分享一本汽车构造专业英语

英语机械笔译 · 发表于 2023-3-7 11:36:21

各位机械行业的老师好，我是一名从事机械类翻译13年的英语笔译，对机械行业的技术资料翻译有一点心得。大家如果有什么翻译方面的问题，可以发给我，一起讨论。后续我也会找一些英语的机械类工具书，陆续发布到必威APP精装版下载，感兴趣的老师可以自取。

补充内容 (2023-3-8 15:54):
之前发的书不知道为什么没有了，在三楼补发一下

游隼 · 发表于 2023-3-7 15:30:09

亲，没看到书啊

一修城 · 发表于 2023-3-7 19:12:14

书呢？

英语机械笔译 · 发表于 2023-3-8 15:54:19

不好意思，我明明发过书了，却没有显示出来，在这里重发一下

汽车构造专业英语.pdf (723.64 KB, 下载次数: 1)

英语机械笔译 · 发表于 2023-3-8 22:39:20

发一段测试文稿，看看能支持公式不

Formulas and Table for Regular Polygons.—A
regular polygon is a many-sided, two-
dimensional figure in which the lengths of the sides are equal. Thus, the angle measures are
also equal. An equilateral (equiangular) triangle is the polygon with the least number of
sides. The following formulas and table can be used to calculate the area, length of side, and
radii of the inscribed and circumscribed circles of regular polygons.

where N= number of sides; S= length of side; R = radius of circumscribed circle; r =

radius of inscribed circle; A = area of polygon; and, a= 180° ÷ N = one-half center angle of one
side. See also Regular Polygon on page 74.

Area, Length of Side, and Inscribed and Circumscribed Radii of Regular Polygons

Example 1: A regular hexagon is inscribed in a circle of 6 inches diameter. Find the area and the
radius of an inscribed circle. Here R = 3. From the table, area A = 2.5981R2 = 2.5981
× 9 = 23.3829 square inches. Radius of inscribed circle, r = 0.866R = 0.866 × 3 = 2.598
inches.

Example 2: An octagon is inscribed in a circle of 100 mm diameter. Thus R = 50. Find the area and
radius of an inscribed circle. A = 2.8284R2 = 2.8284 × 2500 = 7071 mm2 = 70.7 cm2. Radius of
inscribed circle, r = 0.9239R = 09239 × 50 = 46.195 mm.

Example 3: Thirty-two bolts are to be equally spaced on the periphery of a bolt-circle, 16 inches
in diameter. Find the chordal distance between the bolts. Chordal distance equals the side S of a
polygon with 32 sides. R = 8. Hence, S = 0.196R = 0.196 × 8 = 1.568 inch.

Example 4: Sixteen bolts are to be equally spaced on the periphery of a bolt-circle, 250
millimeters diameter. Find the chordal distance between the bolts. Chordal distance equals the side
S of a polygon with 16 sides. R = 125. Thus, S = 0.3902R = 0.3902 × 125 = 48.775 millimeters.
No. of

-A----

--A---

-A---          R---          R---

-S--          S--

-r--

-r-

Sides       S2

R2          r2             S

r          R          r          R          S

3       0.4330    1.2990    5.1962    0.5774 2.0000    1.7321    3.4641    0.5000
0.2887
4       1.0000    2.0000    4.0000    0.7071 1.4142    1.4142    2.0000    0.7071
0.5000
5       1.7205    2.3776    3.6327    0.8507 1.2361    1.1756    1.4531    0.8090
0.6882
6       2.5981    2.5981    3.4641    1.0000 1.1547    1.0000    1.1547    0.8660
0.8660
7       3.6339    2.7364    3.3710    1.1524 1.1099    0.8678    0.9631    0.9010
1.0383
8       4.8284    2.8284    3.3137    1.3066 1.0824    0.7654    0.8284    0.9239
1.2071
9       6.1818    2.8925    3.2757    1.4619 1.0642    0.6840    0.7279    0.9397
1.3737
10       7.6942    2.9389    3.2492    1.6180 1.0515    0.6180    0.6498    0.9511
1.5388
12    11.196    3.0000    3.2154    1.9319 1.0353    0.5176    0.5359    0.9659
1.8660
16    20.109    3.0615    3.1826    2.5629 1.0196    0.3902    0.3978    0.9808
2.5137
20    31.569    3.0902    3.1677    3.1962 1.0125    0.3129    0.3168    0.9877
3.1569
24    45.575    3.1058    3.1597    3.8306 1.0086    0.2611    0.2633    0.9914
3.7979
32    81.225    3.1214    3.1517    5.1011 1.0048    0.1960    0.1970    0.9952
5.0766
48    183.08       3.1326    3.1461    7.6449 1.0021    0.1308    0.1311    0.9979
7.6285
64    325.69       3.1365    3.1441 10.190    1.0012    0.0981    0.0983    0.9988
10.178

A  =  NS2 cot α ÷ 4

=  NR2 sin α cos α

=  Nr2 tan α

r  =  R cos α

=  (S cot α) ÷ 2  =

(A cot α) ⁄ N

R  =  S ÷ (2 sin α)

=  r ÷ cos α =

A ⁄ (N sin α cos α)

S  =  2R sin α

=  2r tan α =  2

(A tan α) ⁄ N

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