The measured resistance in fresh water of an 8-ft ship model moving at 6.50 ft/sec was 9.60 lb. (a) What would be the velocity of the 128-ft prototype? (b) What force would be required to drive the prototype at this speed in salt water?

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Accepted Answer

(a) Since gravity forces predominate, we obtain

[latex]\frac{V_{m}}{V_{p}}=\sqrt{L_{r}}=\sqrt{8/128} [/latex] and [latex]V_{p} =\frac{6.50}{(1/16)^{1/2}}=26.0 [/latex] ft/sec

(b) [latex]\frac{F_{m}}{F_{p}}=\gamma _{r}L^3_{r} [/latex] and [latex]F_{p}=\frac{9.60}{(62.4/64.0)(1/16)^3}=40,330 [/latex] lb

This latter value can be obtained by using the drag formula: Drag force = [latex]C_{f}\rho \frac{A}{2}V^2 [/latex]

For model, [latex]9.60=C_{f}\left(\frac{62.4}{2g} \right)\left(\frac{A}{(16)^2} \right)(6.50)^2 [/latex] and [latex] \frac{C_{f}A}{2g}=\frac{(9.60)(16)^2}{(62.4)(6.50)^2} [/latex] (1)

For prototype, force [latex]= C_{f}\frac{64.0}{2g}A(26.0)^2 [/latex] and [latex] \frac{C_{f}A}{2g}=\frac{force}{(64.0)(26.0)^2} [/latex] (2)

Equating (1) and (2) since the value of [latex]C_{f} [/latex] is the same for model and prototype, we obtain

[latex]\frac{(9.60)(16)^2}{(62.4)(6.50)^2} =\frac{force}{(64.0)(26.0)^2} [/latex] , from which force [latex]=40,330 [/latex] lb, as before

Question 6.33: The measured resistance in fresh water of an 8-ft ship model...

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