Pivot Point

[LIboater]

I have been having a lengthy "discussion" with some friends about the pivot point of a twin engine cruiser. We discussed v-drives and stern drives. To normalize the comparison we assumed that the stern drive uses only forward/reverse thrust to pivot.

Ignoring drag of any sort on hull, wind, current or other external factors we wanted to know the PP. There were three "opinions"

1) Between the props. Thus, no matter where the props were, under boat, off back of transom or even (for purposes of theory) 20 ft behind the transom one of us insisted the pivot is between the spinning props

2) At the point that the force/thrust is delivered to the hull. In this argument it says very simply that the force developed by the props is sent through a shaft to a trans (either a v-drive, a stern drive trans or if you like direct to engine if that is setup) and that the point of connection is the pivot point. Thus, in a v-drive the PP is essentially right at the v-drive.

3) The third opinion was that the PP will always be at the Center of Mass of the boat. It does not matter where the props are or how the driving force is delivered to the boat.

Which is correct?

Thanks!!!

[Hard Rock Candy]

I believe that the center of the swept radius will be located at the balance point between the direction of thrust and the area of drag. Mac

[LIboater]

I'm sorry I don't understand that. Can you clarify? Also, what if we disregard drag.

[Hard Rock Candy]

On a sailboat you can not disregard drag because the balance between drag on the keel and hull versus wind pressure on the sail is what makes the boat move. So when you are sailing along you are actually moving sideways as well as forward. The sail is pushed to the side and the boat leans over because the keel and hull drag in the water holds the boat back. If you wish to cut drag you would have to diminish keel area and hull area as much as is possible. The three main components of a sailboats propulsion work like legs on a three legged stool. One leg is the wind blowing on the sail, another leg is the sail design itself, the third leg Is the underwater shape and area of drag on the hull. Between the three points you can vary the water controlled leg and the sail control will vary the second leg. And the third leg will leg change with the wind direction and force. The sailor is there to control the two legs he/she has control over, which should allow them to keep the wind direction where it is needed to get to someplace they want to go. Sorry about such a drug out letter, but I am poorly equipped in knowledge about words to make an explanation short and too the point. Mac

[Hard Rock Candy]

I tried towing a sailboat for a friend who lost a couple bolts from his drive plate on a Perkins. We hooked it up with about 75 ft of 1/2 inch nylon braid. With him on his boat and me on mine and each of us holding a handheld radio, I proceeded to sail away. NO good. we had hooked the tow rope to a rear cleat on my boat to the anchor roller on his boat. Immediately as the tow rope cam tight, it started to pull the back of my boat around because my boat was sailing sideways and when I turned the rudder of mine to make it straighten up, the drag against my rudder just slowed the whole works down to almost nothing. after a couple tries we eventually found that a bridle between the side center cleat and a front cleat on my boat was workable, but very hard to continually adjust to go in some predetermined direction. Finally I just motored about 20 miles to get him back to his dock, at about 3 knots. Mac

[CaptTom]

LI, I think the question is being misunderstood. Disregard your own hull (especially if it's a monohull sailboat) for a moment.

For simplicity, picture a rectangular hull (like a barge) with two outboard motors. Pretend the hull is so light there's virtually no drag. Put one in reverse and one in forward. Will the axis of spin be between the props, or somewhere else?

What if you moved the outboards out onto a bracket? What if you replaced them with inboards with the shaft log some ways forward of the transom?

Interesting question. My assumption was always that the pivot point "tries" to be between the props, but the drag on the hull (along with momentum, the different slippage between forward and reverse, and probably other factors) will affect the actual result quite a bit.

My theory is based on the premise that the thrust, one forward and one reverse, is being generated at the props and that's where the force is being transferred to the boat. I am not a physicist and I don't play one on TV, so I could easily stand corrected.

[LIboater]

Actually my friends with whom I am having this discussion are in fact a physicist and 2 engineers. However, none of them are boaters.

The physicist insists that the pivot point, disregarding drag as you did, of a boat (a light one as you suggest) is still the Center of Mass of the boat. He cites an example of a square rock in space with 2 forces acting on it (positioned much like props). Then no matter where the CoM is located on the rock it will spin/rotate about that CoM. For some reason I find that example easy to understand, but intuitively I feel there is a flaw when applied to our example boat/barge. However, I do not know what that flaw is, or even if I am correct that there is one.

On my 34 ft SR there are metal tabs along the gunwhale indicating where the slings should be placed for a haul out. I must assume that the point that marks the center point between these 2 tabs is the CoM or very close because you want the boat as perfectly balanced as possible when you haul it out. However, the center point between these sling tab markers is nowhere near the observed pivot point of the boat. Granted drag might affect that, but it is hard to think it makes that great a difference.

[GordMay]

Greetings and welcome aboard the CF, LIboater.

[a64pilot]

I think the pivot point is variable, depending on the amount of thrust from each engine,
but even if it's an equal amount, it's not the boats center of mass, as the center of force is not coincident with that, nor is it the drag center either, but all of those do play into it.

In short, I think CaptTom has it right

[ranger58sb]

I think I understand the question...


Our unrestrained pivot point is usually about 1/3rd to 2/5ths of the way aft from the bow.


When we use spring lines, the pivot becomes the forward or aft mid-ships cleats, though, depending on which we're using.


-Chris

[CaptTom]

Originally Posted by LIboater

He cites an example of a square rock in space with 2 forces acting on it (positioned much like props). Then no matter where the CoM is located on the rock it will spin/rotate about that CoM. For some reason I find that example easy to understand, but intuitively I feel there is a flaw when applied to our example boat/barge. However, I do not know what that flaw is, or even if I am correct that there is one.

I'm guessing the flaw is the applied force. In space, if you apply a momentary rotational force to spin an object, your friend's logic will probably hold and it will spin around its CoM. Same with our barge on the water, assuming no drag.

But in our example, we're applying a steady force in TWO locations. One is pulling the boat back, the other is pushing it forward.

I'm going to fall back on an old trick I use for teaching twin-screw handling: Imagine you're holding a stationary shopping cart, both hands on the handle, near the outer edges. Push one and pull the other. Where is the pivot point?

(BTW, to teach single-screw handling, just turn the cart around and push it backward from the front.)

Originally Posted by ranger42c

Our unrestrained pivot point is usually about 1/3rd to 2/5ths of the way aft from the bow.s

That's correct, going forward and using just the rudders(s) to turn. But in this exercise, we're theorizing no rudders, just twin screws and using split clutches to turn.

C'mon, where are all the experts? I'm sure someone can explain this better than I can!

[ranger58sb]

Originally Posted by ranger42c

Our unrestrained pivot point is usually about 1/3rd to 2/5ths of the way aft from the bow.

Originally Posted by CaptTom

That's correct, going forward and using just the rudders(s) to turn. But in this exercise, we're theorizing no rudders, just twin screws and using split clutches to turn.

That's using no rudder, just gears.

-Chris

[a64pilot]

Originally Posted by CaptTom

I'm guessing the flaw is the applied force. In space, if you apply a momentary rotational force to spin an object, your friend's logic will probably hold and it will spin around its CoM. Same with our barge on the water, assuming no drag.

But in our example, we're applying a steady force in TWO locations. One is pulling the boat back, the other is pushing it forward.

I'm going to fall back on an old trick I use for teaching twin-screw handling: Imagine you're holding a stationary shopping cart, both hands on the handle, near the outer edges. Push one and pull the other. Where is the pivot point?

(BTW, to teach single-screw handling, just turn the cart around and push it backward from the front.)



That's correct, going forward and using just the rudders(s) to turn. But in this exercise, we're theorizing no rudders, just twin screws and using split clutches to turn.

C'mon, where are all the experts? I'm sure someone can explain this better than I can!

That's funny, I use a shopping cart to try to explain to the un-initiated why tailwheel aircraft are so difficult. If you pull the cart from the front, your a tricycle airplane, engine pulling in front, steering wheel just behind that and the fixed wheels in the rear, easy right, cart pulls nice and straight.
Now pull it backwards from the rear, engine in front, non steering wheels right behind that and steering wheel in the rear, now how stable is that cart being pulled backwards? That's a tailwheel airplane


Sent from my iPad using Cruisers Sailing Forum

[Dsanduril]

Here are two training manuals that discuss the pivot point with twin screws, both say it moves depending on boat speed, and that with no speed and opposing thrust it is about 1/3 boat length abaft the bow:

https://books.google.com/books?id=4E...page&q&f=false

http://www.ocontoyachtclub.com/ocont...athandling.pdf

I think you can also review this from practical experience, if the pivot point was at the stern on twin I/O you would only be able to turn the boat 180 degrees in 2x length, and experience shows that you can come pretty close to turning the boat in place.

[CaptTom]

All this real-world stuff is muddying up the argument.

Of course when you factor in drag, momentum, wind, current, rudder angle, prop walk, etc. you won't arrive at the perfect theoretical answer to the question of where the pivot point "should" be if those forces are negligible.

The shopping card analogy largely eliminates all those factors, and the pivot point is directly between the locations where thrust (pushing and pulling) is applied. This is because the wheels more relatively freely. If you prefer, replace the shopping cart with a utility cart that has caster wheels that move freely in any direction. Same result.

Use one of those carts with the extended handle and seats for kids behind the rear wheels. Where's the pivot point now? Still between your hands!

Move the analogy to the water. Hold one of those kickboard swimming aids in both hands and do the same thing (one hand push, one hand pull). It will pivot on a spot between your hands. Do the same thing with a full-sized surfboard. No matter where you put your hands, the pivot point is between them if you do the push-pull motion. You can keep going, up to the barge we originally talked about. The pivot point remains between the applied forces, again assuming we disregard drag and all those other factors.

I'm hoping someone proves me wrong, that would be educational.

My own experience is that the pivot point when using the rudders is nowhere near the pivot point using split clutches alone. I use both forces all the time in docking (I have just inches to spare on all sides at my home slip.)