 mi. an hour would take X seconds for the ship to pass a point from the tip of its nose to the furthest most point of its tail. Likewise in a linear relationship (which this is) it would take a 22 ft. Nautilus X seconds at 37 mph to pass the same point. Where X. equals X. So it is not unreasonable to believe that a truck pulling a 22 ft. sled could achieve a speed of 37 mph instead of 190 mph. For a 7 ft. sub this would be 11 mph. Again as I said these are simple linear relationships where X would-be the constant for both (for all three) subs. Now, there are two questions; 1; if it was at 37 mph or 300 mph (maybe it was 60 mph) would the equivalent speed of 11 mph for a 7 ft. model look the same? 2; I'm not sure what scale or speed they said it was can't find it on the DVD.

I seriously doubt that when they were filming this they took into consideration Froudes law.

You can't apply a linear scale to the subject of scale speed, because the water molecules remain a constant, whilst the size of the vessel changes.

The formula I have presented here, is one shown in Norbert Bruggens 'Model Submarine Technology' book for calculating a scale speed where the boat is required to perform correctly i.e. set up a realistic wake in the water, good hydrovane response etc.

Now in the case of a Sci-fi design like the Disney Nautilus, we have no genuine fullsize counterpart to use as a guideline.

For instance, I am currently building a 1/100 scale model of the British Vanguard class SSBN. Fullsize, this vessel attains a total speed of 25 knots (submerged). Using the formula I presented-

25/squareroot of 100=

25/10= 2.5 knots

2.5 x 1.75 (model factor) = 4.375 knots

Using a linear scale -

25/100= 0.25 knots (a hopeless dribble)

In the case of the nautilus, the only figure we have to go on is the speed the sled was winched through the water.

Now I agree entirely, that the film makers likely didn't give a fig for Froudes law. They no doubt experimeted with different speeds until they got something that looked dramatic enough for cinema audiences.

However, for us modellers wishing to recreate that scene, then we must use some method of scaling that speed to our boats.

If the figure of 37mph is accurate, then to set up the same or at least very similar wake, we must attempt to reach near that speed according to the scale of our respective boats.

To obtain the speed is no doubt possible, controlling the boat in any meaningful manner is another thing entirely!

Andy



Edited By Sub culture on 1086798240

[carcharadon]

25/100= 0.25 knots (a hopeless

25/100= 0.25 knots (a hopeless dribble)


I see what you mean. It would take your sub some 11 seconds to pass by.

[captain nemo]

Well, I've been away from

[color=#000000]Well, I've been away from this thread for a while....a few things.

Regarding the NAUTILUS MINISUB pressure hull]

[tk-7642]

Nautilus Side Keels/chines as Inverted

Nautilus Side Keels/chines as Inverted Undercambered Airfoils

The chines or side keels of the nautilus can be thought of as an inverted under-cambered (curved) airfoil - much like that found on a race car to push it down. The angle of attack is determined by a line reaching from the leading edge of the airfoil to the trailing edge. This line's angle is measured relative to the flow of the water that is the angle of attack. Most airfoils have maximum lift at around 10 deg angle of attack. This would be maximum diving force with the water hitting the top surface of the bow chines at an angle of about 10 deg. to the chord line of the undercambered airfoil. Under cambered (curved) airfoils tend to have the most lift.
There is a point at which a cambered airfoil will create zero lift (diving downforce here) and it would be at a slight Negative angle of attack - that is with the bow trimmed slightly upward as it travels through the water (not considering wheelhouse or salon pods). The water stream would hit under the chines or lateral keels slightly (under the curve a bit at the bow). The hull itself being curved also has a lifting body effect called fuselage lift. These 2 lift variables are STRONGLY affected by angle of attack or trim.
We need to determine what level trim means for a Goff Nautilus. Is it the deck? Or is it the leading and trailing edges of the bow and stern when connected by a line as in an under cambered airfoil (which it is). What do you think is the best way?
The CG also has it's effect in pitching moment. The undercambered airfoils like the Nautilus and its side keels have a large pitching moment - that is to say that a small change in the CG will have a large effect on the pitching of the airfoil (diving here).
Almost none of this applies if the Nautilus is made with a straight hull and straight lateral keels.
I always thought the ram was higher in order for it to be more likely to pierce the hull of lower draft ships.
I have the 32 inch. and am planning to run it trimmed with a slightly negative angle of attack relative to the line made by the leading and trailing edges of the lateral keel chines, so that the airfoil creates no lift (no diving force).
Also I am going to put the CG at the salon to start and move it forward for more stability as needed (I hope I dont have to - as the salon should be at the center of gravity having all those delicate specimens - avoiding the back of the bus bouncing effect when on the surface. Careful thought went into the design I think.
The trimming /waterflow relative to the airfoil shaped sidekeel chines is absolutely critical to the (lift) downforce and diving tendency.
If you adjust the angle of attack properly you should have no diving tendency - if you can keep it that way. How?
By having a properly placed longitudinal CG, and as low as possible in the hull. Maybe I can drill out my keel and fill it with lead.
All your input is saving me alot of post assembly modification in the future!

[captain nemo]

Let me see if I

Let me see if I understand what you're saying here.

Are you suggesting that, when the boat is traveling level through the water, that the upswept bow and rakers generate forces that will induce a dive?

Pat Regan

[bob the builder]

I told you so! I

I told you so! I told you so! I told you so! I told you so! I told you so! I told you so! I told you so! I told you so! I told you so! I told you so! I told you so! I told you so! I told you so! I told you so! I told you so! I told you so! I told you so! I told you so! I told you so! I told you so! I told you so! I told you so! I told you so! I told you so! I told you so!

Not that I'd ever say that, of course.

Wow. Welcome to the wonderful world of hydrodynamics. I thought it had something to do with those side keels, but didn't know for sure.

Interesting, to say the least...

[JWLaRue]

Okay....if the upwards angled forward

Okay....if the upwards angled forward chines of a Goff Nautilus induce a tendency to dive, then how does this differ hydrodynamically from the use of bow planes? ...as in upwards angled bow planes do just the opposite.....

Also, the chine on the Nautilus runs the length of the hull, it seems to me that this would have an effect. Perhaps something creating the effect of an infinitely long surface.

-Jeff

[Guest]

Hmm, it may be worth

Hmm, it may be worth building a set of forward mounted hydrovanes with an undercambered section, and see how they affect the performance of a more, shall we say, conventional boat.

Andy

[tk-7642]

The whole Goff Nautilus is

The whole Goff Nautilus is a diving plane. The small "Diving planes" in the lateral keel chines could be thought of as trim tabs like those found on the rudder or elevator of an airplane. They point the undercambered airfoil Nautilus relative to the water flow thus changing the angle of attack (AOA) causing lift (dive) or steady level cruising. About 2 deg negative angle of attack (relative to the undercambered airfoil lateral keel chines - not the sub) in the Nautilus would be used for cruising underwater. At around negative 2 deg angle of attack the Nautilus would NOT rise or sink AT ALL (not considering the wheelhouse or salon effect). This angle of attack is what matters here. It is the chord line of the airfoil (leading edge of the curve to trailing edge of the curve) relative to the direction of water flow. The lateral keel chines and hull are the airfoil. So a line must be drawn from the bow (from around first serration) to the stern (just before the prop) and this is the chord line. The flow of water relative to this chord line is the angle of attack. This angle of attack must be NEGATIVE so that the Goff Nautilus is not sucked down. This means that the Goff Nautilus must cruise with this CHORD LINE pointing 2 degrees UP. This has nothing to do with the deck being level or not - it is the chord line of the lateral keel chine's airfoil that matters. This happens to be with the deck just about level on the plans (don't know on the different models) and could be checked with a protractor. Goff's is a carefully thought out design.
Naturally, things get more complicated when you consider that the models out there all have gigantic, squared off, paddlewheeler-like salon housings masking the water flow on the entire rear half of the lateral keel chines - destroying the hydrodynamics. This would disturb the water flow over the rear of the inverted airfoil moving the airfoil center of pressure forward and making the sub more unstable. They don't stick out into the water stream that far on the plans, or the 11 ft! I'm still trying to figure out how to sink mine into the sub hull some. Err on the flush and rounded edged, streamlined side with the salon pods - like on the plans!
Jules Verne's Nautilus as described in the book as I recall used only ONE pair of diving surfaces at the CG. This provided downward lift and "sucked" the sub downward. Some very early sub designs had only one set of diving planes. Goff's Nautilus does the same only streched out along the entire sides of the sub. These could be thought of as extremely low aspect ratio wings allowing function at larger AOA than normal. Not only that, but the serrations act as turbulators on the leading edge and may allow still larger AOA as well.
What is different about the Goff Nautilus is that it not only acts as an inverted airfoil - but an UNDERCAMBERED inverted airfoil. This has more lift (at our speeds). This makes it very efficient at diving - AT A POSITIVE ANGLE OF ATTACK ONLY. Also this type of airfoil requires more care in the placement of the CG due to a greater pitching moment. This can be compensated for by the stern diving planes somewhat IF they are allowed to work by having realistically sized, streamlined, authentic salon housings.
The water hitting under the bow curve does push up on the front of the sub. This happens when at a negative AOA, and this in part, is why the airfoil has no lift at negative AOA (no dive). You could think of them as diving planes pointed up against the downward tendency of the overall airfoil if you want. How do you adjust these "diving planes" ? By trimming the whole sub up or down. I think its much simpler to consider the whole sub as an inverted curved wing and just trim it (point it) up or down, as needed for steady cruising under water.
Bottom line is if you use curved up lateral keel chines you need to carefully set the CG (due to pitching moment), and trim with the bow slightly up and the deck more or less level underwater. If you use straight chines almost none of this occurs, but it takes longer to submerge when moving.
This Nautilus would have been great for getting under water quickly to attack ships. Trim the nose down, ramming speed, and its under MUCH faster than ballast tanks alone could do. When at proper depth underwater trim to around negative 2 deg AOA to stop the dive, and go to ramming speed. Have you noticed the GIANT, PROMINENT angle of attack (level) gauge in the wheel house?

[PaulC]

Jeff,

What you are saying is

Jeff,

What you are saying is what strikes me -- the side rakers, if built to angle up from longitudinal center, should act as a lifting force.

Drag from the broken keel at the diving well, occuring aft of the center of rotation, would seem to more likely pitch the boat down forward (I believe someone has already posted this theory).

[captain nemo]

Bob,

In my last post, I

Bob,

In my last post, I wasn't agreeing with TK7642's "inverted airfoil" theory; I was asking him to clarify it. Now that he has, I have to say he's wrong.

Jeff is right. The upswept bow imparts a lifting moment, not a dive, when the boat is in level trim.

If your model is diving uncontrollably when underway in a decks-level attitude, it's not due to negative lift generated by an inverted airfoil effect of the upturned bow; the problem is elsewhere.

Pat

[tk-7642]

I have explained why I

I have explained why I believe my theory to be right. Please let me know where you believe it is mistaken.