Non dimesionalizing ship's moment of inertia

Discussion in 'Boat Design' started by souravsandesh, Jan 21, 2016.

  1. souravsandesh
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    souravsandesh Junior Member

    Hello can anyone tell me how to non dimensionalize ship's moment of inertia??
    I need to use it to run a code in matlab..
  2. DCockey
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    DCockey Senior Member

    That is a basic question. Are you still a student or do you know have a degree? If so is the degree in naval architecture or other engineering?
  3. philSweet
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    philSweet Senior Member

    You want the formula I = M / Omega to work, where M and Omega are both nondimensional. So you need the nondimensional length term for the arm in M, the nondimensioal force term for M, and the nondimensional time for Omega.
  4. fredrosse
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    fredrosse USACE Steam


    In all of engineering and science, there should (or should I say "Must") be dimensional consistancy. Some terms have been developed which are dimensionless numbers, such as Reynolds Number, indicating the flow regimes for fluids, etc., plus many more.

    As an example the Reynolds number, being a dimensionless number, is the same across all units of measure, in the English system of units, the Metric system, etc, all would have exactly the same numerical value of Reynolds number for a given flow configuration. Not so with the length of something, if you say the length of the ship is 274, it has no meaning unless you also state "274 feet" or "274 Meters", or "274 inches". Some physical properties need units to have meaning, and without knowing the units, the information is useless. Yes, in the USA one might assume a ship length would be stated as feet, and other places another set of units, but that is just an assumption unless the units system in use is clearly defined.

    I believe the moment of inertia is clearly defined as a property with dimensioned units, and you cannot have a moment of inertia that is dimensionless. bast tarty formulas are very often cooked up to allow calculation of all sorts of things with various boundary conditions and required knowledge of the units system being used, but that is dangerous territory for those who do not understand the principals of dimensional analysis. Am I missing something here?
  5. daiquiri
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    daiquiri Engineering and Design

    It is a very basic and simple problem, which should be an easy task for an engineering student.

    Anyways, I'll give you a hint: the moment of inertia has the dimension of mass*length^2
    The non-dimensionalized moment of inertia has the dimension of 1
    So, you need to find an X such that:
    mass*length^2 * X = 1
    The rest is up to you.
  6. Samdaman
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    Samdaman Junior Member

    Moment of inertia's units are Length^4. Not as you've stated above
  7. TeddyDiver
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    TeddyDiver Gollywobbler

    No fence to anyone but I find it somewhat disturbing to find answers like ' you should know this' repeadetly. From the question it seems obvious the OP knows the basic formula and asks about something else. Either answer the question or not would be much easier... Just wondering..

    BR Teddy
  8. TANSL
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    TANSL Senior Member

    Teddy, you are quite right in general but in this case, in my opinion, the question of the OP does not make sense so it is necessary, first, to clarify what is being asked.
    I also think that to know his professional category, for now, does not help.
  9. Heimfried
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    Heimfried Senior Member

    You refer to the area moment of inertia (2nd grade) and daiquiri was obviously referring to the mass moment of inertia.
  10. Samdaman
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    Samdaman Junior Member

    Of course. My bad!
  11. fredrosse
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    fredrosse USACE Steam

    non-dimensional moment of inertia??

    daiquiri, I am confused by your answer here (#5), with reference to my reply on the same subject just before your reply.

    Could you please explain what the original poster person was asking, to my understanding of a "dimensionless number" there are no attached dimensions.

    My response says the moment of inertia must be a dimensional number, which I firmly know to be true in my engineering experience, so I am perplexed here. Is there another definition of "dimensionless number" that I am unaware of?
  12. daiquiri
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    daiquiri Engineering and Design

    if an equation or a system of equations which describe a physical system is written in a consistent unit system, they can be rewritten in such way to adimensionalize them.

    For example, consider the simple equation of the steady, uniform horizontal motion of a vessel:
    Thrust = Drag
    Both thrust and drag are forces, and they both have the unit of Newtons [N] in the SI unit system. So if you divide both left and right side of the equation by another force, like weight, you get:
    Thrust/Weight = Drag/Weight
    The units of both left and right side of this new equation is [N]/[N], which formally gives 1. Since 1 is a pure number which does not depend on the unit system in use, by dividing the equation with another force it has been adimensionalised (or non-dimensionalised). The terms of the equation are now dimensionless coefficients, and not dimensional forces.

    The advantage of writing adimensionalised equations is that it is independent of the unit system, thus giving the possibility of analyzing it in more comprehensive and fail-safe (calculation-error safe) way. If the thrust/weight ratio is 0.3, it is 0.3 in both SI and Imperial unit system. It allows you to carry on the calculations all the way to the final result and only then to calculate the value of forces in your preferred unit system.

    It also allows you to normalize plots of forces and to find elegant relationships which would remain hidden if the calcs were done in terms of dimensionalized forces. An example of this: say that you have a keel of area A which travels at speed V through water with density Rho. If you wanted to plot the graph of the lateral force (lift) generated by the keel at various speeds and angles of attack, you would have to plot separate curves for each speed, creating a family of parametric curves (valid for a given unit system) or a 3-D graph.
    But if you divide the lift force by the term (0.5 Rho V^2 A) which has the dimensions of a force (Newton in SI system), you get the dimensionless lift coefficient CL. The messy 3-D plot now becomes a very simple and elegant single-curve plot of CL vs. Alpha, which we all know so well.

    Other equations of body statics and dynamics can be non-dimensionalised in a similar manner, the main trick consists in finding the most appropriate and convenient dividing terms.
    The moment of inertia can be adimensionalized too. For example, the quantity I/(m H^2), where:
    - I is the moment of inertia of the ship,
    - m is the mass of the anchor
    - H is the height of the mast​
    Gives a non-dimensional moment of inertia, or a coefficient.
    However, the mass of the anchor and the mast height immediately appear as a not very smart choice of reference quantities to use in the calculations.
    So the trick consists in finding the adimensionalizing terms which will give both physical and practical utility to the resulting coefficient. Usually, the visual examination of equations reveal the best and most computational cost-efficient parameters to use for the adimensionalization. Other times, the established conventions will be used. See the example of lift (and drag) coefficients, which are widely-adopted non dimensional forces.

    For example: consider the equation of the free fall of a body:
    m dz/dt = mg - Drag
    , which can be also written as:
    ma = mg - D​

    The numerical values given by the above equation will depend on the unit system used. At the time t=0 (when speed is zero), the force acting on the 10 kg (22.1 lb) body will have a value of:
    - 98.1 N in SI units;
    - 22.1 lbf in Imperial units.
    But if we divide both the left and the right side of the equation by weight W=mg, we get:
    (a/g) = 1 - D/W
    the quantity n=a/g is the non-dimensional acceleration (the so-called "g" number - remember when Tom Cruise pulls 6 g's in the Top Gun movie? ;) ) And the quantity d=D/W is the non-dimensional force ([Newton]/[Newton] = [1]), so the equation becomes:
    n = 1 - d
    which is non-dimensional and its numerical results are independent of the unit system used.
    at t=0, the non-dimensional force acting on the body is:
    - n=1​

    in both SI and British units.

    For all the above reasons, scientists work almost always with non-dimensional equations.

    Hope it helps making the process more clear. :)

    Last edited: Jan 26, 2016
  13. fredrosse
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    fredrosse USACE Steam

    ".........the force acting on the 10 kg (4.54 lb) body will have..........."

    Guess you were just making sure I was paying attention here, 22 LB mass?

    I understand your presentation, thanks for your effort. My understanding is somewhat different, as I would assign units to any value of "g" for meaning, never knowing what planet I was on.
  14. philSweet
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    philSweet Senior Member

    And a bit more along the same lines.

    Depending on the sort of methods you are using to try to solve complicated systems of differential equations, it can be very helpful to bound a variables range from 0 to 1, or -1 to 1. So many conventional ways of nondimensionalizing use a characteristic value that is equal to the maximum a quantity can achieve. This ensures terms like X^2 will always be smaller than X.

    Sometimes this bound of !x! always less than 1 isn't important to the solution method, but having all the different variables about the same magnitude is important. In that case, instead of dividing by max values, dividing by average values can work better. Among other things, this lets you say that (sums of) higher order differential quantities are always much smaller than any lower order differential quantity. It also allow you to estimate the error you introduce by dropping these terms.

    This leads to quite a few situations where there are two common references that differ by a factor of 2.

    When you leave statics behind (time as a dimension does not exist) and quasi-statics behind (time exists as a dimension, but all dxi/dt and higher time derivatives = zero), then you need a time divisor to produce a nondimensional time. The same goals apply. But there are additional constraints built in to the physics. The conventions used in statics and quasi-statics are generally such that they will continue to work in a fully dynamic analysis. They may appear to be arbitrary and under-determined, but when you extend into the dynamic realm, they are just what they need to be. That's where that 0.5 comes from in 0.5 rho V^2. It doesn't help or hinder in statics, but it does simplify things in dynamics.

    For a cyclical dynamic, the period for one cycle is set to either 1 or 2Pi. Both are very common. The inverse of each is also used, so you see four different quantities used. For noncyclic events, some other pertinent period is used. The period needed for a plane to travel one wing chord is a common one. The period that the far-stream fluid would need to traverse the diameter of a cylinder is used for Karmen vortex streets.

  15. sharpii2
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    sharpii2 Senior Member

    You could make it a function of the ship's Beam.

    Such as (Beam/x)^4

    This way you should get the same actual amount no matter which unit of measurement you used (not the same number), whether you use feet, cubits, meters, or centimeters.
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