Handy Engine Sizing Formula

Discussion in 'Diesel Engines' started by Bahama, Aug 13, 2010.

  1. Bahama
    Joined: Jun 2010
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    Bahama Junior Member

    I'm reading Dave Gerr's Propeller Handbook and playing around with the formulas. When I combined the formula to estimate the maximum theoretical hull speed with the displacement speed formula, I noticed that I could cancel out the LWL from the formula, which shocked me actually and it gave a very simple formula to estimate the engine HP needed regardless of the LWL. Here is the formula, let me know if you see anything wrong with it:

    D = Displacement in Lbs.
    P = Engine Percentage Used at Maximum Theoretical Speed (calcuated from 1.34 * LWL^0.5)
    T = Engine HP needed so that it uses P% at the Maximum Hull Speed

    T = 1.34^3 * D / (10.665^3 * P)

    For my use, D=44000; P = 60%; and so T = 145.457HP

    This means that my max hull speed is reached at 60% (P) of the 145.457HP (T), which calculates at: 87.2742 HP.

    My LWL is 42' and so my max hull speed calculates at 8.6842 Knots (1.34 * LWL^0.5)

    My max hull speed speed is reached at 87.2742HP (60% of engine), which is 8.6842 Knots; and 10.2962 knots is reached with 100% of the 145.457HP.

    Let me know if you see any errors in what I came up with. I hope it works because this formula is quite simple, and like I said, I was shocked that the LWL cancelled out when I combined the two formulas into each other.
  2. Alik
    Joined: Jul 2003
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    Alik Senior Member

    I made enclosed diagram, originally developed for saiboats. It works for motorsalers and displacement boats, with fixed pitch propellers. Fr=Fn is Froude number.
    Power is in kW per ton of displacement.

    Attached Files:

  3. Easy Rider
    Joined: Oct 2009
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    Easy Rider Senior Member

    "Maximum Theoretical Speed"
    What's maximum about it?
    Do you mean "hull speed"? (1.34x (square root) of WLL?)
    What's the 0.5 for?
    P=engine load at hull speed?
    As to the "T=" sentence would it mean the same thing if you took the word "maximum" out?
    Hull speed is just hull speed ..how can it be "maximum" hull speed?

    I have a rule of thumb for the power needed for a displacement power boat
    arrived at by a great many observations of other boats taking into consideration the fact that most boats are overpowered.
    Rule = 3 to 4hp per ton of displacement. Up for draggy high PC hulls and down for slippery shapes like TAD's yellow Cedar.

    Easy Rider
  4. Bahama
    Joined: Jun 2010
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    Bahama Junior Member

    I should have laid out my work in detail. Here is what I did:

    Dave Gerr's "Propeller Handbook", pages 10-12, presents a "Displacement Speed Formula (Formual 2-1)":

    Speed-Length (SL) Ratio = 10.665 / (DisplacementLbs / ShaftHP)^(1/3)
    The ShaftHP is the shaft HP at the propeller.

    He also shows just below this formula that:

    Speed-Length (SL) Ratio = Knots / (LWL)^(1/2)
    LWL is measured in feet.

    I wanted to rework the formulas so that it solved for ShaftHP rather than it being used as an input.

    I also wanted to assume that the knots speed used was simply the hull-speed formula of: 1.34 * LWL^(1/2)

    And, once I obtained the shaft HP needed for that hull-speed, I wanted my e)ngine to be at 60%, so I'll divide that number by .6 to get the shaft-HP that I want to be targeting for an engine to purchase.

    So, with that in mind, I started playing with the formulas:

    I'll assume the following:

    Shaft HP (SHP) = 87.2739218137377 (I derived this using my new SHP shown at the bottom of this post; but let use the number calculated as a check to prove if my formula works)
    Length of Water Line (LWL) = 42 (feet)
    Displacement (LBS) = 44000 Lbs
    Engine Percent (EP%) = 60% (the percent of the engine power used when running at hull-speed)
    Knots Speed (KS) = 8.684192536 Knots (derived by 1.34 * sqrt(LWL) )

    First, let's just use the formula as is, and plug in my numbers as a check to see if my formulas even work:

    Speed-Length (SL) Ratio = 10.665 / (LBS / SHP)^(1/3)
    Speed-Length (SL) Ratio = 10.665 / (44000 / 87.2739218137377)^(1/3)
    Speed-Length (SL) Ratio = 1.34

    So, using my calculated numbers, from my formulas, come out to a Speed-Length Ratio of 1.34, which is what I set out to do. So, this confirms to me that I didn't make a formula mistake... so, let me show what my new formula was, and how I derived it:

    First, I reworked the formula to solve for SHP:

    SHP = LBS / 10.665^3 * (KS / LWL^(1/2))^3 this can be rewritten as:
    SHP = LBS / 10.665^3 * KS^3 / LWL^(3/2)

    I can take either of these two formulas above (which are the same thing said differently) and simply replace the KS (knot speed) with the hull formula using my knot speed found at 1.34 * the square root of my LWL to give these two formulas (respectively):

    SHP = LBS / 10.665^3 * ( (1.34 * LWL^(1/2)) / LWL^(1/2))^3
    SHP = LBS / 10.665^3 * (1.34 * LWL^(1/2))^3 / LWL^(3/2)

    OK, all of this above takes into account LWL and Displacement Lbs, and it works to give me the Shaft HP for a given LWL and Displacement in order to obtain a speed in knots of 1.34 * sqrt(LWL).

    Those are nice formulas. What surprised me is that I played around this this some more and I was able to eliminate LWL totally from the equasions above because they actually offset themselves in the formula anyway (and that really surprised me) and so that rework of the formula gives the following simplified formula:

    SHP = 1.34^3 * LBS / 10.665^3

    Wow! Look at how simple the formula got. Like I said, it shocked me that LWL was not needed and was removed through basic simplification of the formula.

    So that gives me the Shaft HP that I need to travel at my hull speed ( 1.34 * LWL^(1/2) )

    SHP = 1.34^3 * LBS / (10.665^3 * EP%)

    So that gives me the buffer that I need by dividing the 87.27392181
    SHP by 60% to give me my target SHP of 145.4565364

    So, with a SHP of 145.4565364 HP the engine will be running at my target 60% while traveling at the Hull Speed.

    Does anyone see an error in the simplification that I did? If not, then this would show that these other formulas assumed hulls speed when they devised them, which is why it would work to toss out LWL. It absolutely does not make sense to me that I could simplify away LWL... but I did.
  5. apex1

    apex1 Guest

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  6. Bahama
    Joined: Jun 2010
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    Bahama Junior Member

    I deal with prima donna's in my work all the time and so I'm used to working with them. In fact I work with so many of them that I coined a phrase that "I'll take rude expertise over kind incompetence". But if someone's going to be rude they better have the answers too, otherwise they're just a jerk, rather than a smart prima donna.

    I heard the hot air from you, but didn't shed any light on the subject.
  7. CDK
    Joined: Aug 2007
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    CDK retired engineer

    "I'll assume the following:

    Shaft HP (SHP) = 87.2739218137377"

    Bahama, I suggest you use a slide rule, not a calculator.
    1 person likes this.
  8. wardd
    Joined: Apr 2009
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    wardd Senior Member

    where do I get a 145.4565364 hp engine, and can I get it in a cheap Chinese clone?
  9. apex1

    apex1 Guest

    That was the expected reaction.........

    Now it would be fine if you just start at the beginning of the design spiral instead of reinventing the wheel.
  10. larry larisky

    larry larisky Previous Member

  11. Bahama
    Joined: Jun 2010
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    Bahama Junior Member

    All I did, was play around with the math from some formulas that I read to better understand them. while doing so, I noticed a odd, and surprising thing, where through "simplification" (a math term) LWL was actually eliminated from the formula. This shocked me since LWL is about as fundamental as it gets for HP to hull speed as it gets.

    So, I throw my math findings out there to see if anyone can figure out why such a thing would have occured and you'd think that I'd said the world was flat. I'm more surprised by the reaction that I'm getting. Rather than having someone who is competent in math comment, I get some wise cracks about slide rules and HP motors.

    When I saw that LWL was elimiated from the formula I can't justifiy it. I can't figure out why it happened, but it did. So either my math was wrong, the formula given in the book is wrong, or there is some odd explaination as to why LWL can be tossed out if you assume Hull Speed for your knots.

    I like math, I assumed that engineers here like math, so I tossed it out. I was wrong.
  12. Ad Hoc
    Joined: Oct 2008
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    Ad Hoc Naval Architect

    First question, are you a naval architect?

    Well, you have answered that:

    So, you’re quoting an over generalised simplified formulae that carries significant caveats, but treat them as absolutes. Then you start by quoting expected speeds, from the results of such formulae, to the tune of 1/10,000th of a knot!!!

    The problem is, as has been pointed out to you above by Apex et al, you need to understand what your objective is. You seem to prefer to immerse yourself into a world of abstract numbers, yet criticise others that are trying to point you in the right direction. But, because it was not what you expected nor in the manner you expected, you appear to assume the fault lies with others, not your approach.

    You are destined to remain a mathematician rather than understand design/naval architecture with your current MO. Your choice, but designing a boat is more than just maths and quoting answers to endless decimal places.

    Your should start by understanding what it is you are trying to achieve and why…and then ask the question.
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  13. SheetWise
    Joined: Jul 2004
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    SheetWise All Beach -- No Water.

    "I like math, I assumed that engineers here like math, so I tossed it out. I was wrong."

    You haven't eliminated LWL, you've simply eliminated the proof. You're using a constant that was derived from LWL.
  14. Easy Rider
    Joined: Oct 2009
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    Easy Rider Senior Member

    Bahama is just experimenting w numbers and wonders if his conclusions have any validity. He threw way too many numbers out there for me to swallow.
    However you (bahama) didn't eliminate WLL from your formula as S/L ratio is included and that is a function of WLL. How did you get a S/L ratio of 10? Shouldn't it be around .9? I am not a NA but I don't think a full displacement vessel should be driven a hull speed, SD yes but not FD. Remember my rule of thumb? 3 to 4hp per ton. 4hp per ton would be 88hp for your boat. If you were to drive your boat w 3hp per ton (66hp) you'd probably loose less than a knot and burn 1.5 gph. With a 100hp engine you'd have an engine load of 72% and about 9 knots top speed. And all this is guesses and if you have a SD hull it's rubbish.

    Easy Rider

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    FAST FRED Senior Member

    In rational terms , sailors (free wind) will go hull speed,

    tug boats deadheading with CO fuel will run over "hull speed" , an extra 2000hp makes nice waves,

    the Navies of the world long ago learned its cheaper with a L/B ratio of over 6-1 can go hull speed ,
    for a while on taxpayer fuel, with a re-supply boat nearby.

    Most of the rest of the world chooses to cruise at .9 to 1.15 times the sq rt of the lwl.

    This is the spot you should be aiming to cruise , unless you are a Saudi Prince.

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