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clc;
clear;
mprintf('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-2.2 Page 28 ') //Example 2.2
G=3.6; //Diffential ratio
N=3500/G; //[rpm] Axle rotational speed
d=30; //[in] Diameter of tire
dist=N/(60)*(%pi*d) //[in] Distance traveled in 1 sec
dist=dist/12; //[ft] Distance traveled in 1 sec
t=1; //[sec] Time period
F=864; //[lb] Force exerted by tire on road surface
W=F*dist; //[ft*lb] Workdone in 1 sec
P=F*dist/t; //[ft*lb/sec] Power
hp=P/550; //[hp] Power in horse power 1hp=550 ft*lb/sec
mprintf('\n\n Power to do work %f hp',hp);
//Comparing it to motor output:
Tm=300*12; //[in*lb] Engine torque
Nm=3500; //[rpm] Engine speed
Pm=Tm*Nm/63000;
mprintf('\n Motor output %f hp',Pm);
mprintf('\n The power output equaled the power at tire/road surface.');
//Note: The deviation of answer from the answer given in the book is due to round off error.(In the book values are rounded while in scilab actual values are taken)
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