Weight of boat with moorings

[Franklin]

I come here to ask this question because I want to see if anybody can poke a hole into my theory that weight of the boat on a mooring is not all that important when calculating how heavy a mooring block needs to be.

Here is my thinking: There is no energy associated with weight. The dynamic energy (energy by wind is considered static) created by the waves get it's energy from the waves, not the weight of the boat. Think of a heavy boat offshore, it doesn't move nearly as violently as a lighter weight boat does, all because it takes a lot more energy to move it. Therefore, if a heavy boat moves less than a light boat, both should have approx the same dynamic energy applied to the mooring because the energy comes from the wave, not the boat. Sure, a heavier object moving at the same speed as a lighter weight will take more energy to stop it from moving, but they don't move the same in the water.

So, if you can poke a hole into my theory that weight is mostly meaningless, but windage is the primary concern of a boat tied to a mooring, please do because I hear so many people ask the weight of a boat when trying to determine if the mooring can hold it and not the windage.

BTW: a little physics lesson: Force = the amount of energy it takes to move and object at a given acceleration, we are concerned with the amount of energy at impact which is this formula: K.E. = 1/2 M x V squared. Mass in units of kilograms and velocity in meters per second. Where force is just F=MxA (acceleration). Just thought I would clear that up because so many get that wrong.

 

[justsomeguy]

So, if you can poke a hole into my theory that weight is mostly meaningless,

This ought to be interesting...

 

[Franklin]

This ought to be interesting...

Yes, I am never one to take people's word for anything, I need to understand the physics of it

 

[shemandr]

Yeah, I think windage is way underrated in designing a anchor or mooring system. In part because it's difficult to quantify.
I've taken my share of physics and really have no idea how that lesson relates to forces on an anchor or mooring. I've stood over Sir Isaac Newton's tomb and begged understanding or at least intervention for my finals - to no avail. He might still be pissed about Quantum Mechanics.

 

[weinie]

 

[Franklin]

Yeah, I think windage is way underrated in designing a anchor or mooring system. In part because it's difficult to quantify.
I've taken my share of physics and really have no idea how that lesson relates to forces on an anchor or mooring. I've stood over Sir Isaac Newton's tomb and begged understanding or at least intervention for my finals - to no avail. He might still be pissed about Quantum Mechanics.

The formula has to do with the dynamic energy created by the waves (like in case of a hurricane with substantial waves). The formula to calculate the static pressure of the wind: lbs of pull = .00256 x wind knots squared x sq footage towards the wind x drag (flat plate = 2 a pole = 1.6 -- this is the big guestimation variable when dealing with a boat). If a monohull, for sq footage I use 30% of hull length because they swing and that is more windage that when pointing dead into it.

 

[kloudie1]

I think that the pull on the line in a seaway is related to the weight of the boat and the sine of the angle of the wave passing under it.. example.. pulling a car up a hill is harder if the car is heavier.. the mooring is "pulling" the boat up the wave.. kind of a simplistic view, and there are lots of other things going on, but I think the weight of the boat is a significant factor in pull on the mooring.. in addition to windage.. If the anchorage is protected and there is little fetch for the waves to build up, then windage is going to be the major force to be dealt with.

 

[Franklin]

I think that the pull on the line in a seaway is related to the weight of the boat and the sine of the angle of the wave passing under it.. example.. pulling a car up a hill is harder if the car is heavier.. the mooring is "pulling" the boat up the wave.. kind of a simplistic view, and there are lots of other things going on, but I think the weight of the boat is a significant factor in pull on the mooring.. in addition to windage.. If the anchorage is protected and there is little fetch for the waves to build up, then windage is going to be the major force to be dealt with.

but...a car is on solid ground where as a boat floats and parts the water. Not exactly the same concept.

 

[weinie]

Conservation of Momentum is a function of mass and is what is in play here.

The wave imparts energy to the boat getting it moving at with momentum p = mv
Integrating force w/ respect to time is momentum.... i.e. when the rope pulls on the boat at the end when it is out of slack it exerts a force for a small time period. That integral is momentum which is a function of mass.

 

[Franklin]

... If the anchorage is protected and there is little fetch for the waves to build up, then windage is going to be the major force to be dealt with.

Well, when I rode out Ike in Galveston back in 2008 I had 6' waves hitting the boat. I am now in Tonga with a similar anchorage size but deeper water. They tell me sometimes the waves get to 2 meters. The size is about 1/2 a mile by 1/5 a mile.

 

[Franklin]

Conservation of Momentum is a function of mass and is what is in play here.

The wave imparts energy to the boat getting it moving at with momentum p = mv
Integrating force w/ respect to time is momentum.... i.e. when the rope pulls on the boat at the end when it is out of slack it exerts a force for a small time period. That integral is momentum which is a function of mass.

Ok...but the V in the formula is effected by the weight. If it is a heavy boat, the V is much smaller compared to a lighter boat -- the wave moves a heavier boat slower than a lighter boat. Energy still comes from the wave.

 

[Franklin]

...when the rope pulls on the boat at the end when it is out of slack it exerts a force for a small time period. That integral is momentum which is a function of mass.

Not a small period of time, but the exact instance the pressure is applied like when a truck hits a wall. The formula is K.E. = 1/2 M x V squared. Then the K.E. measured in Jules needs to be converted to pressure and I forget that formula.

 

[weinie]

dude... put down the bong.

 

[Franklin]

dude... put down the bong.

Please come with facts, not insults. I am trying to have an intelligent conversation here. If you can't then get off my thread.

 

[weinie]

KE is measured in Joules. Not jules. And why are units important here. Energy is NOT pressure.
Not one of your statements in any of your posts is remotely accurate with the exception of the definition of KE.

 

[shemandr]

Exactly. Sorry not to be up to this discussion. Carry on!

 

[Franklin]

KE is measured in Joules. Not jules. And why are units important here. Energy is NOT pressure.
Not one of your statements in any of your posts is remotely accurate with the exception of the definition of KE.

Try googling Joules to PSI or Ft lbs like this site: http://www.dlflange.com/calculators/joules-to-ft-lbs/

[ Joules to ft-lbs conversion formula: Joules / 1.3558 ]

 

[LeoLambert310]

Found this table
Minimum Dor-Mor Mooring Anchor Size Requirements

Boat Beam Wind Load Dor-Mor Wind Load Dor-Mor
Lenght Sail/Power 64 Knots Size 100 Knots Size
20' 8/9' 1,600lb. 200lb. 3,600lb. 400lb.
25' 8/9' 2,200lb. 300lb. 5,000lb. 500lb.
30' 9/11' 3.200lb. 400lb. 7,000lb. 700lb.
40' 11/14' 5,400lb. 700lb. 12,000lb. 2,000lb.
50' 13/16' 7,300lb. 1,000lb. 16,000lb. 2,000lb.
60' 15/18' 9,100lb. 1,000lb. 20,000lb. 2,000lb.
80' 19/22' 13,000lb. 2,000lb. 31,000lb. 4,000lb.

Based on wind loading data at min. 3:1 scope, with suitable bottom conditions for anchor to embed. Wind loads based on data from ABYC.

 

[weinie]

Not wave wind...

but i'll accommodate you:

2 boats... They go up a wave to the same height. They now have different POTENTIAL energies equal to mgh. They will ACCELERATE at the same rate g. Therefore, they will have the same velocity at the bottom of the wave.

So both boats have the SAME velocity at the bottom of the wave but DIFFERENT kinetic energies.
The boat with greater mass will have greater momentum than the boat with less mass. (momentum p =mv)
That momentum is going to give the heavier boat greater inertia which will put a greater force on the rope.

 

[Franklin]

Not wave wind...

but i'll accommodate you:

2 boats... They go up a wave to the same height. They now have different POTENTIAL energies equal to mgh. They will ACCELERATE at the same rate g. Therefore, they will have the same velocity at the bottom of the wave.

So both boats have the SAME velocity at the bottom of the wave but DIFFERENT kinetic energies.
The boat with greater mass will have greater momentum than the boat with less mass. (momentum p =mv)
That momentum is going to give the heavier boat greater inertia which will put a greater force on the rope.

I do not agree with that. Let me go to the extremes: a big ship doesn't go up at all when a 5' wave hits it's bow, but my boat will go up quiet a bit. So, with that in mind, a heaver boat doesn't go up as high as a lighter boat.

I think this is the source of our disagreement. You are saying a heavier boat reacts to waves the same as a lighter boat. It is pretty much accepted that heavier displacement boats are more comfortable at sea because they are less jerky. The motion is less violent and it is because of the weight resisting the energy of the wave. Same goes while in an anchorage.