meteorology-第23章

按键盘上方向键 ← 或 → 可快速上下翻页，按键盘上的 Enter 键可回到本书目录页，按键盘上方向键 ↑ 可回到本页顶部！
————未阅读完？加入书签已便下次继续阅读！




the　luminous　body　is　rising　or　setting　the　segment　of　the　circle　above



the　earth　which　is　cut　off　by　the　horizon　will　be　a　semi…circle；　if



the　luminous　body　is　above　the　horizon　it　will　always　be　less　than　a



semicircle；　and　it　will　be　smallest　when　the　luminous　body　culminates。



First　let　the　luminous　body　be　appearing　on　the　horizon　at　the　point



H；　and　let　KM　be　reflected　to　H；　and　let　the　plane　in　which　A　is；



determined　by　the　triangle　HKM；　be　produced。　Then　the　section　of　the



sphere　will　be　a　great　circle。　Let　it　be　A　（for　it　makes　no　difference



which　of　the　planes　passing　through　the　line　HK　and　determined　by



the　triangle　KMH　is　produced）。　Now　the　lines　drawn　from　H　and　K　to　a



point　on　the　semicircle　A　are　in　a　certain　ratio　to　one　another；　and



no　lines　drawn　from　the　same　points　to　another　point　on　that



semicircle　can　have　the　same　ratio。　For　since　both　the　points　H　and



K　and　the　line　KH　are　given；　the　line　MH　will　be　given　too；



consequently　the　ratio　of　the　line　MH　to　the　line　MK　will　be　given



too。　So　M　will　touch　a　given　circumference。　Let　this　be　NM。　Then　the



intersection　of　the　circumferences　is　given；　and　the　same　ratio　cannot



hold　between　lines　in　the　same　plane　drawn　from　the　same　points　to　any



other　circumference　but　MN。



　　Draw　a　line　DB　outside　of　the　figure　and　divide　it　so　that



D：B=MH：MK。　But　MH　is　greater　than　MK　since　the　reflection　of　the



cone　is　over　the　greater　angle　（for　it　subtends　the　greater　angle　of



the　triangle　KMH）。　Therefore　D　is　greater　than　B。　Then　add　to　B　a　line



Z　such　that　B＋Z：D=D：B。　Then　make　another　line　having　the　same　ratio　to



B　as　KH　has　to　Z；　and　join　MI。



　　Then　I　is　the　pole　of　the　circle　on　which　the　lines　from　K　fall。　For



the　ratio　of　D　to　IM　is　the　same　as　that　of　Z　to　KH　and　of　B　to　KI。　If



not；　let　D　be　in　the　same　ratio　to　a　line　indifferently　lesser　or



greater　than　IM；　and　let　this　line　be　IP。　Then　HK　and　KI　and　IP　will



have　the　same　ratios　to　one　another　as　Z；　B；　and　D。　But　the　ratios



between　Z；　B；　and　D　were　such　that　Z＋B：D=D：　B。　Therefore



IH：IP=IP：IK。　Now；　if　the　points　K；　H　be　joined　with　the　point　P　by　the



lines　HP；　KP；　these　lines　will　be　to　one　another　as　IH　is　to　IP；　for



the　sides　of　the　triangles　HIP；　KPI　about　the　angle　I　are



homologous。　Therefore；　HP　too　will　be　to　KP　as　HI　is　to　IP。　But　this



is　also　the　ratio　of　MH　to　MK；　for　the　ratio　both　of　HI　to　IP　and　of



MH　to　MK　is　the　same　as　that　of　D　to　B。　Therefore；　from　the　points



H；　K　there　will　have　been　drawn　lines　with　the　same　ratio　to　one



another；　not　only　to　the　circumference　MN　but　to　another　point　as



well；　which　is　impossible。　Since　then　D　cannot　bear　that　ratio　to



any　line　either　lesser　or　greater　than　IM　（the　proof　being　in　either



case　the　same）；　it　follows　that　it　must　stand　in　that　ratio　to　MI



itself。　Therefore　as　MI　is　to　IK　so　IH　will　be　to　MI　and　finally　MH　to



MK。



　　If；　then；　a　circle　be　described　with　I　as　pole　at　the　distance　MI　it



will　touch　all　the　angles　which　the　lines　from　H　and　K　make　by　their



reflection。　If　not；　it　can　be　shown；　as　before；　that　lines　drawn　to



different　points　in　the　semicircle　will　have　the　same　ratio　to　one



another；　which　was　impossible。　If；　then；　the　semicircle　A　be



revolved　about　the　diameter　HKI；　the　lines　reflected　from　the　points



H；　K　at　the　point　M　will　have　the　same　ratio；　and　will　make　the



angle　KMH　equal；　in　every　plane。　Further；　the　angle　which　HM　and　MI



make　with　HI　will　always　be　the　same。　So　there　are　a　number　of



triangles　on　HI　and　KI　equal　to　the　triangles　HMI　and　KMI。　Their



perpendiculars　will　fall　on　HI　at　the　same　point　and　will　be　equal。



Let　O　be　the　point　on　which　they　fall。　Then　O　is　the　centre　of　the



circle；　half　of　which；　MN；　is　cut　off　by　the　horizon。　（See　diagram。）



　　Next　let　the　horizon　be　ABG　but　let　H　have　risen　above　the



horizon。　Let　the　axis　now　be　HI。　The　proof　will　be　the　same　for　the



rest　as　before；　but　the　pole　I　of　the　circle　will　be　below　the　horizon



AG　since　the　point　H　has　risen　above　the　horizon。　But　the　pole；　and



the　centre　of　the　circle；　and　the　centre　of　that　circle　（namely　HI）



which　now　determines　the　position　of　the　sun　are　on　the　same　line。　But



since　KH　lies　above　the　diameter　AG；　the　centre　will　be　at　O　on　the



line　KI　below　the　plane　of　the　circle　AG　determined　the　position　of



the　sun　before。　So　the　segment　YX　which　is　above　the　horizon　will　be



less　than　a　semicircle。　For　YXM　was　a　semicircle　and　it　has　now　been



cut　off　by　the　horizon　AG。　So　part　of　it；　YM；　will　be　invisible　when



the　sun　has　risen　above　the　horizon；　and　the　segment　visible　will　be



smallest　when　the　sun　is　on　the　meridian；　for　the　higher　H　is　the



lower　the　pole　and　the　centre　of　the　circle　will　be。



　　In　the　shorter　days　after　the　autumn　equinox　there　may　be　a



rainbow　at　any　time　of　the　day；　but　in　the　longer　days　from　the　spring



to　the　autumn　equinox　there　cannot　be　a　rainbow　about　midday。　The



reason　for　this　is　that　when　the　sun　is　north　of　the　equator　the



visible　arcs　of　its　course　are　all　greater　than　a　semicircle；　and　go



on　increasing；　while　the　invisible　arc　is　small；　but　when　the　sun　is



south　of　the　equator　the　visible　arc　is　small　and　the　invisible　arc



great；　and　the　farther　the　sun　moves　south　of　the　equator　the



greater　is　the　invisible　arc。　Consequently；　in　the　days　near　the



summer　solstice；　the　size　of　the　visible　arc　is　such　that　before　the



point　H　reaches　the　middle　of　that　arc；　that　is　its　point　of



culmination；　the　point　is　well　below　the　horizon；　the　reason　for



this　being　the　great　size　of　the　visible　arc；　and　the　consequent



distance　of　the　point　of　culmination　from　the　earth。　But　in　the　days



near　the　winter　solstice　the　visible　arcs　are　small；　and　the



contrary　is　necessarily　the　case：　for　the　sun　is　on　the　meridian



before　the　point　H　has　risen　far。







　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　6







　　Mock　suns；　and　rods　too；　are　due　to　the　causes　we　have　described。



A　mock　sun　is　caused　by　the　reflection　of　sight　to　the　sun。　Rods　are



seen　when　sight　reaches　the　sun　under　circumstances　like　those　which



we　described；　when　there　are　clouds　near　the　sun　and　sight　is



reflected　from　some　liquid　surface　to　the　cloud。　Here　the　clouds



themselves　are　colourless　when　you　look　at　them　directly；　but　in　the



water　they　are　full　of　rods。　The　only　difference　is　that　in　this



latter　case　the　colour　of　the　cloud　seems　to　reside　in　the　water；



but　in　the　case　of　rods　on　the　cloud　itself。　Rods　appear　when　the



composition　of　the　cloud　is　uneven；　dense　in　part　and　in　part　rare；



and　more　and　less　watery　in　different　parts。　Then　the　sight　is



reflected　to　the　sun：　the　mirrors　are　too　small　for　the　shape　of　the



sun　to　appear；　but；　the　bright　white　light　of　the　sun；　to　which　the



sight　is　reflected；　being　seen　on　the　uneven　mirror；　its　colour



appears　partly　red；　partly　green　or　yellow。　It　makes　no　difference



whether　sight　passes　through　or　is　reflected　from　a　medium　of　that



kind；　the　colour　is　the　same　in　both　cases；　if　it　is　red　in　the



first　case　it　must　be　the　same　in　the　other。



　　Rods　then　are　occasioned　by　the　unevenness　of　the　mirror…as



regards　colour；　not　form。　The　mock　sun；　on　the　contrary；　appears



when　the　air　is　very　uniform；　and　of　the　same　density　throughout。　This



is　why　it　is　white：　the　uniform　character　of　the　mirror　gives　the



reflection　in　it　a　single　colour；　while　the　fact　that　the　sight　is



reflected　in　a　body　and　is　thrown　on　the　sun　all　together　by　the　mist；



which　is　dense　and　watery　though　not　yet　quite　water；　causes　the　sun's



true　colour　to　appear　just　as　it　does　when　the　reflection　is　from



the　dense；　smooth　surface　of　copper。　So　the　sun's　colour　being



white；　the　mock　sun　is　white　too。　This；　too；　is　the　reason　why　the



mock　sun　is　a　surer　sign　of　rain　than　the　rods；　it　indicates；　more



than　they　do；　tha