Jon boat water ballast

Discussion in 'Stability' started by curttampa, Apr 6, 2020.

  1. jehardiman
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    jehardiman Senior Member

    Barry, you are missing the whole concept of buoyancy, which is why I didn't want to go there. Everything below the waterline is buoyant volume, so yes, putting a large object that has volume below the waterline causes buoyancy, but in order to fit that item below the waterline you had to remove an equal volume of buoyancy (even if it is just air), therefore the total buoyant volume didn't change. BUOYANCY IS NOT A FUNCTION OF WEIGHT OR DENSITY, IT IS A FUNCTION OF VOLUME!

    And yes, hydrostatic pressure is due to the depth of the water, but it is reduced by the head of the water directly above it....lets say I have a 1 foot cube, the pressure on the bottom of the cube is rho*g*h where rho =1.988 slug/ft^3, g = 32.2 ft/sec^2, and h = depth in feet of the bottom of the cube. The pressure on the top of the cube rho*g*(h-1)...so when h =1 the sum of the pressure (P) is P = (rho*g*h)-(rho*g*(h-1)) = rho*g*1 = 64 lb/sqft...and when h=10: P=64 lb/sqft, and when h=100: P =P=64 lb/sqft…. this is because the buoyancy of 1 cubic foot of volume in seawater is always equal to 64 lbs (note: there are others that may try to confuse this with the effects of temperature and pressure, but for all main purposes 1 cubic foot of seawater is a buoyancy of 64 lbs)

    Now lets take the cases you object to... lets say I take a hull shape with internal water ballast; I cut it off horizontally at the waterline (as we both agree that everything above the waterline is just a weight) and throw it into a special Veg-O-Matic (TM) that cuts it into 1 foot x 1 foot vertical columns. I sort out all the 1x1 columns. Each column has a hull depth of h below the waterline, and a water ballast tank depth of Db below the waterline and a top of the water ballast tank Dt below the waterline. I examine the different cases of each column, and there are 5.
    Case 1 There is no water ballast
    Case 2 The water ballast is against the hull and extends to the waterline. (i.e. Db =h and Dt=0)
    Case 3 The water ballast is against the hull and does not extend to the waterline (i.e. Db =h and Dt is not = 0)
    Case 4 the water ballast is above the hull and extends to the waterline (i.e. Db< h and Dt=0)
    Case 5 the water ballast is above the hull and does not extend to the waterline.(i.e. Db< h and Dt is not = 0)

    Case 1 The effective vertical pressure is P= (rho*g*h)-(rho*g*0)- {(rho*g*Db)-(rho*g*Dt)} => (rho*g*h)-(rho*g*0)- {(rho*g*0)-(rho*g*0)} = rho*g*h which is what we expect

    Case 2 The effective vertical pressure is P= (rho*g*h)-(rho*g*0)- {(rho*g*Db)-(rho*g*Dt)} => (rho*g*h)-(rho*g*0)- {(rho*g*h)-(rho*g*0)} = 0 which again is what we expect

    Case 3 The effective vertical pressure is P= (rho*g*h)-(rho*g*0)- {(rho*g*Db)-(rho*g*Dt)} = rho*g (h -(h-Dt)) => rho*g*(h-(depth of water ballast tank))

    Case 4 The effective vertical pressure is P= (rho*g*h)-(rho*g*0)- {(rho*g*Db)-(rho*g*Dt)} = rho*g (h -(Db-0)) => rho*g*(h-(depth of water ballast tank))

    Case 5 The effective vertical pressure is P= (rho*g*h)-(rho*g*0)- {(rho*g*Db)-(rho*g*Dt)} = rho*g (h -(Db-Dt)) => rho*g*(h-(depth of water ballast tank))

    As you can see in Cases 3-5, all the water ballast does is reduce the effective hydrostatic pressure on the bottom plate directly under it equal to the depth of the water ballast above it. The buoyant volume is effectively negated by the volume of the water ballast.
    Please don't make me get out The Potato Of Science! (and all you engineers who had the Meriam textbooks will know what I mean). Really...this is the way it works. I know it seems counter intuitive, but this is the real physics behind it.
     
  2. latestarter
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    latestarter Senior Member

    I am puzzled, the above seems so out of line with accepted science from Archimedes onwards.

    Archimedes' principle - Wikipedia https://en.wikipedia.org/wiki/Archimedes%27_principle

    "We can express this relation in the equation:
    F b = ρ g V
    where F b denotes the buoyant force applied onto the submerged object, ρ denotes the density of the fluid, V represents the volume of the displaced fluid and g is the acceleration due to gravity"

    Looking at it another way using dimensional analysis
    Dimensional analysis - Wikipedia https://en.wikipedia.org/wiki/Dimensional_analysis#Checking_equations_that_involve_dimensions

    Volume is length x length x length it has no dimensions that give rise to a force whereas density is mass / length x length x length when multiplied by volume (length x length x length) the lengths cancel out leaving mass which is related to force.
     
  3. Mr Efficiency
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    Mr Efficiency Senior Member

    Amazing this turned into a technical argument, common sense says ballast chambers in jon boats would be akin to fitting portholes in a coffin. Almost !
     
  4. jehardiman
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    jehardiman Senior Member

    func·tion
    /ˈfəNG(k)SH(ə)n/
    noun
    Mathematics
    1. a relationship or expression involving one or more variables.
      "the function (bx + c)"
    While I don't expect everyone on this forum to have had the benefit of a complete education in Naval Architecture, I will admit the flaw of using technical term and concepts that maybe not everyone has become conversant in. In the definition example the implicit understanding of "the function (bx + c)" is Fx (a function of x) = (bx+c) where x is the variable, and b and c are constants. Similarly "F b = ρ g V" means Fb (a function of buoyancy) = rho * g * V. Notice there is no variable for temperature, or pressure in the equation. Therefore, for the purposes of that equation, at that time, at that location (g does vary by where you are), and in that fluid (rho) at the then present temperature and pressure; the force of buoyancy (Fb) is only dependent on the volume (which is the only variable at that time, at that place, and in that fluid). While we could get into a meaningless discussion of all the ways g and rho can vary over the surface of this and other planets, or striated density layers in seawater runoff areas, for the most part I think we can, and should, limit g to the normally assigned gravitational constant, and rho to your preferred values for fresh and salt. Again, leaving volume as the only variable.

    So let us, as Mr Efficiency suggested, put this to bed by saying that for the surface of the earth, in the static near field water of constant density, the force of buoyancy on a body is only effected by,and therefore a function of, the submerged volume.
     
  5. DCockey
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    DCockey Senior Member

    The net vertical force of a volume submerged in salt water will be greater than the same volume submerged in fresh water.

    A boat will float higher with less submerged volume in saltwater than in fresh water (assuming the weight of the boat is unchanged).

    If the density of the fluid is constant then the net vertical force will depend only on the submerged volume.
     
  6. gonzo
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    gonzo Senior Member

    I think that there is a mistake in the pressure definition. The pressure is a function of the head of water (or other fluid). However, the force against the hull will change if it has water inside. If the water level inside and outside are equal, then there is a balanced force on the hull plating.
    Also, most people confuse buoyancy with flotation. An object floats when the buoyancy is greater than its weight. Adding foam, or anything inside the hull will not increase buoyancy and will decrease flotation.
     
  7. Barry
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    Barry Senior Member

    JH comments was that the Hydrostatic pressure is reduced, which is not true. Hydrostatic force is the pressure say in psi acting over a specific area on the hull bottom. In a static situation, the pressure at any
    constant depth is the same. If you disagree, perhaps you could explain how the hydrostatic force against the hull to water interface can be zero while the surrounding water is say 10 psig
     
    Last edited: Apr 25, 2020
  8. Ad Hoc
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    Ad Hoc Naval Architect

    What on earth is a "slug", as a unit??

    Im so glad we no longer use old Imperial units. :p

    Sadly it seems it is not only volume/buoyancy some are confused with. Basic hydrostatic pressure too.
     
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  9. jehardiman
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    jehardiman Senior Member

    Unfortunately, SI units are also f**ked up real bad by most people/engineers also....why do many people insist that displacement is in kilograms or Tonnes when it should be in Newtons or kiloNewtons....And don't get me started on the stupidity of kilogram force or poundals… leave that to the people who had to go to university for 4 years to learn the black water flows downhill on concrete. Slugs is the proper mass density unit...or should I have used gamma (weight density) and thrown g out.

    I'm fed up with this thread because though I try to help peoples understanding of a fairly subtle and snarky effect of buoyancy it seems some people 1) cannot read and comprehend, and 2) have very little understanding of actual physics. Berry and others cannot get their head around what buoyant pressure actually is and simply parrot degenerate, over simplified, equations made for the masses to learn by rote but never to be understood by them. I disliked my first Thermodynamics professor, but he taught me to write out the entire energy equation then start throwing out the stuff that didn't change.

    And Berry, you have failed to grasp the concept. The terms I used were "effective vertical pressure" and "effective hydrostatic pressure" and psig is meaningless without about 10 other qualifiers. Better to use psia so you get the hydrostatic pressure of the air also.

    And FWIW, Archimedes' principle is not "the force of buoyancy = ρ g V", that is an incomplete formulation of a much longer equation. Archimedes' principle is the statement that the buoyant force on an object is equal to the weight of the fluid displaced by the object, and as well as it served the old greek and those to this day, it to is a simplification of a much more complex problem....the whole reason I didn't want to go down this rabbit hole. The hard concept that few get about water ballast is that any of the "fluid" contained inside the buoyant perimeter has not been displaced, it is still there and is effectively not part of the displacement as Archimedes understood and formulated it. And F**K NO!, I don't see the need or have the inclination to prove this with differential calculus that most won't be able to follow anyway.

    I should not have been kind and should have never risen to the troll bait.
     
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  10. Mr Efficiency
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    Mr Efficiency Senior Member

    Well beyond my ken, and probably 99.9% of people, that which jehardiman speaks of in this instance, but it is an interesting little insight into the difference between what is the case in strict actuality, and that which suffices for the vast majority of practical purposes.
     
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