Small displacement craft resistance

[mc_rash]

Hey all,

I am designing a flat bottom boat with following particulars:

Loa = 3.3 m
Lwl = 3.02 m
Bwl = 1.24
D = 0.177 m
Displ = 385 kg
LCB -4.2% of Lwl aft Lwl/2

For my own interest i did a CFD resistance analysis in Star CCM+. The analysis was performed in calm water, at a speed of 2.2 m/s (Fn = 0.4) and fixed draught, so no movement possible (translation and rotation). The analysis resulted in a total resistance of 140 N and thus a theoretical shaft power of R*v = 140 N * 2.2 m/s = 308 W. Although this was a very basic simulation (no trim, no sinkage, no engine modeled, probably coarse mesh, etc.) 308 W still sound a bit low for me to move a boat at it's theoretical hull speed. A 2.9 kW outboard would be mouch more than required as a propulsor.

What do you guys think?

 

[skaraborgcraft]

Me thinks i need to change from dragging a model hull across the pond, and into the matrix. Sorry, beyond me, but i will be looking at that programme.

 

[philSweet]

That is the computed propulsive power. Now you need to include all the losses to get to the powerhead power. Prop efficiency is probably about 50% . Shaft losses on tiny outboard legs are going to be at least 10% (higher at part throttle). So 308W*1/0.5 * 1/0.9 = 684W as a first guess. The friction, wave, and interference losses of the drive leg are also significant.

 

[mc_rash]

That is the computed *propulsive power*. Now you need to include all the losses to get to the powerhead power. Prop efficiency is probably about 50% . Shaft losses on tiny outboard legs are going to be at least 10% (higher at part throttle). So 308W*1/0.5 * 1/0.9 = 684W as a first guess. The friction, wave, and interference losses of the drive leg are also significant.

That's what I meant with "very basic simulation", so the bare hull resistance without any added resistance by the outboard itself nor I didn't count with efficiency of prop and outboard.

But even with your efficiency losses, do 700W make sense? I mean there are many boats like my design with a 3kW outboard - do they actually need the power or would a smaller engine be sufficient to reach hull speed? I'm gonna simulate higher speeds above the theoretical hull speed to see the resistance rising.

@skaraborgcraft I can make use of the software via my university but I have no chance to make real world tests on a pond although I would like to.

 

[philSweet]

My dink planes quite nicely on 3 hp. It will push 500 pounds to about 12 knots. People row at about 60-80 Watts, and they can go at four knots at that power on a bit longer boat. Four people can paddle a 28' canoe weighing 2 tons at 4 knots. Displacement speed was usually achievable by 2 hp per ton, and that includes all losses - windage, fouling allowance, maneuvering allowance, everything. It was a practical level of power. I ran about 3 hp on average on the intercoastal waterway (3 gallons per day) on my 14,000#, 38' sailboat. Installed power was about 18hp and I ran a 14" two-bladed folding Martec prop that would fit in your pocket.

 

[fastwave]

You are modeling half the hull. Did you remember to double your forces?
In addition since you are not leeting the boat heave your displacement is likely to be significantly too low

 

[mc_rash]

You are modeling half the hull. Did you remember to double your forces?
In addition since you are not leeting the boat heave your displacement is likely to be significantly too low

Yes, only half hull is modelled due to symmetry. For the half hull it was 70 N so yes I doubled the value.

I was waiting for someone to notice this ;D

 

[messabout]

Seat of my pants (and lots of experience with little boats)..... displacement of 385kg on a short very fat boat will need a lot more thrust than might be provided by 308 watts to achieve hull speed. At a displacement of 100 kg then I think 308 watts might work.

 

[wallabycreek]

Short answer: your 308 W is tow-power (effective power), not shaft power. Once you add realistic efficiencies and some missing physics, the number lands much closer to ~0.8–1.2 kW at the prop shaft to hold 2.2 m/s (4.3 kt, Fn≈0.40) in flat water. A 2.9 kW outboard is plenty (headroom for chop/wind), but you don’t need that much to make “hull speed.”

Why 308 W looks low
Your CFD is fixed-attitude (no sinkage/trim) and likely light on wave resolution. At Fn≈0.4 the boat sits in the “hump” where residuary (wave-making) drag grows fast. Free trim/sinkage typically increases wetted area and wave drag on flat bottoms, sometimes by 20–60% vs. fixed attitude.

Also, 308 W = R×V is effective power (EP) at the tow point. What the motor must deliver is shaft power (P_S):

• EP = 308 W (from your 140 N × 2.2 m/s)
  
  
• Overall propulsive efficiency (η₀) for a small outboard is commonly 35–55% (prop + losses + hull-prop interaction).
  
  
• P_S = EP / η₀ ≈ 308 / 0.45 ≈ 680 W (and easily >800 W once you add margins).
  

Add sea margin 15–25% for wind, steering, growth, and you’re around 0.8–1.2 kW to hold 4.3 kt most days.

Quick reality-check with simple friction math
Using your particulars (LWL 3.02 m, BWL 1.24 m, T 0.177 m, Δ 385 kg):

• Estimate wetted surface S ≈ 3.5 m² (flat bottom + vertical sides; bottom area ≈ Δ/T ≈ 2.18 m², side area ≈ perimeter×T ≈ 1.3 m²).
  
  
• Re ≈ 6.6×10⁶, ITTC-57 C_F ≈ 0.00323.
  
  
• With a conservative form factor (1+k) ≈ 1.3 for a flat-bottom chine hull, viscous + form drag gives ~35 N at 2.2 m/s.
  
  
• That leaves ~100 N as residuary/wave drag to reach your 140 N total—plausible for Fn≈0.4, but likely under once you let the hull trim/sink.
  

Letting the boat pitch down a degree or two and squat a bit can add tens of newtons, pushing total resistance into the 170–220 N range. That’s EP 375–485 W, i.e., ~0.8–1.3 kW at the shaft at 45–50% efficiency—right where practical experience lands.

What to do next (fast ways to tighten the number)

1. Re-run CFD with heave & pitch free (6-DOF not necessary; heave/pitch is enough).
   
   
2. Refine free-surface (finer cells around the bow and near the transom; capture the shoulder wave).
   
   
3. Model the transom as wet/dry with ventilation if applicable; it matters at Fn≈0.4.
   
   
4. Try a simple empirical check in parallel (e.g., ITTC friction + a residuary estimate from a similar pram series) to bracket results.
   
   
5. Tow test: a spring scale on a calm morning at steady 4.3 kt will tell you the truth in an hour.
   

Sizing the motor (practical take)

• For 4–4.5 kt cruise in flat water: ~1 kW class shaft power is usually enough.
  
  
• For headwinds/chop/accel and happier prop options: 1.5–2.0 kW is a sweet spot.
  
  
• 2.9 kW gives lots of margin (and higher sprint speed before the hump wall), but you’ll spend most of the time well below that.
  

If you want, share a sketch/sections (or the SOR for loading), and I’ll run a quick resistance bracket and a prop/gear ratio suggestion so you can pick an outboard size with confidence.