曲線部分の凹面の中心の場合、半径rの曲率半径を持つ点Cから距離r1/n1の点は、点Cから距離n1r1の仮想イメージP1を生み出します。
- 凹面の中心の場合、半径rの曲率半径を持つ点Cから距離r1/n1の点は、点Cから距離n1r1の仮想イメージP1を生み出します。
- 凹面の中心と曲率半径について、点Cから距離r1/n1の点があります。この点からは、点Cから距離n1r1の仮想イメージP1が生成されます。
- 凹面の中心では、曲率半径rの点Cから距離r1/n1の点があります。この点には、点Cから距離n1r1の仮想イメージP1が現れます。
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以下の文がきれいに訳せません。 If C is the center of curvature of the curved portion of the lens, and r the radius of curvature, then a point at distance r1/n1 from C will give rise to a virtual image P1 at distance n1r1 from C, the imaging between the two points being aplanatic. 特に「, and r the radius of curvature,」の部分と「, the imaging between the two points being aplanatic.」の部分を日本語にするとき困ってしまいます。どう考えればいいのでしょうか。
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「もし C がレンズの局面部分の曲率の中心で、r を曲率の半径とすれば。C から r1/nl の距離の点に、この２不遊点の間に像が出来、バーチャル像 P１ が C から n1r1 の距離に出来る。」 the imaging between the two point being aplanatic 「二つの点の間の像は、不遊（無収差）だから」でしょうかね。
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We have two kinds of the final stage of the particle orbit: One is the scattering case and another is the collisional case. For the scattering case, after the passage near the planet a particle orbit settles again to the Keplerian at the region far from the planet. The final orbital elements (b_f~, e_f~, ε_f, δ_f), of course, are different from the initial owing to the gravitational interaction with the planet. On the other hand, we regard the case as the collision of the particle with the planet when the distance between the particle and the center of the planet becomes smaller than the planetary radius, r_p, which is given in the units adopted here by r_p=R_p/R=(3M/4πρ)^(1/3)/R, =4.57×(10^－3)h/(R/1AU), (2・12) where R_p and R are the radius of the planet in ordinary units and the distance between the planet and the Sun, respectively. どうかよろしくお願いします。
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In the above, x, y, and z are the Hill coordinates and r is the distance between two particles. As shown in Paper I, a solution of Keplerian motion satisfies Eq. (6) if the mutual interaction term 3r/r^3 can be neglected. よろしくお願いします。
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Fig. 3a and b. An example of recurrent non-collision orbits. The orbit with b=2.4784, e=0, and i=0 is illustrated. To see the orbital behavior near the protoplanet, the central region is enlarged in b; the circle shows the sphere of the two-body approximation and the small one the protoplanet. Fig. 4a and b. Same as Fig. 3 but b=2.341, e=0, and i=0. This is an example of recurrent collision orbits. Fig.5. Minimum separation distance r_min between the protoplanet and a planetesimal in the case of (e, i)=(0, 0). By solid curves, r_min in the first encounter is illustrated as a function of b. The level of protoplanetary radius is shown by a thin dashed line. The collision band in the second encounter orbits around b=2.34 is also shown by dashed curves. Fig. 6. Minimum distance r_min in the chaotic zone near b=1.93 for the case of (e, i)=(0, 0); it changes violently with b. Fig. 7a-c. Contours of minimum separation distance r_min in the first encounter for the case of (e, i)=(1.0, 0.5); b=2.3(a), 2.8(b), and 3.1 (c). Contours are drawn in terms of log_10 (r_min) and the contour interval is 0.5. Regions where r_min>1 are marked by coarse dots. Particles in these regions cannot enter the Hill sphere of the protoplanet. Fine dots denotes regions where r_min is smaller than the radius r_cr of the sphere of the two-body approximation (r_cr=0.03; log_10(r_cr)=-1.52). Fig. 3a and b. および Fig. 4a and b. ↓ http://www.fastpic.jp/images.php?file=3994206860.jpg Fig.5. および Fig. 7a-c. ↓ http://www.fastpic.jp/images.php?file=2041732569.jpg Fig. 6. ↓ http://www.fastpic.jp/images.php?file=2217998690.jpg お手数ですが、よろしくお願いします。
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These come from the fact that for small values of b~ and μ, as considered here, Eqs.(2・1) and (2.2) are approximately symmetric with respect to the y-axis. Now, we will see features of collision of a particle with the planet. It should be noticed that, as mentioned in §2, the radius of the planet changes with the distance between the Sun and the planet is assigned. Here we will concentrate on the collision near the orbit of the present Earth. よろしくお願いします。
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It should be noticed that the collision probability is, of course, a function of the radius of the planet, which depends on the distance between the Sun and the planet in the system of units adopted here ( see Eq.(2・12)). Hence, using the results of our orbital calculations, we can evaluate Pc(bi~) individually for the various regions from the Sun. Now, the differential cross section (or, more exactly, cross length in our two-dimensional case), dσc(bi~), is defined such that the number of particles which collide with the planet per unit time is given by Fdσc(bi~). Here, F is a flux( per unit time and per unit length) of particles which come near the planet with an impact parameter, bi~. よろしくお願いします。
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A total of 12,000 of the available 16,000 Ottoman soldiers were moving west, to be in position to launch an attack by nightfall on the day of battle. The main Ottoman force of between two and a half and three divisions, estimated between 6,000 and 16,000 rifles, were deployed at Tel el Negile and Huj with detachments at Tel esh Sheria, Jemmameh, Hareira, Beersheba, and Gaza, to prevent the EEF from out-flanking Gaza. The rear of the EEF was to be attacked by the Ottoman 16th Division, at a point where the road from Khan Yunis to Gaza crossed the Wadi Ghuzze, and by the Beersheba Group which was to advance via Shellal, to attack Khan Yunis. The 22,000-strong attack force consisted of 12,000 infantry and 11,000 mounted troops, supported by between 36 and 96 field guns and 16 howitzers. The mounted units were to stop the Ottoman reinforcements from Tel el Sheria, Jemmameh, Hareira, Negile, Huj, and Beersheba, from reinforcing the Gaza garrison while the infantry captured the town.
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フィリピンに滞在中です。 子どもの宿題の質問の意味を理解段階で悩んでいます。 問題は次の通りです。 Some friends were on a trip from A to B via P. Walking from A to P at 6 km/h and the rest of the way at 3 km/h took 1 hour and 40 minutes. If they had walked the first stretch at 3 km/h and the rest of the way at 2 km/h, it would have taken 3 hours. Find the distance between A and B. 質問です。 Walking from A to P at 6 km/h and the rest of the way at 3 km/h took 1 hour and 40 minutes. この一節は、「徒歩でAからPまで毎時6km、毎時3kmで道の休憩、（これらの合計）で1時間40分」と和訳するのでしょうか？ the rest of the way at 3 km/hの意味自体が理解できません。 よろしくお願いいたします。
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As mentioned above, there are two peaks in R(e,i) in the e-i diagram: one is at e≒1 (i<1) and the other at i≒3 (e<0.1). The former corresponds directly to the peak in R(e,0) at e≒1 found in the two-dimensional case. The latter is due to the peculiar nature to the three-dimensional case, as understood in the following way. Let us introduce r_min (i, b,ω) in the case of e=0, which is the minimum distance during encounter between the protoplanet and a planetesimal with orbital elements i, b, and ω. In Fig. 16, r_min(i,b)=min_ω{r_min(i,b,ω)} is plotted as a function of b for various i, where r_min<r_p (=0.005) means “collision”; there are two main collision bands at b≒2.1 and 2.4 for i=0. For i≦2, these bands still exist, shifting slightly to small b. This shift is because a planetesimal feels less gravitational attracting force of the protoplanet as i increases. As i increases further, the bands approach each other, and finally coalesce into one large collision band at i≒3.0; this large band vanish when i≧4. In this way, the peculiar orbital behavior of three-body problem makes the peak at i≒3 (e<0.1). Though there are the peaks in R(e,i), the peak values are not so large: at most it is as large as 5. This shows that the collisional rate is well described by that of the two-body approximation <P(e,i)>_2B except for in the vicinity of v, i→0 if we neglect a difference of a factor of 5. Now we propose a modified form of <P(e,i)>_2B which well approximates the calculated collisional rate even in the limit of v, i→0. We find in Fig.14 that <P(e,i)> is almost independent of i, i.e., it behaves two-dimensionally for i≦{0.1 (when e≦0.2), {0.02/e (when e≧0.2). (35) This transition from three-dimensional behavior to two-dimensional behavior comes from the fact that the isotropy of direction of incident particles breaks down for the case of very small i (the expression <P(e,i)>_2B given by Eq. (29) assumes the isotropy). In other words, as an order of magnitude, the scale height of planetesimals becomes smaller than the gravitational radius r_G=σ_2D/2 (σ_2D given by Eq. (3)) and the number density of planetesimals cannot be uniform within a slab with a thickness σ_2D for small i. Table 4. Numbers of three-dimensional collision events found by orbital calculations for the representative sets of e and i. In the table r_p is the radius of the protoplanet. Table 5. The three-dimensional collision rate <P(e,i)> for the case of r_p=0.005 (r_p being the protoplanetary radius), together with two-dimensional <P(e,i=0)> Fig. 14. Contours of the evaluated <P(e,i)>, drawn in terms of log_10<P(e,i)> Fig. 15. Contours of the enhancement factor R(e,i) Table 4.↓ http://www.fastpic.jp/images.php?file=1484661557.jpg Table 5.↓ http://www.fastpic.jp/images.php?file=6760884829.jpg Fig. 14. ＆Fig. 15.↓ http://www.fastpic.jp/images.php?file=8798441290.jpg 長文になりますが、よろしくお願いします。
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On 25 May, three German columns attacked a salient north-west of Bray-en-Laonnois and gained a footing in the French first trench, before being forced out by a counter-attack. On 26 May German attacks on salients east and west of Cerny were repulsed and from 26–27 May, German attacks between Vauxaillon and Laffaux Mill broke down. Two attacks on 28 May at Hurtebise were defeated by French artillery-fire and on the night of 31 May – 1 June, attacks by the Germans west of Cerny also failed. On the morning of 1 June after a heavy bombardment, German troops took some trenches north of Laffaux Mill and were then pushed out in the afternoon. On 2 June a bigger German attack began, after an intensive bombardment of the French front, from the north of Laffaux to the east of Berry-au-Bac. On the night of 2/3 June, two German divisions made five attacks on the east, west and central parts of the Californie Plateau and the west end of the Vauclerc Plateau. The Germans attacked in waves, at certain points advancing shoulder-to-shoulder, supported by flame-thrower detachments and gained some ground on the Vauclerc Plateau, until French counter-attacks recovered the ground. Despite the French holding improvised defences and the huge volumes of German artillery-fire used to prepare attacks, the German organised counter-attacks (Gegenangriffe) met with little success and at Chevreux north-east of Craonne, the French had even pushed further into the Laon Plain.
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According to Hayashi et al., We will adopt the system of units where the distance between the planet and the Sun, the sum of their masses and the angular velocity of the rotation of the planet are all unity. When the coordinate is chosen such that the (x,y) plane coincides with the rotational plane of the planet, i.e., the ecliptic plane, the Sun and the planet are at rest on the x-axis and the planet is at the origin, then the equations of motion are given by (Szebeheley^10)), x’’－2y’=－∂U/∂x, (2・1) y’’+2x’=－∂U/∂y, (2・2) where U is the effective potential, described as U=－(μ/r_1)－((1－μ)/r_2)－((1/2)r^2)+U_0. (2・3) よろしくお願いします。
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回答ありがとうございました。 「r the radius ～」の部分は「ｒ＝半径」と考えればいいのですね。 「the imaging ～」の部分は being を使っているのにbe動詞のように考えるのですか？ 理由を表しているとご推測のようですが・・・。