The “crossed-strings” method of Hottel [13] provides a simple means to calculate view factors between surfaces that are of infinite extent in one direction. For two such surfaces (a) with unobstructed views of one another, the view factor is of the form F 12 = 1 2 w 1 [ ( a c + b d ) − ( a d + b c ) ] Use this method to evaluate the view factors F 1 2 for sketches ( b ) through ( e ). Compare your results with those from using the appropriate graphs, analytical expressions, and view factor relations.
The “crossed-strings” method of Hottel [13] provides a simple means to calculate view factors between surfaces that are of infinite extent in one direction. For two such surfaces (a) with unobstructed views of one another, the view factor is of the form F 12 = 1 2 w 1 [ ( a c + b d ) − ( a d + b c ) ] Use this method to evaluate the view factors F 1 2 for sketches ( b ) through ( e ). Compare your results with those from using the appropriate graphs, analytical expressions, and view factor relations.
Solution Summary: The author explains that the view factor between the surface that area of infinite extent in one direction is 0.78.
The “crossed-strings” method of Hottel [13] provides a simple means to calculate view factors between surfaces that are of infinite extent in one direction. For two such surfaces (a) with unobstructed views of one another, the view factor is of the form
F
12
=
1
2
w
1
[
(
a
c
+
b
d
)
−
(
a
d
+
b
c
)
]
Use this method to evaluate the view factors F12for sketches (b) through (e). Compare your results with those from using the appropriate graphs, analytical expressions, and view factor relations.
For the walking-beam mechanism shown in Figure 3, find and plot the x and y coordinates of the
position of the coupler point P for one complete revolution of the crank O2A. Use the coordinate
system shown in Figure 3. Hint: Calculate them first with respect to the ground link 0204 and
then transform them into the global XY coordinate system.
y
-1.75
Ꮎ
Ꮎ
4
= 2.33
0242.22
L4
x
AP = 3.06
L2 = 1.0
W2
31°
B
03 L3 = 2.06
P
1
8
5
.06
6
7
P'
The link lengths, gear ratio (2), phase angle (Ø), and the value of 02 for some geared five bar
linkages are defined in Table 2. The linkage configuration and terminology are shown in Figure
2. For the rows assigned, find all possible solutions for angles 03 and 04 by the vector loop
method. Show your work in details: vector loop, vector equations, solution procedure.
Table 2
Row
Link 1 Link 2
Link 3
Link 4
Link 5
λ
Φ
Ө
a
6
1
7
9
4
2
30°
60°
P
y 4
YA
B
b
R4
R3
YA
A
Gear ratio:
a
02
d
05
r5
R5
R2
Phase angle: = 0₂-202
R1
05
02
r2
Figure 2.
04
X
Problem 4
A .025 lb bullet C is fired at end B of the 15-lb slender bar AB. The
bar is initially at rest, and the initial velocity of the bullet is 1500 ft/s
as shown. Assuming that the bullet becomes embedded in the bar,
find (a) the angular velocity @2 of the bar immediately after impact,
and (b) the percentage loss of kinetic energy as a result of the impact.
(c) After the impact, does the bar swing up 90° and reach the
horizontal? If it does, what is its angular velocity at this point?
Answers: (a). @2=1.6 rad/s; (b). 99.6% loss
=
(c). Ah2 0.212 ft. The bar does not reach horizontal.
y
X
4 ft
15 lb
V₁
1500 ft/s
0.025 lb
C
30°7
B
A
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