Savitsky Power Prediction

Discussion in 'Software' started by guest12020101217, Sep 23, 2003.

  1. Dingo
    Joined: Dec 2003
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    Dingo Junior Member

    Translation

    Sorry everyone, but I will not be making a translation of the spreadsheet. It will take awhile and my custom with my programs is the user interface is in English, but the background calcs and notes are written in Dutch or Afrikaans. The input parts are in English as well as the information in the "help" tab, but the calculations use Dutch for the names etc. in the calculations. If one has the papers by Savitsky and others that I reference on the right-hand side of the calculations, it would be easy to follow the process. The headings under "Berekenings" (Calculations) are as follows: aanhangsels (appendages), luchtweerstand (air resistance), rolstabiliteit (roll stability), zeewaardigheid (seaworthiness).
     
  2. Flev
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    Flev New Member

    Dear Dingo,
    is it correct the formula for M-factor in your sheet?

    I think the 1/2 factor is wrong....

    Flev
     
  3. Dingo
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    Dingo Junior Member

    I have somewhere in my papers that the factor 1/2 was added to more closely resemble what actually happens. I could not find where I got this value, but it is not a typo. I have found that it more accurately predict the pre-planing hump resistance. If I can track down where I found this, I will let you know. If anyone else knows of where this factor was referenced from, please let me know so I can add it into the references in the spreadsheet.
     
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  4. Olav
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    Olav naval architect

    Dingo is right. There's a paper called "Design of Propulsion Systems for High-Speed Craft" by Donald L. Blount and Robert J. Bartee in which they feature this particular factor:

     
    Last edited: Nov 3, 2010
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  5. Ricardo Rios
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    Ricardo Rios New Member

    I have the sheet Dingo

    "M"-Dingo=SI( 0,98 + 0,5*( 2*(LZP/B)^1,45*EXP(-2*(Fn_v - 0,85)) - 3*(LZP/B)*EXP(-3*(Fn_v - 0,85)) ) >= 0,5; 0,98 + 0,5*( 2*(LZP/B)^1,45*EXP(-2*(Fn_v - 0,85)) - 3*(LZP/B)*EXP(-3*(Fn_v - 0,85)) ); 0,5 )


    I have two comments

    1. By limiting factor "M"> = 0.5?,
    2. the factor "M" original "blount and fox" is:

    M=0,98 + 2*( 2*(LZP/B)^1,45*EXP(-2*(Fn_v - 0,85)) - 3*(LZP/B)*EXP(-3*(Fn_v - 0,85)

    on this factor "M" can apply that "Olav" says

    PD:sorry for my English
     
  6. Dingo
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    Dingo Junior Member

    Question?

    Sorry Ricardo, was there a question?
     
  7. Ricardo Rios
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    Ricardo Rios New Member

    Answer to Dingo

    I have the publication "Small-craft power prediction" I read it and said that M:

    M= 0.98+2*(LCG/Bpx)^1.45*e^(-2(Fv-0.85))-3(LCG/Bpx)*e^(-3(Fv-0.85))

    I have de publication "Design of propulsion systems for high-speed craft" I read it and said:

    the following variation of the original "M" factor is now recommended

    M'=K(M-1)+1 for 0=< K=<1

    K is a correlation facto for refining the hull resistance at hump speed. The following indicates the influence of varying K:

    K=0 for original simplified Savitsky resistance prediccion
    K=0.5 for realistic hump speed resistance for craft of normal proportions and loading
    k=1 for original Blount/Fox resistance prediction

    in the spreadsheet "prestasie"is not expressed in this way, but the numerical meaning is the same, good job Dingo I was a little confused.

    The coefficient M is multiplied by the bare hull resistance, Why? in "prestasie" appears multiplied by the resistance of the hull with appendages?

    PD:please excuse my bad grammar, do not speak English.
     
  8. Dingo
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    Dingo Junior Member

    Corrected Spreadsheet

    Ricardo, thank you for that excellent catch. You should play cricket for Australia! I have attached a corrected spreadsheet here (the values did not vary much, but I believe to keep errors down whenever possible). If anyone sees another error, please let me know.

    Does anyone know how to update the spreadsheet in the BoatDesign.net library?
     

    Attached Files:

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  9. tangi
    Joined: Oct 2011
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    tangi New Member

    Hello , i'm new on the forum
    i'm from FRANCE , not very in english..,
    i have made my own boat for pleasure, small fast boat : Rib , with aluminium hull, 6,95m, 900kg with engine and fuel, the engine : mercury optimax 250XS ( 270hp at prop shaft ).
    actual max speed : 62kt.
    constant V hull 24° with 3 raws of strake.
    http://pneuboat.com/frames.php?url_frame=/serv-bibl/modules/pages_perso/aff_pp.php&num_pp=1349


    Hello Dingo!
    so , i've tried your excel savitsky model , to test my hull: it's seem's to be ok = 60kt for 262 effective HP. !! ( but for each speed , comments is :"Note: not planing" , it's stange ? )

    i have somme question about the savitsky model :
    -
    1-the model don't take the influence of stake for the lift force ? important , not important ?

    2-in your excel model , i' have test a boat we have at office , 11m, 6000kg, V hull : 21°at transom, 26° at mid lenght, with two 315HP + zdrive ( Nb of prop = 2 ( in motorisation section)
    the excel result : 37kt for 328hp effective , the boat run à 37-38kt, so ok!
    -> to know how many HP i need if i mount only one engine , i've pass the value " Nb of prop " at 1 ( for one engine ?, right ?) --> calculate --> result effective power would be 325HP !! i ve imagned a single 500HP or something around that !!

    could you said me where i've do a mistake?

    I hope you could help me.

    thank's and best regards

    Tangi
     
  10. Dingo
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    Dingo Junior Member

    Info Needed

    Tangi, what is the LCG of the boat? Also waterline length and beam? Otherwise post the spreadsheet to me at Dingo at Engineer dot com so I see all the parameters.
     
  11. Dingo
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    Dingo Junior Member

    Effective Power

    Tangi,

    Effective power is defined as P[eff] = D x V. It is not the value of the engine power P[eng], but the amount of power required to tow the boat due to its drag. When you enter prop = 1, then the drag decreases since there is only one prop and shaft in the water, and the resistance is lower (what you would expect). To work out the engine power required, P[eng] = P[eff] / propulsive efficiency (due to propeller, shaft and wake losses). If you are using two engines, then each engine would only have to have roughly half of what you worked out for one engine.

    Regarding the waterjet in your e-mail, if the waterjet is not matched to the engine or desired speed of the vessel, then it will have more losses compared to a propeller. Waterjets (depends on the model) sometimes are only efficient at higher speeds like +30 knots.

    I hope I explained to your satisfaction.

    Dingo
     
  12. tangi
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    tangi New Member

    hello thank's.

    so , if i have understood your explications :

    for a calculate total effpower of 328hp ( 2 zdrive )

    the power engine needed will be : 328 / 0.65 ( example of total efficiency of transmission, prop included ) = 504hp so 2 engine with around 250hp at crankshaft, It's ok ?.

    what is the common efficiency used for a " standard prop "?
    ----------------------------------------" standard waterjet" ?
    on our waterjet boat we have the good motor rpm for the jet + impeller, but with the same motor+Z-drive vs jet we have a différence of 6kt ( around +15% of efficiency loose for waterjet vs prop !!)..

    thank's for all.

    Tangi
     
  13. J Feenstra
    Joined: Jan 2012
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    J Feenstra Junior Member

    Hello Dingo,

    I am currently designing an planing craft, and the problem I’m having is that my design doesn’t fit some of the test parameters used by savitksy or other methods you use for the sheet. Does youre sheet interpolate without regard for applicability or how does this work.
    For example, Savitsky & Brown, "Procedures for Hydrodynamic Evaluation of Planing Hulls in Smooth and Rough Water", Marine Technology, Okt. 1976
    This document has a derived formula from fridsma, that you use in Zeewaardigheid.
     
  14. Captain Sunset
    Joined: Jun 2012
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    Location: Belgium

    Captain Sunset Junior Member

    could anybody please help me out with planing surface needed?

    Hi all,
    I am trying to build a twinhull boat for inland waterways, so there will be a lot of really flat water.
    I am thinking of putting a huge ski underneath that i can push down with a big lever.

    So my 2 questions:

    I wonder what the surface area of the ski has to be to planing at 10 km/h with a boat weight of 2000 kg (max)?

    How fast does the boat have to go if it 's weight is 2000 kg (max) and the planing surface is 20m² ?

    (the lengt of the boat will be about 12meters)

    I would be really very thankfull if somebody knows how to calculate this stuff. The template is too advanced for me :(

    If I have room for such a "ski" (or bi-ski for stability?) and I build it, I will happily post pictures :)
     

  15. Yellowjacket
    Joined: May 2009
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    Yellowjacket Senior Member

    At 10 kph you aren't going to be planing with a 36 foot boat. You are just about at hull speed and trying to plane isn't a good idea, you are at displacement boat speeds and if that is where you want to cruise, you need a displacement hull. If you want to go much faster the displacement hull will get very inefficient, but where you are now, the planing hull is very inefficient.
     
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