# Savitsky Power Prediction

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

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### Captain SunsetJunior Member

I have pulled a kid on a rope on two very small ski's and it could stand up at 5 km/h

when i was about 60 kg i could stand up at 5 km/h as well on a surface of less then a m² (we would waterski on a upside down small table etc for the fun)

but my only question is how much surface area would you need theoretically to keep a 2 ton boat planing at 10 km/h?

but my only question is how much surface area would you need theoretically to keep a 2 ton boat planing at 10 km/h, for a completely flat and horizontal surface?

I don't know the maths, hope somenone can help me out.

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### YellowjacketSenior Member

You don't understand. The speed that it takes to plane is a function of your waterline length. You basically have to climb up on top of the bow wave. Technically it is a function of Froude number (Froude number greater than .9 to 1.5), weight and power, but more simplistically, it can be described relative to hull speed and the fact that you have to get over the bow wave to plane.

Hydrocomp has a rule of thumb to apply, where planing speed, V (in knots) is equal to 4 X the length of the cg forward of the transom, divided by the square root of the beam. For a typical 36 foot hull with a 10 foot beam and a CG 12 feet from the transom you are looking at 4x36/3.16 (the 3.16 is the square root of 10), or about 15 knots.

So your boat won't plane, or even start to plane until you are going close to 15 knots. In reality, if you have a very light boat, it can plane at lower speeds and the transition to planing would be relatively seamless, but most boat with motors and a lot of weight don't behave that way.

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### Captain SunsetJunior Member

Thanks

I understand that the climbing over the bow wave is the power consuming part.

I intend not to climb the bow wave.

I want to lower a "gigantic waterski" with a pull-lever, to push this ski against the water as soon as i attain enough speed. And then push up the boat so the twin hulls are pushed out of the water...

That is why i want to know how big the surface area (minimum length/width?) of a ski has to be for about 2000 kg and at a speed of perhaps 10 km/h

I hope you love crazy ideas. But this way i don 't climb the bow wave...

The ski would be under the deck between the hulls, with the lever attached to the deck, if i use a big enough lever or winch, i can manually lift the 2tons...

and again, I really appreciate that you like to think about stuff like this...

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### YellowjacketSenior Member

Good luck...

I'm sure a Satisvsky analysis will tell you what you want to know, but you have to make some assumptions and do some serious calculation. Not many folks have the time or the inclination to want to do it for you. There are planing hull analysis calculators on line that could do it with a few plug in numbers, but really, you are the only one that can do it. Time to brush up on your math.

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### kroskrisBoat designer

Thank you!

I have searched the whole day for a way, to predict two different planing hulls, and here you are!

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### DingoJunior Member

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### kroskrisBoat designer

Many Thanks to you good sir!

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### Barba_BNew Member

Hi Dingo,
I have designed and build small power boat 4,5m based upon your spreadsheet. With Mercury 40 Hp outboard it goes max. 25kn according to gps, but it should go 30-35kn (calculated). After the boat was build, I remeasured it (weight) an run the calculation again and was 10% diff. It should go 25kn with 21HP. Mercury is not new, but it sounds and runs good. Am I that bad designer?

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### SukiSoloSenior Member

First thing would be check the propellor is correct for the engine and application - size and pitch. Often a source of mis match, especially when it has been replaced, or badly damaged/repaired.

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### JimboatSenior Member

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### mdmkJunior Member

Is this spreadsheet applicable to outboard engine?

MMK

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### AqualungNew Member

Can I have acces to the password?

Hi Dingo I'm a naval engineering student from brazil and I'm doing my conclusion work in university about applying the method of savitsky in planing boats. Wondering if we can enter in contact by email so I can understand a few things in your paper and also if you have published this excel in some place so I can reference you in my work.

Thanks!

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### DingoJunior Member

Translations

This is not directed to Aqualung (this post just happens to come after his post), but to students that always request an English translation of my spreadsheet. The input/output page is in English, the user instructions (2nd tab) is in English, but the calculations and notes (which I originally hid many moons ago) are in Dutch (but the technical symbols are standard). I put this spreadsheet up to help students and others calculate the basic power etc. required; I am willing to help with technical issues, errors or how to enter data or reading the results, but I will not translate it since some effort is required from the user regarding research and minor tweaks.

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### Ilan VoyagerSenior Member

Applicable to any planing hull that enters in the given shapes of Savitsky paper whatever the motorization.

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### YellowjacketSenior Member

Dingo,

I'm probably not working with the latest copy of your spreadsheet, but I did have a question relative to the trim tab calculation.

I had wanted to use negative trim tab angles to try to lift the bow of a small planing craft. When I use your spreadsheet and put in a negative trim tab angle, I see a message that a negative number is an invalid number.

I'm assuming that this is because you haven't looked at the effects of negative trim tab angle, but when I ignore your warning, for small trim tab angles I do get changes that are expected, in that trim angle increases. At the same time the depth of planing does seem to decrease, but in response to trim angle increases I would expect that.

I guess the real question is, does inputting a negative trim angle add a downward force on the trim tab and do you have any idea as to the validity of the calculation?

While I'm not purely interested in a perfectly accurate assessment, I will go out and try it to see how it works in the real world, but I was just wondering if you had thought about the workings. In reality, for small angles (say 3 to 5 degrees) I might think that the calculations should work, if we are just talking about a momentum change at the transom.

Don't know how you calculated but I remain curious and was wondering if it is would be valid for small trim tab angles.

Thanks,

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