Super high speed ship of big payload

 

High speed ship of big payload – revisi

Author: David Judbarovski, systems engineering, principle inventor, 1939dmj@gmail.com 

 

A front side of the ship looked as a wedge, and as a wedge it is lifting up the undersurface water in the air. when the ship goes ahead.

Supposed

V– the ship speed, m/s, A- length of the wedge, m;, H – hight of it, m; B – width of it, m, so W – the power to go it,

W = A  * B * H * 9.8  * H / (A/V) = 9.8 * B * H^2 * V, kW

The hull of the ship uses the air as a lubricator. 

Power loses on two sides of the ship are W1 = 2  * L * H * 2 m/s * 0.5 * (H + 2) = 2 L * H * (H +2)  watt. L - length of the ship.

Power loses on bottom of the ship W2 = L * B * 2m/s * ( H +2) Watt.

If A = 12m, B =10 m, H = 6 m, v = 180 km*h = 50 m/s, L = 100 m, so W =   180,000 kW, W1 + W2 = 26,0 kW – negligible.  

100 * 6 * 10 = 6000 ton).  180,000 kWh / (6000 ton * 180 km/h) = 0.17 kWh/ ton * km . 1.0 kg oil = 4.4 kWh = 50 cents. 10 t*km =1.7/4.4 = 20 cents.
10,000 km* 6000 ton = 60 * 10^6 ton *km  * 0.17/ 4.400 kWh/ton oil = 2,300 ton oil