How is it that motorboats lean into a tight turn

Off the top of my head; the typical approach is to start with the stability derivatives. These are the sensitivity of the forces and moments to changes in the motion, controls, or propulsion.

For example the derivative Ybeta is the change in side force, Y, per degree of sideslip (leeway), beta. Nbeta would be the change in yawing moment per degree of sideslip. Nr is the change in yawing moment due to yaw rate (rate of turn). Etc. There's a stability derivative for each force and moment, for each motion variable.

If you want to figure out how the boat will roll in a turn, you'd have to estimate each of the derivatives and then solve for the solution that simultaneously satisfies each of the equations.

For the example above, in which adding a skeg reduced the roll, here's what was happening. For a given speed, a certain amount of side force is necessary to turn the boat at a given rate, which is equal to m*V*R (mass times velocity times turn rate). This has to be supplied by sideslipping the hull, so that (ignoring rudder forces for the moment) Ybeta * beta = m*V*R. So beta = m*V*R/Ybeta.

There's also a rolling moment derivative, Lbeta, which tends to roll the boat away from the direction of the sideslip. And there's a rolling moment from the boat's buoyancy, which I'll call Lphi (and naval architects would call GM), that is the rolling moment due to heel. For the boat to be in equilibrium in heel, all the moments have to sum to zero; Lbeta*beta + Lphi*phi = 0. From this you can solve for the heel angle: phi = -Lbeta*beta/Lphi.

When you substitute in the result of the side force equation, you get

phi = -m*V*R*Lbeta/(Ybeta*Lphi)

But wait! There's also a yawing moment from beta, Nbeta, which tends to make the boat track straight, and a yawing moment from heel (Nphi) due to the change in underwater shape, and a yawing moment from the rudder (Nrudder). Which you solve to get the rudder required. Unfortunately, there's also roll due to rudder (Lrudder) because the rudder is below the center of gravity, and sideforce due to rudder deflection (Yrudder), too. Not to mention yaw moment due to turn rate (Nr) that gives the boat its yaw damping, and maybe rolling moment due to yaw rate (Lr) and side force due to yaw rate (Yr), too. In particular, the sideforce due to rudder will be to the outside of the turn, requiring more sideslip, and more roll, than indicated above.

So you have three equations (sideforce balance, roll moment, yaw moment) and three unknowns (phi, beta, rudder) to solve simultaneously. The only difficulty is getting the values of those pesky derivatives. They are determined by the hull characteristics. For example, the roll moment due to sideslip, Lbeta, will depend on the deadrise angle of the hull, with a V'd hull tending to roll more with sideslip than a flat bottom hull.

In the example cited, the boat rolled excessively in a turn because the ratio (Lbeta/Ybeta) was too high or the roll stiffness (Lphi) was too small. Adding the skeg produced more sideforce per degree of sideslip, increasing Ybeta. In addition, the skeg was below the c.g., so its contribution to the rolling moment was opposite of that the hull (tending to "trip" the boat and roll to the outside of the turn). So the skeg lowered the ratio (Lbeta/Ybeta) and reduced the roll.

So that's how you predict how a boat will roll in a turn.
 
Mobo's banking in turns

There is some sense in the propulsion unit assistance theory; vectored thrust from a propeller underwater and with a lever force under the hull will give a banking effect into turns.
 
Lakesailor said:
Off the top of my head; the typical approach is to start with the stability derivatives. These are the sensitivity of the forces and moments to changes in the motion, controls, or propulsion.

For example the derivative Ybeta is the change in side force, Y, per degree of sideslip (leeway), beta. Nbeta would be the change in yawing moment per degree of sideslip. Nr is the change in yawing moment due to yaw rate (rate of turn). Etc. There's a stability derivative for each force and moment, for each motion variable.

If you want to figure out how the boat will roll in a turn, you'd have to estimate each of the derivatives and then solve for the solution that simultaneously satisfies each of the equations.

For the example above, in which adding a skeg reduced the roll, here's what was happening. For a given speed, a certain amount of side force is necessary to turn the boat at a given rate, which is equal to m*V*R (mass times velocity times turn rate). This has to be supplied by sideslipping the hull, so that (ignoring rudder forces for the moment) Ybeta * beta = m*V*R. So beta = m*V*R/Ybeta.

There's also a rolling moment derivative, Lbeta, which tends to roll the boat away from the direction of the sideslip. And there's a rolling moment from the boat's buoyancy, which I'll call Lphi (and naval architects would call GM), that is the rolling moment due to heel. For the boat to be in equilibrium in heel, all the moments have to sum to zero; Lbeta*beta + Lphi*phi = 0. From this you can solve for the heel angle: phi = -Lbeta*beta/Lphi.

When you substitute in the result of the side force equation, you get

phi = -m*V*R*Lbeta/(Ybeta*Lphi)

But wait! There's also a yawing moment from beta, Nbeta, which tends to make the boat track straight, and a yawing moment from heel (Nphi) due to the change in underwater shape, and a yawing moment from the rudder (Nrudder). Which you solve to get the rudder required. Unfortunately, there's also roll due to rudder (Lrudder) because the rudder is below the center of gravity, and sideforce due to rudder deflection (Yrudder), too. Not to mention yaw moment due to turn rate (Nr) that gives the boat its yaw damping, and maybe rolling moment due to yaw rate (Lr) and side force due to yaw rate (Yr), too. In particular, the sideforce due to rudder will be to the outside of the turn, requiring more sideslip, and more roll, than indicated above.

So you have three equations (sideforce balance, roll moment, yaw moment) and three unknowns (phi, beta, rudder) to solve simultaneously. The only difficulty is getting the values of those pesky derivatives. They are determined by the hull characteristics. For example, the roll moment due to sideslip, Lbeta, will depend on the deadrise angle of the hull, with a V'd hull tending to roll more with sideslip than a flat bottom hull.

In the example cited, the boat rolled excessively in a turn because the ratio (Lbeta/Ybeta) was too high or the roll stiffness (Lphi) was too small. Adding the skeg produced more sideforce per degree of sideslip, increasing Ybeta. In addition, the skeg was below the c.g., so its contribution to the rolling moment was opposite of that the hull (tending to "trip" the boat and roll to the outside of the turn). So the skeg lowered the ratio (Lbeta/Ybeta) and reduced the roll.

So that's how you predict how a boat will roll in a turn.
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No, but seriously Lakey ..

My neighbour made himself a little one-person runabout from a couple of sheets of plywood and set off around the bay in it - and it figured out all of the above - by itself! ( at least to the extent that it didn't immediately flip as he turned).
 
Lakey, what's in the water up your way? My brain turned off reading that...

It is the thrust of the prop at a lower level than the water line. If a planing boat, usually they have plenty of power, hence the lever arm makes them lean in. If displacement, as said, the inertia of the boat makes them lean out. Quite easy to flip a small over-powered dinghy if not carefull. e.g. 8ft pram with Seagull long shaft Silver Century. Don that!
A
If out board or outdrive the thrust is heavily vectored. Rudders do similar but a bit less so.
 
Off the top of your head!!

That said, Lakey is correct. For a displacement boat/ship there are 3 phases to the turn. When the rudder is applied, because it is acting below the centre of gravity it creates a heeling moment in addition to the turning moment. This heeling moment causes the Ship to lean inwards on the turn.

Having heeled over with the rudder now steady, the centre of buoyancy will have moved and therefore there will be a lever to bring the Ship upright and because it is now turning the centrifugal force will cause the ship to heel outwards. As Lakey says however, it is turning but also slipping partly sideways in the turn.

The third phase of the turn is the steady state turning circle for a given rudder angle and Ship speed. The photo of the Warship is apt, as these 3 phases of the turn are easily distinguishable and can be felt when carrying out a turn at speed.

With planing vessels I suspect that heeling moment created by the rudders/outdrive is not counteracted in the same way by the immerison of the inboard side of the hull because the vessel is planing. But there I run out of my off the top of my head knowledge.
 
My just thought of theory:

If a planing hull turns to starboard, the starboard hull is going slower through the water thus not producing as much lift, so sinks a bit.

A displacement hull doesn't produce lift therefore would lean to port because of a very big word that makes me sound clever.
 
affinite said:
Im not a mobo owner but I suspect that if you pulled the kill cord during a hard turn in a planing boat, it would drop of the plane, bite the water and she could barrel roll outwards because of the above
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I was in a speed boat towing a skier on a hard fast turn, when it left the water for a moment. As soon as it did, it started to roll outwards, and when it landed, we thought it was going to flip.

Reckon the skier was the only thing prevented it!! :eek:
 
I would have thought that the sideslipping of planing hulls and of high powered small vessels, erm, how do you describe it?

Right, wee fast vessel turns hard to port, vessel will sideslip to starboard due to centrifugal force. Starboard portion of the hull "slides up" the static water, lifting the starboard side of the hull. Simultaniously the port ( inner) side of the hull is sliding sideways "away" from the water and will drop in consequence. This combination of more lift on the outside of the turn will tilt the hull inwards to the turn. If the boat is moving too fast then the flow of water across the hull will break up and the boat starts to slip sideways and "highsiding" becomes a possibility.

Large vessels like ferries and warships have a relatively high centre of gravity, The high centroid will be pulling outwards well above water level, the immersed hull is still sliding away from the turn but it is still changing direction so as described by others above the vessel will heel outwards, the high centroid overcoming the tendency for a hull shape to lean into a turn.

That just "feels" right to my old brain.
 
The reason that a fast boat with outdrives leans inwards, is that they are almost all V-hulls that are quite happy to tip one way or the other at the drop of a hat anyway, and it doesn't take much vectored thrust below the waterline to act around the keel (as in the bottom), and tip the boat into the turn.

We tested this out the hard way a few weeks back, with a full-lock turn at 20kts.
A few more gentle turns to warm up.
Warning to all to hold on and brace.
Countdown from 5 ... 4 ... 3 ... 2 ... 1
... and the 7 yr old disappears off the seat and face-plants onto the cockpit floor (carpeted).
Gets up from the floor.
Then hits me on the arm really quite hard and sits down again saying nothing but still glaring at me.
 
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Anything in the fact that the part of the hull inside the turn is going slower than the part on the outside of the turn?

More lift the further to the outside of the turn you move on the hull and less on the inside. Natural tendency to roll into the turn.

With displacement hulls you are looking at centrifugal force acting on a C/G that's way above the centre of lateral resistance.

Turn-the-aircraft-carrier-01.jpg
 
Apropos not very much at all, once upon a time I used to do rescue duty at a dinghy club. The rescue boat was an outboard dory, a bit like a Dell Quay. When people weren't falling in, I'd amuse myself by locking the outboard steering on the centreline, standing up, and steering the boat by shifting my not inconsiderable weight. Tip her to port and she'd turn to port. And vice-versa. That was most definitely at displacement speeds.
 
Twister_Ken said:
Apropos not very much at all, once upon a time I used to do rescue duty at a dinghy club. The rescue boat was an outboard dory, a bit like a Dell Quay. When people weren't falling in, I'd amuse myself by locking the outboard steering on the centreline, standing up, and steering the boat by shifting my not inconsiderable weight. Tip her to port and she'd turn to port. And vice-versa. That was most definitely at displacement speeds.
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As soon as you tip a boat the centre of drag moves over to that side as more of the hull is immersed. Your drive from the propellor is still passing along the centre line of the hull so boat turns around the drag centre. Sailing dinghies can also be steered by tipping, but not too far.
 
agurney said:
The same way that motorbikes do, oodles of grip from the propulsion system.
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Cycles and motorcycles lean inwards because they have to in order to remain balanced on two in-line wheels - the displaced c. of g. balancing out the centrifugal force from the turn. The rider first flicks the front wheel to the right, say, to create the imbalance which permits a turn to the left. It's all done automatically as part of the process of learning to ride a bike. Don't know the answer re. boats though.
 
ProDave said:
Motorbikes lean in to a turn, because the rider leans that way to keep the COG directly over the wheels.

I's perfectly possible to turn a motorbike without leaning it, but it would fall over, so you don't do that do you?

Same as riding a pushbike, you keep the ballance. Something you have to learn, which is why kids fall off their bikes until they have mastered the art of keeping the ballance. Something that once learned, you don't forget.
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Only at very low speeds........ once about 10-15 mph is reached then you countersteer and the bike leans....... there's a whole DVD on it by Keith Code of the California Superbike School showing everything in slow motion. Very clever and we humans do it automatically.
 
Red Admiral said:
As soon as you tip a boat the centre of drag moves over to that side as more of the hull is immersed. Your drive from the propellor is still passing along the centre line of the hull so boat turns around the drag centre. Sailing dinghies can also be steered by tipping, but not too far.
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except you tip dinghies the other way, to steer to port without a rudder you lean to starboard.
 
nimbusgb said:
Anything in the fact that the part of the hull inside the turn is going slower than the part on the outside of the turn?

More lift the further to the outside of the turn you move on the hull and less on the inside. Natural tendency to roll into the turn.

With displacement hulls you are looking at centrifugal force acting on a C/G that's way above the centre of lateral resistance.

Turn-the-aircraft-carrier-01.jpg
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This is what I was explaining, except the water flow on the outside is faster, hence the pressure drops. The water flow on the inside is slower hence the pressure rises.

The high pressure inside pushes the submerged hull outwards and the top sides lean inwards. In big ships though the height of the mass above the water causes a bigger force and she leans to the outside.

All boats try and lean outwards but some boats will lean inwards because the water pressure exerts a greater force than centrifugal thingamy bob.

It is a fact that there will be a pressure differential on any shaped object that is turning in a fluid with side slip i.e. the direction of water flow is not dead ahead.

Similar principle when squat happens in shallow water i.e. the water accelerates to fit between the hull and seabed, pressure drops, boat sits deeper.

All off the top of my head, prove me wrong.
 
peterb said:
In order to turn the hull must develop a sideways force. It does this by putting itself at an angle to the local water flow under the hull. Planing boats have vee hulls. The sideways skidding causes the side of the boat on the outside of the turn to develop more 'lift' than that on the inside of the turn, so the boat 'banks' inwards. It's exactly the same effect as that given by dihedral on an aircraft wing.
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An aircraft has to be "banked" by the pilot in order to induce a turn. The rudder is used to provide secondary corrections only. If you tried to turn an aircraft using rudder alone it would simply yaw. In a way it's like a bike where the rider creates the banking. In the case of a boat though the banking follows automatically from rudder movement. Personally I think that the rudder movement induces yaw which scrapes up a wall of water along the outer side of the boat which leaves a lower water level along the inner side. The boat then adapts an attitude based on this local sloping water surface. This wall will always tend to level itself out and in the case of displacent boats they aren't fast enough to create enough of a wall to matter and centrifugal effects dominate.
 
reeac said:
An aircraft has to be "banked" by the pilot in order to induce a turn. The rudder is used to provide secondary corrections only. If you tried to turn an aircraft using rudder alone it would simply yaw. In a way it's like a bike where the rider creates the banking. In the case of a boat though the banking follows automatically from rudder movement. Personally I think that the rudder movement induces yaw which scrapes up a wall of water along the outer side of the boat which leaves a lower water level along the inner side. The boat then adapts an attitude based on this local sloping water surface. This wall will always tend to level itself out and in the case of displacent boats they aren't fast enough to create enough of a wall to matter and centrifugal effects dominate.
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Exactly. Use the rudder on its own and the plane just yaws. But that means that it is side-slipping, and the effect of dihedral on a side-slipping wing is that the two sides have different angles of attack, and hence different amounts of lift. That in turn makes the aircraft bank; the inclined lift vector then accelerates the aircraft sideways to make it change its course.

In a vee-bottomed planing boat you get exactly the same effect. The initial effect of rudder is to yaw the hull (change the heading), without causing any change of course. The hull is now crabbing through the water; this gives the two sides of the bottom different angles of attack which in turn cause the boat to bank (as in the aircraft case). The crabbing effect also makes the hull generate a sideways force which accelerates the boat towards the centre of the turn, thus changing the course.

Incidentally, my old uni prof always came down hard on anyone who talked about centrifugal or centripetal forces. He insisted that we talked about central forces, and reckoned that anything else could be ambiguous.
 
peterb said:
Exactly. Use the rudder on its own and the plane just yaws. But that means that it is side-slipping, and the effect of dihedral on a side-slipping wing is that the two sides have different angles of attack, and hence different amounts of lift. That in turn makes the aircraft bank; the inclined lift vector then accelerates the aircraft sideways to make it change its course.

In a vee-bottomed planing boat you get exactly the same effect. The initial effect of rudder is to yaw the hull (change the heading), without causing any change of course. The hull is now crabbing through the water; this gives the two sides of the bottom different angles of attack which in turn cause the boat to bank (as in the aircraft case). The crabbing effect also makes the hull generate a sideways force which accelerates the boat towards the centre of the turn, thus changing the course.

Incidentally, my old uni prof always came down hard on anyone who talked about centrifugal or centripetal forces. He insisted that we talked about central forces, and reckoned that anything else could be ambiguous.
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Well the only time that I flew a plane I was told to move the joystick sideways to move the ailerons in opposite directions [one up, the other down] in order to effect the banking necessary to make a turn so as far as I know it's the pilot who initiates the banking but maybe the dihedral angle helps.

I've seen inflatables [not RIBs] planing and they often have flat bottoms
... do they bank on turns? Anyone know?

I know about centrifugal and centripetal and always found the distinction a bit nit-picking so maybe I agree with the prof.
 
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