Textbook of Engineering Drawing, Second Edition

4.26 Textbook of Enginnering Drawing--------.,-..,...----------

Eccentricity e = VI FI / VI A = VI F2 / VIB

therefore VI F 2 -VIFI / VIB-VIA = FI F2/ V 1 V 2

From the triangle FI CO

OC = 40 mm (half of minor axis)

FIe = 60 mm (half of major axis)

Thus FlO =~602 -40^2 = 44.7mm

Hence Fl2 = 2FIO = 89.4 mm

on substitution e = 89.4 = 0.745Also,eccentricity e = VI V JAB, Hence, the distance between

120 -

the directrices AB = VIV /e = 161mm.

Problem : A fountain jet is dicharged from the ground level at an inclination of 45°. The jet

travels a horizontal distance of 10m from the point of discharge and falls on the ground.

'Trace the path of the jet.

Construction (Fig. 4.39)

o

....

M+-~~------~-----------4B

c

Fig. 4.39

1. Draw the base AB of 10m long and locate its mid-point C.


2. Through C draw a line perpendicular to AB forming the axis.


3. Through A and B, draw lines at 45°, to the base intersecting the axis at D.


4. Divide AD and BD irtto the same number of equal parts and number the points as shown.


5. Join 1-1' , 2-2' ,3-3' etc., forming the tangents to the required path of jet.


6. A smooth curve passing through A and B and tangential to the above lines is the required
   path of the jet which is parabolic in shape..