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Below are the calculations for your triangular seastead concept using a 1/10 linear scale Froude model.
| Item | Feet | Inches |
|---|---|---|
| Side length (2 sides, front-to-back) | 80 ft | 960 in |
| Back width (side-to-side) | 40 ft | 480 in |
| Item | Feet | Inches |
|---|---|---|
| Length | 19 ft | 228 in |
| Chord | 10 ft | 120 in |
| Width | 3 ft | 36 in |
| Submerged length at rest (50%) | 9.5 ft | 114 in |
| Above water length at rest (50%) | 9.5 ft | 114 in |
For a 1:10 Froude model, every linear dimension is divided by 10.
| Item | Feet | Inches |
|---|---|---|
| Side length (2 sides) | 8.0 ft | 96.0 in |
| Back width | 4.0 ft | 48.0 in |
| Item | Feet | Inches |
|---|---|---|
| Length | 1.9 ft | 22.8 in |
| Chord | 1.0 ft | 12.0 in |
| Width | 0.3 ft | 3.6 in |
| Submerged length at rest (50%) | 0.95 ft | 11.4 in |
| Above water length at rest (50%) | 0.95 ft | 11.4 in |
Using the simple box volume:
V_leg = 19 × 10 × 3 = 570 ft³
Since each leg is 50% submerged:
Submerged volume per leg = 570 × 0.5 = 285 ft³
With 3 legs total:
Total submerged volume = 3 × 285 = 855 ft³
Typical seawater density is about 64 lb/ft³. Caribbean seawater is close to this, so I used:
γ_seawater = 64 lb/ft³
W_full = 855 × 64 = 54,720 lb
Volume scales as (1/10)^3 = 1/1000, so:
W_model = 54,720 / 1000 = 54.72 lb
| Item | Value |
|---|---|
| Total submerged volume, full scale | 855 ft³ |
| Target displacement weight, full scale | 54,720 lb |
| Target displacement weight, 1/10 scale model | 54.72 lb |
Each full-scale leg is 570 ft³, so one 1/10 scale leg has:
V_scale_leg = 570 / 1000 = 0.57 ft³
Foam density given: 2 lb/ft³
W_foam_leg = 0.57 × 2 = 1.14 lb
1 ft³ = 7.48052 gallons = 119.688 cups
0.57 ft³ × 119.688 = 68.22 cups expanded foam
If the foam system is mixed 1:1 by volume and assuming final expanded volume is the stated expanded volume, then per batch:
Total mixed liquid needed ≈ final expanded volume / expansion ratio
But you did not give the liquid-to-expanded expansion ratio, only the final expanded density. So from density alone, we can calculate the final foam volume and final foam weight, but not the exact starting cups of Part A and Part B unless we know the supplier’s expansion ratio or mixed liquid density.
| Item | Per scale leg |
|---|---|
| Expanded foam volume required | 0.57 ft³ |
| Expanded foam volume required | 68.22 cups |
| Expanded foam weight | 1.14 lb |
Then the total starting liquid volume depends on the manufacturer expansion ratio:
Starting liquid volume (cups) = 68.22 / (expansion ratio)
Part A = Part B = 34.11 / (expansion ratio)
| Expansion Ratio | Total Mixed Liquid per Scale Leg | Part A | Part B |
|---|---|---|---|
| 10× | 6.82 cups | 3.41 cups | 3.41 cups |
| 15× | 4.55 cups | 2.27 cups | 2.27 cups |
| 20× | 3.41 cups | 1.71 cups | 1.71 cups |
| 25× | 2.73 cups | 1.36 cups | 1.36 cups |
To get the exact cups of each part, please use the expansion ratio from your foam supplier’s datasheet.
5 knots = 5 × 1.68781 = 8.439 ft/s
For 1/10 scale:
V_model = V_full / √10 = 8.439 / 3.1623 = 2.668 ft/s
| Item | Value |
|---|---|
| Full-scale speed | 5 knots = 8.439 ft/s |
| 1/10 scale model tow speed | 2.67 ft/s |
Under Froude scaling, force scales as:
F_full = F_model × λ³
where λ = 10, so:
F_full = F_model × 1000
Mechanical power:
P = F × V
Using full-scale speed at 5 knots:
P_full(ft·lb/s) = (F_model × 1000) × 8.439
P_full(ft·lb/s) = F_model × 8439
Convert ft·lb/s to watts:
1 ft·lb/s = 1.35582 W
P_full(W) = F_model × 8439 × 1.35582
P_full(W) = F_model × 11,442
If your sensor on the scale model reads drag in lb-force, then:
Full-scale ideal mechanical watts at 5 knots = (model drag in lb) × 11,442
So the constant is:
11,442
Use:
Electrical watts = (model drag in lb) × 11,442 / η
Example efficiencies:
| Overall Thruster Efficiency η | Multiply Sensor Reading By |
|---|---|
| 0.70 | 16,346 |
| 0.60 | 19,070 |
| 0.50 | 22,884 |
If you do not yet know thruster efficiency, use 11,442 for the ideal hydrodynamic power and then divide by your estimated total propulsion efficiency later.
Under Froude scaling, time scales as the square root of length scale:
T_full = T_model × √10
T_full = T_model × 3.1623
So the multiplier is:
3.1623
| Quantity | Result |
|---|---|
| Scale ratio | 1:10 |
| Scale triangle side length | 8.0 ft = 96 in |
| Scale triangle back width | 4.0 ft = 48 in |
| Scale leg length | 1.9 ft = 22.8 in |
| Scale leg chord | 1.0 ft = 12.0 in |
| Scale leg width | 0.3 ft = 3.6 in |
| Scale submerged leg length | 0.95 ft = 11.4 in |
| Full-scale target displacement | 54,720 lb |
| Scale-model target displacement | 54.72 lb |
| Expanded foam volume per scale leg | 0.57 ft³ = 68.22 cups |
| Expanded foam weight per scale leg | 1.14 lb |
| Scale tow speed to simulate 5 knots full scale | 2.67 ft/s |
| Power constant for ideal full-scale watts | 11,442 × (model drag in lb) |
| Roll-period scaling multiplier | 3.1623 |
If you want, I can do a second-pass refined version using a more realistic estimate for the actual volume of a NACA-shaped leg instead of a full rectangular block. That would give you much better numbers for: