配石







上越新幹線の建設に伴い発掘された深沢遺跡配石遺構（群馬県みなかみ町）。縄文時代後期の集団墓地である[1]。


配石（はいせき）とは、表面のなめらかな河原石などの自然石を目的をもって配置したり、組合せたりした遺構である。ピラミッドのような石組みの建造物はふつう含まれない。



概要


配石の全体の形はさまざまあり、その性格も集落内の土地区画など日常生活に関わる標識から、埋葬や祭祀など信仰に関わるものまで多岐にわたると考えられる。単独のものは組石（くみいし）とも呼ぶ。配石のうち、石を円形に（ドーナツ状）に配置させるものは環状配石や環状列石と呼ぶ。立石をともなう場合もある。礫を土坑の中や外に配置した埋葬施設を特に配石墓と呼び、配石によって作られた住居を敷石住居と呼ぶ。


ヨーロッパには巨石を並べた配石遺跡があり、ストーンヘンジはその代表例である。日本では縄文時代前期に顕著となり、中期には急増する。環状列石をともなう著名な遺跡には、秋田県鹿角市の大湯環状列石や青森県青森市の小牧野遺跡、秋田県北秋田市の伊勢堂岱遺跡があり、これらはいずれも縄文時代後期の遺跡である。また、石材資源の豊富な中部地方の山麓地域においても盛んに作られる。



脚注


[ヘルプ]




  1. ^ 現地案内板より。




関連項目



  • 立石 (考古学)

  • 列石

  • 巨石記念物

  • ストーンサークル




Popular posts from this blog

Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

Image
28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
Read more

Does disintegrating a polymorphed enemy still kill it after the 2018 errata?

Image
up vote 13 down vote favorite 1 After the 2018 errata, the disintegrate spell description now reads: A creature targeted by this spell must make a Dexterity saving throw. On a failed save, the target takes 10d6 + 40 force damage. The target is disintegrated if this damage leaves it with 0 hit points. If you were to polymorph an enemy into a rat and then disintegrate it, would the enemy be disintegrated or would it just return to its original form? I know that for Druids, it’s not an instant kill anymore, but is this the case for polymorph as well? dnd-5e spells polymorph errata share | improve this question edited 18 hours ago
Read more

A Topological Invariant for $pi_3(U(n))$

Image
3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
Read more