Tech Manual for Erjavec's Automotive Technology: A Systems Approach
6th Edition
ISBN: 9781133933731
Author: ERJAVEC, Jack, Thompson, Rob
Publisher: Delmar Cengage Learning
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Chapter 12, Problem 6RQ
True False? Fiats MultiAir system relies on
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Problem 9.11 An 80 mm long line MN has its end M 15 mm in front of the V.P. The distance between
the ends projector is 50 mm. The front view is parallel to and 20 mm above reference line. Draw the
projections of the line and determine its inclination with the V.P. Also, locate the traces.
Interpretation Front view of a line is parallel to xy, therefore,
1. The line is parallel to the H.P.
2. The top view of the line has true length.
3. The front view has projected length equal to the distance be-
tween the projectors.
Construction
Refer to Fig. 9.11.
1. Draw a reference line xy. Mark point m' 20 mm above xy and
point m 15 mm below xy.
2. Draw a 50 mm long line m'n' parallel to xy.
3. Draw an arc with centre m and radius 80 mm to meet projec-
tor from point n' at point n. Join mn to represent the top view.
Determine its inclination with xy as the inclination of line MN
with the V.P. Here = 51°.
4. Traces Extend line mn to meet xy at point v. Project point v
to meet m'n' produced at…
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30
20
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и
D
P
60
60
80
⑤
b
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m
ab
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40
Chapter 12 Solutions
Tech Manual for Erjavec's Automotive Technology: A Systems Approach
Ch. 12 - Explain staged and continuously variable value...Ch. 12 - Define valve margin.Ch. 12 - What usually causes warpage in an aluminum cylin...Ch. 12 - What are the two ways pushrods can be checked for...Ch. 12 - Which of the following is not true of knurling? It...Ch. 12 - True False? Fiats MultiAir system relies on...Ch. 12 - To allow the engine to run on the Atkinson cycle,...Ch. 12 - Prob. 8RQCh. 12 - To ensure proper seating of the valve, the valve...Ch. 12 - When checking valve fit, contact between the valve...
Ch. 12 - If the valve face and valve seat do not contact...Ch. 12 - Many engines with WT have a phaser mounted to the...Ch. 12 - Multiple valve engines tend to be more efficient...Ch. 12 - Which of the following is not a type of valve...Ch. 12 - Which of the following statements is not true?...Ch. 12 - While discussing the reasons for resurfacing a...Ch. 12 - While inspecting the timing gears of an engine:...Ch. 12 - Technician A says that positive valve stem seals...Ch. 12 - When fitting a freshly ground valve into a freshly...Ch. 12 - Technician A says off-square springs may cause...Ch. 12 - Technician A says it is not necessary to measure...Ch. 12 - Technician A uses a 30-degree stone for topping a...Ch. 12 - While discussing the Valvetronic system:...Ch. 12 - While checking a hydraulic phaser after the...Ch. 12 - While servicing a pair of aluminum cylinder heads:...
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- Problem 10.16 An isosceles triangle of base 40 mm and altitude 54 mm has its base in the V.P. The surface of the plane is inclined at 50° to the V.P. and perpendicular to the H.P. Draw its projections. Construction Refer to Fig. 10.17. An isosceles triangle has its base in the V.P., so con- sider that initially the triangle ABC is placed in the V.P. with base AB perpendicular to the H.P. 1. First stage Draw a triangle a'b'c' keeping a'b' perpendicular to xy to represent the front view. Project the corners to xy and obtain ac as the top view. 2. Second stage Reproduce the top view of first stage keeping ab on xy and ac inclined at 50° to xy. Obtain new points a', b' and c' in the front view by joining the points of intersection of the vertical projectors from a, b and c of the second stage with the corresponding horizontal locus lines from a', b' and c' of the first stage. Join a'b'c' to represent the final front view. Here, the front view is an equilateral triangle of side 40 mm. X 54…arrow_forward%9..+ ۱:۱۹ X خطأ عذرا ، الرقم الذي أدخلته خاطئ. يرجى إدخال رقم بطاقة الشحن الصالحة والمحاولة مرة أخرى. رصيد هاتفك قم بمسح الرمز = رقم بطاقة التعبئة 7794839909080 رمز مكون من 13 او 14 رقماً طريقة إعادة التعبئة قم باعادة تعبئة الرصيد إعادة تعبئة الإنترنت إعادة تعبئة الرصيد O >arrow_forwardProblem 10.14 A hexagonal plane of side 30 mm has a corner in the V.P. The surface of the plane is inclined at 45° to the V.P. and perpendicular to the H.P. Draw its projections. Assume that the diagonal through the corner in the V.P. is parallel to the H.P. d' a 2 b b.f C' c.e b 'C' H.P. (a) V.P E HEX 30 e' O' d' a a' b' C' b' X y a b,f c,e d b,f (b) c,earrow_forward
- Problem 10.18 A 60° set-square has the shortest edge of 40 mm lying in the V.P. The surface is in- clined to the V.P. and perpendicular to the H.P. such that the front view appears as an isosceles triangle. Draw the projections of the set-square and determine its inclination with the V.P. Construction Refer to Fig. 10.18. A 60° set-square inclined to the V.P. and per- pendicular to the H.P. can appear as an isosceles triangle in the front view, when the shorter edge is in the V.P. 1. First stage Draw a right angled triangle 40 a' a' b' c' b' X 40 C' a,b C a,b a'b'c' keeping 40 mm long a'b' perpen- dicular to xy. Project the corners to xy and obtain ac as the top view. 2. Second stage Draw another right angled triangle a'b'c' on the horizontal locus line from points a', b' and c' of the first stage such that length of b'c' is equal to that of a'b'. Project a'b' to meet xy at ab. Draw an arc with centre a and radius equal to ac of the first stage to meet the vertical projector of c' at…arrow_forwardProblem 9.13 A 70 mm long line PQ does not have H.T. and V.T. One end of the line is 30 mm in front of the V.P. and 20 mm above the H.P. Draw its projections. Interpretation As the line PQ does not have H.T. and V.T., it is parallel to both H.P. and V.P. Construction Refer to Fig. 9.13. 1. Draw a reference line xy. Mark point p' 20 mm above xy and point p 30 mm below xy. 2. Draw a 70 mm long line p'a' parallel to xy to repre- sent the front view. X 20 p Fig. 9.13 3. Also, draw a 70 mm long line pq parallel to xy to represent the top view. 70 q yarrow_forwardProblem 10.19 A square lamina ABCD of side 40 mm is suspended from a point O such that its surface is inclined at 30° to the V.P. The point O lies on the side AB 12 mm away from A. Draw its projections. Construction Refer to Fig. 10.19. 1. First stage Draw a square a'b'c'd keeping a'd' parallel to xy. Mark a point o' on a'ď at a distance 12 mm from end a' as the point of suspension. Also, mark the centre of the square g' to represent the centre of gravity. 2. Second stage Reproduce the front view of first stage such that o'g' is perpendicular to xy. Project corners and obtain bd as the top view. 3. Third stage Reproduce the top view keeping bd inclined at 30° to xy. Obtain new points a', b', c' and ď' in the front view by joining the points of intersection of the vertical projectors drawn from points a, b, c and d of the third stage with the corresponding horizontal locus lines drawn from points a', b', c' and ď of the second stage. Join new a'b'c'd to represent the final front view.arrow_forwardProblem 10.15 A circular plane of diameter 50 mm is resting on a point of the circumference on the V.P. The plane is inclined at 30° to the V.P. and the centre is 35 mm above the H.P. Draw its projections.arrow_forwardYou are asked to manufacture 10 kg of polyester with a number-average molecular weight of 1000 by polymerizing butane-1,4-diol(HO(CH2)4OH) with adipic acid (HOOC-(CH2)4-COOH).a) What weight of diol and diacid do you need, respectively? To whatextent, p, should the reaction be carried out to? Assume a stoichiometricbalance.b) What are the number and weight fractions of dimer, trimer and tetramerat this point in the reaction?c) Because of the polymerization by dehydration to olefin, 3 mol% of thediol will be lost. What would be the number-average molecular weightwhen the reaction is carried out to the same extent? How could you offsetthis loss so that the desired molecular weightarrow_forward9.4. A PID temperature controller is at steady state with an output pressure of 9 psig. The set point and process temperature are initially the same. At time = 0, the set point is increased at the rate of 0.5°F/min. The motion of the set point is in the direction of lower temperatures. If the current settings are PART 3 LINEAR CLOSED-LOOP SYSTEMS Ke = 2 psig/°F Ti = 1.25 min TD = 0.4 min plot the output pressure versus time.arrow_forward9.6. A PI controller has the transfer function Determine the values of K, and T. 5s + 10 Ge Sarrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
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