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Full-scale leg diameter = 4 ft = 48 in
Model leg diameter available = 5 in
Length scale factor (model/full):
λ = L_model / L_full = 5 / 48 = 0.1041667
So the scale is: 1 : 9.6 (model : full), i.e. full/model = 9.6.
| Quantity | Scale law (model/full) | With λ = 1/9.6 |
|---|---|---|
| Length | L_m/L_f = λ |
0.10417 |
| Area | A_m/A_f = λ^2 |
0.01085 |
| Volume / Displacement / Mass (same density) | V_m/V_f = λ^3 |
0.001130 (≈ 1/884.736) |
| Time / Period | T_m/T_f = √λ |
0.32275 |
| Velocity (to match Froude number) | U_m/U_f = √λ |
0.32275 |
Note: Froude similarity matches gravity-wave effects. Reynolds number will not match at this scale unless you use special methods.
| Item | Full-scale |
|---|---|
| Body (living area) length | 60 ft = 720 in |
| Body width | 14 ft = 168 in |
| Body height | 8 ft = 96 in |
| Leg count | 4 legs (2 front, 2 back) |
| Leg diameter | 4 ft = 48 in |
| Leg length | 35 ft = 420 in |
| Leg submergence (given) | ~60% of leg length/volume underwater |
| Item | Full-scale | Model-scale (divide by 9.6) |
|---|---|---|
| Body length | 720 in | 75.00 in |
| Body width | 168 in | 17.50 in |
| Body height | 96 in | 10.00 in |
| Leg diameter | 48 in | 5.00 in (your cylinders) |
| Leg length | 420 in | 43.75 in |
| Leg portion underwater (60%) | 0.60 × 420 in = 252 in | 26.25 in |
Assuming each leg is a right circular cylinder, and seawater density ≈ 64 lb/ft³. (For freshwater use 62.4 lb/ft³.)
π r² L = π × (2 ft)² × 35 ft = 140π = 439.823 ft³V_sub = 0.6 × 439.823 = 263.894 ft³| Quantity | Per leg | All 4 legs |
|---|---|---|
| Buoyancy @ 60% submerged (seawater, 64 lb/ft³) | 16,889 lb | 67,557 lb |
| Buoyancy @ 60% submerged (freshwater, 62.4 lb/ft³) | 16,467 lb | 65,868 lb |
π r² L = 0.4975 ft³ (≈ 859 in³)V_sub = 0.2985 ft³| Quantity | Per leg | All 4 legs |
|---|---|---|
| Buoyancy @ 60% submerged (seawater, 64 lb/ft³) | 19.10 lb | 76.4 lb |
| Buoyancy @ 60% submerged (freshwater, 62.4 lb/ft³) | 18.62 lb | 74.5 lb |
Interpretation: if (approximately) only the legs provide buoyancy and you want them at ~60% submergence, then the total model weight (including ballast) would be on the order of ~76 lb in seawater (or ~75 lb in freshwater). If your model body/barrels also displace water, the required ballast weight will be lower.
You can convert whatever your barrel dimensions actually are by multiplying by 9.6.
Typical 55-gallon drum dimensions (common, but verify yours):
Diameter ≈ 23 in, length/height ≈ 34.5 in.
If you connect two barrels end-to-end:
2 × 34.5 = 69.0 in23.0 inFull-scale represented (multiply by 9.6):
69.0 × 9.6 = 662.4 in = 55.2 ft23.0 × 9.6 = 220.8 in = 18.4 ft| Two-barrel “body” (typical barrels) | Model | Represents at full scale (×9.6) |
|---|---|---|
| Length | 69.0 in | 662.4 in (55.2 ft) |
| Diameter | 23.0 in | 220.8 in (18.4 ft) |
55 gal × 231 in³/gal = 12,705 in³ = 7.35 ft³14.70 ft³14.70 × 64 = 941 lb (model-scale)Because barrels add a lot of buoyancy, your model may need substantial ballast (or you may need to intentionally flood/vent the barrels) to achieve the desired ~60% leg submergence.