Ab initio calculation of Li7 photodisintegration

a r X i v :n u c l -t h /0406080v 1 29 J u n 2004

MKPH-T-04-09

Ab initio calculation of 7Li photodisintegration

Sonia Bacca 1,2,Hartmuth Arenh¨o vel 2,Nir Barnea 3,Winfried Leidemann 1and Giuseppina Orlandini 1

1

Dipartimento di Fisica,Universit`a di Trento and INFN (Gruppo Collegato di Trento),

via Sommarive 14,I-38050Povo,Italy

2

Institut f¨u r Kernphysik,Johannes Gutenberg-Universit¨a t Mainz,Johann-Joachim-Becher-Weg 45,D-55099Mainz,Germany

3

Racah Institute of Physics,Hebrew University,91904,Jerusalem,Israel

(Dated:February 8,2008)

Abstract

The 7Li total photoabsorption cross section is calculated microscopically.As nucleon-nucleon interaction the semi-realistic central AV4’potential with S-and P-wave forces is taken.The inter-action of the ?nal 7-nucleon system is fully taken into account via the Lorentz Integral Transform (LIT)method.For the calculation of the LIT we use expansions in hyperspherical harmonics (HH)in conjunction with the HH e?ective interaction (EIHH)approach.The convergence of the LIT expansion is discussed in detail.The calculated cross section agrees quite well with the available experimental data,which cover an energy range from threshold up to 100MeV.

PACS numbers:21.45.+v,24.30.Cz,25.20.Dc,31.15.Ja

The electromagnetic response of an A-body nucleus is a basic property in nuclear physics. It contains important information about the dynamical structure of the system.For this reason microscopic calculations are needed in order to investigate the details of the reaction mechanism and the underlying dynamics.Traditionally such studies could only be made for few-body systems with A≤3.The explicit calculation of?nal state wave functions consti-tutes the main limitation of standard approaches.On the other hand these di?culties can be avoided in the Lorentz Integral Transform(LIT)method[1],where only a bound-state-like problem has to be solved although the full?nal state interaction(FSI)is rigorously taken into account.First applications of the method were made for electromagnetic responses of 4He[2].Enormous progress has also been achieved in the calculation of bound-state wave functions of systems with A>4(No Core Shell Model[3]and E?ective Interaction Hyper-spherical Harmonics(EIHH)approach[4]).The novel methods of LIT and EIHH combined with the help of modern computational resources enabled us to carry out calculations of elec-tromagnetic cross sections of nuclei with A=6[5,6].Therefore,systems with4

In this paper we present the?rst microscopic calculation of a photoabsorption reaction on a nucleus with A=7,namely7Li.In this calculation we use as nucleon-nucleon interaction the Argonne potential AV4’[7],which is a semirealistic central interaction that contains S-and P-wave forces.In recent work[6]we found that the P-wave interaction has a signi?cant in?uence on the photodisintegration cross section of P-shell nuclei,like6He and6Li,and leads to a considerably better agreement with experimental data than a central S-wave interaction only.Thus we have chosen this potential model for the present ab initio calculation of the 7Li total photoabsorption cross section.The calculational procedure is described in detail in[5,6].In the following we will brie?y review the main steps only.

The inclusive unpolarized response function R(ω)is de?ned by the total photoabsorption cross section

σ(ω)=4π2αωR(ω),(1) whereαdenotes the?ne structure constant andωthe photon energy.In the LIT method[1] one obtains R(ω)from the inversion of the integral transform

L(σR,σI)= dωR(ω)

In the unretarded dipole approximation one has

L(σR,σI)=1

2S0+1 M L0,M S01

H?σR?iσI

?D

z

|Ψ0(M L0,M S0) },(3)

where?D z denotes the dipole operator,H the nuclear Hamiltonian,and L0and S0are the angular momentum and spin of the ground state|Ψ0(M L0,M S0) ,with projections M L0and M S0,respectively(since we work with central forces,L0and S0are good quantum numbers). The transform L(σR,σI)is evaluated by inserting a complete set of projection operators C,M L,M S|ΨC(M L,M S) ΨC(M L,M S)|≡ C,M L,M S?P C,M L,M S,where C={L,S,T,T z,π} stands for the quantum numbers characterizing the channels(angular momentum,spin, isospin and its projection,and parity,respectively),M L and M S are third components of angular momentum and spin.In the sum only the channels allowed by the dipole selection rules need to be considered and since the dipole operator does not depend on spin,we do not need to average over the initial spin projections(M S0),neither to sum over M S.Therefore one obtains

L(σR,σI)=1

σI C| ΨC|?D z|Ψ0 |2Im{ ΨC|1

2L0+1 M L,M L0| ΨC|?D z|Ψ0 |2=1

σI Im{ ΨC|

1

TABLE I:Good quantum numbers for the channels|ΨC allowed by the dipole selections rules. L

1

21

2

1

2

T

23

2

3

2

3

?1

2?1

2

?1

2

is evaluated as continuous fraction via the Lanczos coe?cients.This means that the LIT of the total response is the sum of the Lorentz transforms of the individual channels

L(σR,σI)= C L C(σR,σI),(7) where for every channel one has L C(σR,σI)=N2C F C.

In the case of7Li one has total angular momentum and parity Jπ=3 2

with projection T z0=?1

2

.There are six di?erent channels allowed by the dipole selection rules corresponding to angular momentum L=L0?1,L0,L0+1,spin S=S0and isospin T=T0,T0+1with isospin projection conserved T z=T z0.In Table I we show the good quantum numbers of these channels.

For every channel the LIT is calculated by expanding|Ψ0 and|ΨC in terms of the7-body anti-symmetrized hyperspherical harmonics(HH)up to a maximum HH hyperangular momentum[9],i.e.K0max for the ground state|Ψ0 and K max for the channels|ΨC .The convergence of the HH expansion is improved by using the EIHH approach.Our procedure consists in?xing?rst K0max such that convergence for the binding energy is reached,and then we study the behavior of the LIT with increasing K max.As already pointed out in[6], the rate of convergence for a potential that includes a central P-wave interaction,like the AV4’,can be slower than that of a purely central S-wave force.Nevertheless,in the case of 7Li we have obtained a satisfactory convergence in terms of hyperangular momentum,both for ground state energy and LIT.For the ground state of7Li good convergence is reached with K0max=9with a binding energy of45.28MeV.Further increase to K0max=11leads only to a small change of the binding energy by0.05MeV.Because of the dipole selection rule K=K0±1between states with hyperangular momenta K and K0,an expansion of

TABLE II:Reduction of basis states for channel C=3for K max=10.Each symmetry is represented by a Young tableau.The total number of states for each symmetry is listed as N sym, lists the actual number of states in the calculation.

while N sym

used

Symmetry

N sym

00000069307801515742 the ground state up to a certain K0max implies that in|ΨC only states with hyperangular momentum K≤K max=K0max+1contribute to the LIT.Thus it is expected that for su?ciently high K0max a further increase of K max beyond K0max+1will not result in a signi?cant change.In this case a check of the convergence with respect to K0max only will be su?cient.

In the present work,the best calculation of the LIT corresponds to a?nal state with K max=10.In this case the number of HH-basis states to be included becomes quite large, especially for the channels with L C=1and2.For example,already for channel C=3 and K max=10one has6348hyperspherical states.This number has to be multiplied by the number of hyperradial states,about30,to obtain the total number of states needed in the expansion.Therefore,it is desirable to discard those HH states which give only negligible contributions to the LIT.To this end we have studied the importance of the HH states according to their spatial symmetry and found that quite a few of them can be safely neglected(see Table II).We have checked this approximation by performing calculations with the complete set of states for lower values of K max(6and8)and compared the results with those using a truncated set.Whenever the di?erences between results with the reduced and the full basis were negligible(below0.5%)we concluded that the omitted states were negligible also for higher K max.In this way we accomplished for K max=10a sizable reduction from N=190440to N=111900basis functions.In an analogous way we carried out the calculations for all the other channels.The estimated error introduced by these truncations is of the order of0.5%.

We are also able to check the error introduced by the symmetry truncation in a second way.In fact a good check of the quality of the calculation is obtained by considering the

sum over the norms N 2

C de?ned in (5).Using completeness one ?nds

C

N 2

c =

1

2L 0+1

M L 0

Ψ0|?D ?z ?D z |Ψ0 ,

(8)

where the last expression is nothing else than the mean expectation value of the operator ?D ?z ?D z in the ground state,that can be easily calculated (see Ref.[5]).With respect to

Eq.(8)we obtained 1.877[fm 2]for the ground state expectation value with K 0max =9,while

using a symmetry truncated expansion for |ΨC up to K max =10we get 1.871[fm 2].The small di?erence of 0.3%re?ects the small error introduced by the symmetry truncation.

?45

?30?1501530020

40

60

80

100

R (%)

R (%)

C=6

K K K 00

m a x m a x max 7 9

5?60

?45?30?150

15020

406080

100

[MeV]

σC=6

K 0

max 11 7

9R (b)

(a)

(+)

(?)

FIG.1:Upper panel (a):Relative change R K 0max (+)C

for the LIT of channel C=6(σI =10MeV).

Lower panel (b):R K 0

max

(?)

C for the same channel (see text for details).

As next point we address the quality of convergence of the LIT with respect to the HH expansion.For this reason,we ?rst introduce the notation

L K 0max (±)C

for the LIT calculated

with an expansion up to K 0max for the ground state and up to K ±max =K 0

max

±1for |ΨC ,

respectively.That means L K0max(+)

C

represents the LIT of channel C calculated with a max-

imal hyperangular momentum for those states which can be reached by a dipole transition

from the ground state,whereas for L K0max(?)

C

the expansion of channel C is taken up to one step below this value.Then we de?ne

R K0max(±) C =

L K0max(±)

C

?L(K0max?2)(+)

C

2

0 1

2

0 20 40 60 80 100

σ [m b ]

7

Li

T=1/2C=1C=3C=5

0 1

2

20 40 60 80 100

σ [m b ]

7

Li

T=3/2C=2C=4C=6

0 1 2 3 4 0

20

40

60

80

100

σ [m b ]

ω [MeV]

7

Li

Total T=1/2T=3/2

(c)

(b)

(a)

FIG.2:Contribution of various channels to the total cross section.Panels (a)and (b)show the separate contributions of the di?erent channels and their sum for T =1/2and T =3/2,respectively.Panel (c)shows again the T =1/2and T =3/2contributions and the total cross section.

associated with the reaction 7Li +γ→4He +t with a threshold of 2.47MeV (the theoretical threshold obtained with the AV4’potential is 4.70MeV),whereas the lowest open T =

3

1 2 3

4

σ [m b ]

ω [MeV]

FIG.3:Comparison of the theoretical photoabsorption cross section calculated with AV4’potential with experimental data from [12].

and falling o?only slowly.Channel 5is the next in importance rising only slowly above 10MeV with a maximum near 33MeV and becoming then comparable in size to channel 3.Only in the very near threshold region channel 1is dominant but then becomes much smaller than the other two channels.In Fig.2(b)the T =3/2channels have two dominant contributions,almost similar in size.Channel 4is slightly larger showing also a steep rise at threshold and slow fall-o?with a maximum near 23MeV whereas the second in importance,channel 6,shows a slow rise and a peak around https://www.wendangku.net/doc/5a45272.html,pared to these two channels,the remaining channel 2appears quite marginal.In view of the two maxima of almost equal height with a separation by about 17MeV the total T =3/2-contribution exhibits a broader distribution than the T =1/2-contribution with a shoulder on the low-energy side.The maxima of both contributions have about the same size but are separated by about 20MeV.Thus,the resulting total cross section in Fig.2(c)shows also a broad distribution with a steep rise right above threshold,a slight shoulder above the maximum and a slow fall-o?at higher energies.

This characteristic behavior is indeed exhibited by the experimental data on 7Li in Fig.3where we show a comparison of the theoretical result to experimental data from [12].Note that the theoretical cross section is shifted here from the theoretical threshold to the ex-perimental one.One readily notes that the gross properties of the data,steep rise,broad maximum and slow fall o?,are very well reproduced quantitatively over the whole energy

region by the theory.It is worthwhile to emphasize that this result is based on an ab initio calculation in which the complicated?nal state interaction of the7N-system is rigorously taken into account by application of the LIT method.No adjustable parameters were used, the sole ingredient being the AV4’NN potential model.It remains to be seen whether the slight variation of the data near and above the maximum will also be found in an experi-ment with improved accuracy.Therefore,a new measurement of the total cross section with a higher precision would be very desirable.In particular,this could clarify the question whether a simple semi-realistic potential like the AV4’model is su?cient for an accurate theoretical description of this reaction or whether a more realistic nuclear force including a 3N-force is needed.

Acknowledgments

A large part of the numerical calculations have been performed at the computer centre CINECA(Bologna).This work was partially supported by the Deutsche Forschungsgemein-schaft(SFB443).The work of N.B.was supported by the ISRAEL SCIENCE FOUNDA-TION(grant no202/02).We furthermore would like to thank J.Ahrens for providing us with the experimental data.

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[2]V.D.Efros,W.Leidemann,and G.Orlandini,Phys.Rev.Lett.78,432(1997);V.D.Efros,

W.Leidemann,and G.Orlandini,ibid.78,4015(1997).

[3]P.Navr′a til and B.R.Barrett,Phys.Rev.C59,1906(1999);P.Navr′a til J.P.Vary,and B.R.

Barrett,ibid.62,054311(2000).

[4]N.Barnea,W.Leidemann,and G.Orlandini,Phys.Rev.C61,054001(2000);N.Barnea,W.

Leidemann,and G.Orlandini,Nucl.Phys.A693,565(2001).

[5]S.Bacca,M.A.Marchisio,N.Barnea,W.Leidemann,and G.Orlandini,Phys.Rev.Lett.89,

052502(2002).

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[8]V.D.Efros,W.Leidemann,and G.Orlandini,Phys.Lett.B408,1(1997).

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数学快速计算法

数学快速计算法 二位数乘法速算总汇 1、两位数的十位相同的,而个位的两数则是相补的(相加等于10)女口:78 X 72= 37 X 33= 56 X 54= 43 X 47 = 28 X 22 46 X 44 (1) 分别取两个数的第一位,而后一个的要加上一以后,相乘。 (2) 两个数的尾数相乘,(不满十,十位添作0) 78X 72=5616 37 X 33=1221 56 X 54= 3024 43 X 47= 2021 (7+1) X 7=56 (3+1) X 3=12 (5+1) X 5=30 (4+1) X 4=20 8X 2=16 7 X 3=21 6 X 4=24 3 X 7=21 口决:头加1,头乘头,尾乘尾 2、两个数的个位相同,十位的两数则是相补的 如:36 X 76= 43 X 63= 53 X 53= 28 X 88= 79 X 39 (1) 将两个数的首位相乘再加上未位数 (2) 两个数的尾数相乘(不满十,十位添作0) 36X 76=2736 43 X 63=2709 3X 7+6=27 4 X 6+3=27 6X 6=36 3 X 3=9 口决:头乘头加尾,尾乘尾 3、两位数的十位差1,个位的两数则是相补的。 如:48 X 52 12 X 28 39 X 11 48 X 32 96 X 84 75 X 65

即用较大的因数的十位数的平方,减去它的个位数的平方。

48 X 52=2496 12 X 28 = 336 39 X 11= 819 48 X 32=1536 2500-4=2496 400-64=336 900-81=819 1600-64=1536 口决:大数头平方 —尾平方 4、一个乘数十位加个位是 9,另一个乘数十位和个位是顺数 X 78 = 81 X 23 = 27 X 89 = 5 23 2 如:12 X 13= 13 X 15= 14 X 15= 16 X 18= 17 X 19= 19 X 18= (1) 尾数相乘 ,写在个位上 (满十进位 ) (2) 被乘数加上乘数的尾数 12X 13=156 13 X 15= 195 14 X 15=210 16 X 18= 288 2X 3=6 3 X 5=154X 5=20 6 X 8=48 12+3=15 13+5=18 14+5=19 16+8=24 口决:尾数相乘 ,被乘数加上乘数的尾数 （满十进位 ） 6、任何二位数数乘于 11 如 :36 X 45 = 72 X 67 = 45 1 、解 : 3+1=4 4 X 4 = 1的6补5 数是 4X 5=20所以 36 X 45= 1620 2、解: 7+1=8 8 X 6 = 4的8补7 数是 8X 3=24所以 72 X 67 = 4824 3、解: 4+1=5 5 X 7=3的5补8 数是 5X 2=10所以 45 X 78 = 3510 5、10-20 的两位数乘法

数学快速计算方法_乘法速算

一.两个20以内数的乘法 两个20以内数相乘,将一数的个位数与另一个数相加乘以10,然后再加两个尾数的积,就是应求的得数。如12×13＝156,计算程序是将12的尾数2,加至13里,13加2等于15,15×10＝150,然后加各个尾数的积得156,就是应求的积数。 二.首同尾互补的乘法 两个十位数相乘,首尾数相同,而尾十互补,其计算方法是:头加1,然后头乘为前积,尾乘尾为后积,两积连接起来,就是应求的得数。如26×24＝624。计算程序是:被乘数26的头加1等于3,然后头乘头,就是3×2＝6,尾乘尾6×4＝24,相连为624。 三.乘数加倍,加半或减半的乘法 在首同尾互补的计算上,可以引深一步就是乘数可加倍,加半倍,也可减半计算,但是:加倍、加半或减半都不能有进位数或出现小数,如48×42是规定的算法,然而,可以将乘数42加倍位84,也可以减半位21,也可加半倍位63,都可以按规定方法计算。48×21＝1008,48×63＝3024，48×84=4032。有进位数的不能算。如87×83＝7221,将83加倍166,或减半41.5,这都不能按规定的方法计算。 四.首尾互补与首尾相同的乘法 一个数首尾互补,而另一个数首尾相同,其计算方法是:头加1,然后头乘头为前积,尾乘尾为后积,两积相连为乘积。如37×33＝1221,计算程序是(3＋1)×3×100＋7×3＝1221。 五.两个头互补尾相同的乘法

两个十位数互补,两个尾数相同,其计算方法是:头乘头后加尾数为前积,尾自乘为后积。如48×68＝3264。计算程序是4×6＝24 24＋8＝32 32为前积,8×8＝64为后积,两积相连就得3264。 六.首同尾非互补的乘法 两个十位数相乘,首位数相同,而两个尾数非互补,计算方法:头加1,头乘头,尾乘尾,把两个积连接起来。再看尾和尾的和比10大几还是小几,大几就加几个首位数,小几就减掉几个首位数。加减的位置是:一位在十位加减,两位在百位加减。如36×35＝1260,计算时(3＋1)×3＝12 6×5＝30 相连为1230 6＋5＝11,比10大1,就加一个首位3,一位在十位加，1230＋30＝1260 36×35就得1260。再如36×32＝1152,程序是(3＋1)×3＝12,6×2＝12,12与12相连为1212,6＋2＝8,比10小2减两个3,3×2＝6,一位在十位减,1212－60就得1152。 七.一数相同一数非互补的乘法 两位数相乘,一数的和非互补,另一数相同,方法是:头加1,头乘头,尾乘尾,将两积连接起来后,再看被乘数横加之和比10大几就加几个乘数首。比10小几就减几个乘数首,加减位置:一位数十位加减,两位数百位加减,如65×77＝5005,计算程序是(6＋1)×7＝49，5×7＝35,相连为4935,6＋5＝11,比10大1,加一个7,一位数十位加。4935＋70＝5005 八.两头非互补两尾相同的乘法 两个头非互补,两个尾相同,其计算方法是:头乘头加尾数,尾自乘。两积连接起来后,再看两个头的和比10大几或小几,比10大几就加几个尾数,小几就减几个尾数,加减位置:一位数十位加减,两位数百位加减。如67×87＝5829,计算程序是:6×8＋7＝55,7×7＝49,相连为5549,6＋8＝14,比10大4,就加四个7,4×7＝28,两位数百位加,5549＋280＝5829

梁弯矩图梁内力图(剪力图与弯矩图)

简单载荷 梁内力图（剪力图与弯矩图） 梁的简图 剪力Fs 图 弯矩M 图 1 l a F s F F l a F l a l -+ - F l a l a ) (-+ M 2 l e M s F l M e + M e M + 3 l a e M s F l M e + M e M l a l -e M l a + - 4 l q s F + -2 ql 2 ql M 8 2ql + 2 l 5 l q a s F + -l a l qa 2) 2(-l qa 22 M 2 228)2(l a l qa -+ l a l qa 2) (2 -l a l a 2)2(- 6 l q s F + -3 0l q 6 0l q M 3 92 0l q + 3 )33(l - 7 a F l s F F + Fa -M

8 a l e M s F + e M M 9 l q s F ql + M 2 2ql - 10 l q s F 2 l q + M 6 20l q - 注：外伸梁 = 悬臂梁 + 端部作用集中力偶的简支梁 表2 各种载荷下剪力图与弯矩图的特征 某一段梁上的外力情况 剪力图的特征 弯矩图的特征 无载荷 水平直线 斜直线 或 集中力 F 突变 F 转折 或 或 集中力偶 e M 无变化 突变 e M 均布载荷 q 斜直线 抛物线 或 零点 极值 表3 各种约束类型对应的边界条件 约束类型 位移边界条件 力边界条件 （约束端无集中载荷） 固定端 0=w ,0=θ — 简支端 0=w 0=M

梁的剪力、弯矩方程和剪力、弯矩图

5.4.1 梁的剪力、弯矩方程和剪力、弯矩图 梁在外力作用下，各个截面上的剪力和弯矩一般是不相等的。若以横坐标表示横截面沿梁轴线的位置，则剪力Q 和弯矩M 可以表示为坐标的函数，即 它们分别称为梁的剪力方程和弯矩方程。 与绘制轴力图或扭矩图一样，可用图线表明梁的各截面上剪力和弯矩沿梁轴线的变化情况。作图时，取平行于梁轴线的直线为横坐标轴，值表示各截面的位置；以纵坐标表示相应截面上的剪力、弯矩的大小及其正负，这种表示梁在各截面上剪力和弯矩的图形，称为剪力图和弯矩图。 例5-1 简支梁AB 承受承受均布荷载作用，如图 5 - 10a 所示。试列出剪力方程和弯矩方程，并绘制剪力图和弯矩图。 解：(1) 计算支反力以整梁为研究对象，利用平衡条件计算支反力。由于简支梁上的载荷对于跨度中央截面是对称的，所以 A 、 B 两端的支反力应相等，即 (1) 方向如图。 (2) 建立剪力、弯矩方程以梁左端A 为的坐标原点，取坐标为的任意横截面的左侧梁段为研究对象。设截面上的剪力Q () 、弯矩M () 皆为正，如图5-10b 所示。由平衡方程

将(1) 式代入上面两式，解得 （ 2 ） （ 3 ） (2) 、(3) 两式分别为剪力方程和弯矩方程。 (3) 绘制剪力图、弯矩图由式(2) 可知，剪力图为一直线。只需算出任意两个截面的剪力值，如A 、B 两截面的剪力，即可作出剪力图，如图5 - 10c 所示。 由式(3) 可知，弯矩图为一抛物线，需要算出多个截面的弯矩值，才能作出曲线。例如计算下列五个截面的弯矩值：当时, M =0 ；当 时，；当时，。由此作出的弯矩图，如图5-10d 所示。 由剪力图和弯矩图可知，在靠近A 、B 支座的横截面上剪力的绝对值最大，其值为 在梁的中央截面上，剪力Q ＝0 ，弯矩为最大，其值为 例5-2 简支梁AB 承受集中力偶M0作用，如图 5 - 11a 所示。试作梁的剪力图、弯矩图。

工程量快速计算的基本方法经验

工程量快速计算的基本方法经验 本章所述工程量快速计算的基本方法包括：练好“三个基本功”；合理安排工程量计算顺序；灵活运用“统筹法”计算原理；充分利用“工程量计算手册”等四项内容。在实际工作中，只要能够熟练掌握，充分利用以上“基本方法”，就可以快速提高工程量计算业务水平。 第一节练好“三个基本功” 练好“三个基本功”包括：提高看图技能；熟悉常用标准图做法；熟悉工程量计算规则，等三个方面。 一、提高看图技能 工程量计算前的看图，要先从头到尾浏览整套图纸,待对其设计意图大概了解后，再选择重点详细看图。在看图过程中要着重弄清以下几个问题： （一）建筑图部分 1、了解建筑物的层数和高度（包括层高和总高）、室内外高差、结构形式、纵向总长及跨度等。 2、了解工程的用料及作法，包括楼地面、屋面、门窗、墙柱面装饰的用料及法。 3、了解建筑物的墙厚、楼地面面层、门窗、天棚、内墙饰面等在不同的楼层上有无变化（包括材料做法、尺寸、数量等变化），以便采用不同的计算方法。 （二）结构图部分 1、了解基础形式、深度、土壤类别、开挖方式（按施工方案确定）以及基础、墙体的材料及做法。 2、了解结构设计说明中涉及工程量计算的相关内容，包括砌筑砂浆类别、强度等级，现浇和预制构件的混凝土强度等级、钢筋的锚固和搭接规定等，以便全面领会图纸的设计意图，避免重算或漏算。 3、了解构件的平面布置及节点图的索引位置,以免在计算时乱翻图纸查找，浪费时

间。 4、砖混结构要弄清圈梁有几种截面高度，具体分布在墙体的那些部位，圈梁在阳台及门窗洞口处截面有何变化，内外墙圈梁宽度是否一致，以便在计算圈梁体积时，按不同宽度进行分段计算。 5、带有挑檐、阳台、雨篷的建筑物，要弄清悬挑构件与相交的连梁或圈梁的连结关系，以便在计算时做到心中有数。 目前施工图预算和工程量清单的编制主要是围绕工程招投标进行的，工程发标后按照惯例，建设单位一般在三天以内要组织有关方面对图纸进行答凝，因此，预算（或清单）编制人员在此阶段应抓紧时间看图，对图纸中存在的问题作好记录整理。在看图过程中不要急于计算，避免盲目计算后又有所变化造成来回调整。但是对“门窗表”、“构件索引表”、“钢筋明细表”中的构件以及钢筋的规格型号、数量、尺寸，要进行复核，待图纸答凝后，根据“图纸答凝纪要”对图纸进行全面修正，然后再进行计算。 计算工程量时，图中有些部位的尺寸和标高不清楚的地方，应该用建筑图和结构图对照着看，比如装饰工程在计算天棚抹灰时，要计算梁侧的抹灰面积，由于建筑图中不标注梁的截面尺寸，因此，要对照结构图中梁的节点大样计算。再如计算框架间砌体时，要扣除墙体上部的梁高度，其方法是按结构图中的梁编号，查出大样图的梁截面尺寸，标注在梁所在轴线的墙体部位上，然后进行计算。 从事概预算工作时间不长,而又渴望提高看图技能的初学人员，在必要时应根据工程的施工进度，分阶段深入现场了解情况，用图纸与各分项工程实体相对照，以便加深对图纸的理解，扩展空间思维,从而快速提高看图技能。 二、熟悉常用标准图做法 在工程量计算过程中，时常需要查阅各种标准图集，实在繁琐，如果能把常用标准图中的一些常用节点及做法，留在记忆里，在工程量计算时，不需要查阅图集就知道其工程内容和做法，这将节省不少时间，从而可以大大提高工作效率。 工程中常用标准图集基本上为各省编制的民用建筑及结构标准图集，而国标图集以采用

快速计算方法

快速计算方法？ 1.十几乘十几口诀：头乘头，尾加尾，尾乘尾。例：12×14=？解: 1×1=1 2＋4＝6 2×4＝8 12×14=168 注：个位相乘，不够两位数要用0占位。 2.头相同，尾互补(尾相加等于1 0)：口诀：一个头加1后，头乘头，尾乘尾。例：23×27=？解：2＋1＝3 2×3＝6 3×7＝21 23×27=621 注：个位相乘，不够两位数要用0占位。 3.第一个乘数互补，另一个乘数数字相同：口诀：一个头加1后，头乘头，尾乘尾。例：37×44=？解：3+1=4 4×4=16 7×4=28 37×44=1628 注：个位相乘，不够两位数要用0占位。 4.几十一乘几十一：口诀：头乘头，头加头，尾乘尾。例：21×41=？解：2×4=8 2+4=6 1×1=1 21×41=861 5.11乘任意数：口诀：首尾不动下落，中间之和下拉。例：11×23125=？解：2+3=5 3+1=4 1+2 =3 2+5=7 2和5分别在首尾11×23125=254375 注：和满十要进一。 6.十几乘任意数： 口诀：第二乘数首位不动向下落，第一因数的个位乘以第二因数后面每一个数字，加下一位数，再向下落。例：13×326=？解：13个位是3 3×3+2=11 3×2+6=12 3×6=18 13×326=42 38 注：和满十要进一。 快速计算方法？ 数学快速计算方法 第一讲加法速算 一.凑整加法 凑整加法就是凑整加差法,先凑成整数后加差数,就能算的快。8+7=15 计算时先将8凑成10 8加2等于 10 7减2等于5 10＋5=15

如17＋9=26 计算程序是17+3=20 9-3=6 20+6=26 二 .补数加法 补数加法速度快,主要是没有逐位进位的麻烦。补数就是两个数的和为10 100 1000 等等。8+2=10 78+22=100 8是2的补数，2也是8的补数，78是22的补数，22也是78的补数。利用补数进行加法计算的方法是十位加1，个位减补。例如6+8=14 计算时在6的十位加上1，变成16，再从16中减去8的 补数2就得14 如6+7=13 先6+10=16 后16-3=13 如27+8=35 27+10=37 37-2=35 如25+85=110 25+100=125 125-15=110 如867+898=1765 867+1000=1867 1867-102=1765 三.调换位置的加法 两个十位数互换位置，有速算方法是：十位加个位，和是一位和是双，和是两位相加排中央。例如61＋16

工程量快速计算方法

工程量快速计算方法 工程量是施工企业编制工程形象进度统计报表，向工程建设投资方结算工程价款的重要依据。今天我们总结了几方面工程量估算的便捷方法，一起来看吧。 平整场地 计算规则： 1、清单规则：按设计图示尺寸以建筑物首层面积计算。 2、定额规则：按设计图示尺寸以建筑物首层面积计算。 计算方法： 1、清单规则的平整场地面积：清单规则的平整场地面积＝首层建筑面积。 2、定额规则的平整场地面积：定额规则的平整场地面积＝首层建筑面积。 注意事项：

1、有的地区定额规则的平整场地面积：按外墙外皮线外放2m计算。计算时按外墙外边线外放2m的图形分块计算，然后与底层建筑面积合并计算；或者按“外放2m 的中心线×2=外放2m面积”与底层建筑面积合并计算。?为什么夫妻“雲雨”，女性很难达到“癫峰”？与这几点有关这样的话计算时会出现如下难点： 1）划分块比较麻烦，弧线部分不好处理，容易出现误差； 2）2m的中心线计算起来较麻烦，不好计算； 3）外放2m后可能出现重叠部分，到底应该扣除多少不好计算。 2、清单环境下投标人报价时候可能需要根据现场的实际情况计算平整场地的工程量，每边外放的长度不一样。 开挖土方 计算规则： 1、清单规则：挖基础土方按设计图示尺寸以基础垫层底面积乘挖土深度计算。

2、定额规则：人工或机械挖土方的体积应按槽底面积乘以挖土深度计算。槽底面积应以槽底的长乘以槽底的宽，槽底长和宽是指混凝土垫层外边线加工作面，如有排水沟应算至排水沟外边线。排水沟的体积应纳入总土方量内。当需要放坡时，应将放坡的土方量合并于总土方量中。 计算方法： 1、清单规则： 1）计算挖土方底面积： 方法一：利用底层的建筑面积+外墙外皮到垫层外皮的面积。外墙外边线到垫层外边线的面积计算（按外墙外边线外放图形分块计算或者按“外放图形的中心线×外放长度”计算）。 方法二：分块计算垫层外边线的面积（同分块计算建筑面积）。 2）计算挖土方的体积： 土方体积＝挖土方的底面积×挖土深度。 2、定额规则： 利用棱台体积公式计算挖土方的上下底面积。 V=1/6×H×(S上+ 4×S中+ S下)计算土方体积（其中，S上为上底面积，S中为中截面面积，S下为下底面面积）。

快速算法大全

内部函授教材（全套二十六讲） 第一讲：1、十几乘十几速算法——将前边的数加后边尾数，然后两个尾数再相乘。（注：满10进1）。 例：12×14=（12+4）连接（2×4）=168。2、十几乘几十几一将被乘数的个位乘以 乘数的十位，再加到乘数、最后加上它们的个位乘积。例：14×72=[（4×7）+72]连 接（4×2）=1008。3、一百零几乘一百零几一将一个数加上另一个数的个位数，最后加上它们个位数乘积。例：104×108=(104+8)连接(4×8)=11232。4、如果十位相同，个位之和为10的两个两位数相乘，其速算法一将十位加上1后再乘以十位，最后加上它们个位乘积。例：63×67=(6+1) ×6连接（3×7）=4221。5、十位数相同，个位不同且之和不等10的两个两位数相乘，只要将其中一个数加上另一个数的个位数，并乘以十位，最后加上它们个位乘积。例：63×69=(63+9) ×6连接（3×9）=4347。6、 一百零几乘几十几，方法是一将一百零几分成两段计算，将1乘以乘数，然后又用零 几乘以乘数（注：满10进1）。例：102×24=2448。说明：1×24=24。02×24=48。这时，只需将两段之乘积加以排列即是2448。 第二讲：求九十几与九十几的积。 方法：用一个数减去另一个数的补数，在差的后接着写两个数的补数积，如果补数积不满10，就在它前面添一个“0”此数就是得数。例：97×96=(97－4)连接(3× 4)=9312 第三讲：求两个九百九十几的数的积。 方法：在一个数减去另一个数的补数的差的后面，添一个“0”，再添上两个数的补 数的积。如果补数积不满10，就在它前面再添一个“0”，此数就是得数。例：994×992=(994－8) ×1000+8×6=986048 (994的补数6，992补数8) 第四讲：求两个连续数的积。方法：用较小数的平方加上较小数，或用较大数的平方减去较大数，皆可。 例1：35×36=1225+35=1260(1225即352) 例2：49×50=2500－50=2450（2500即502） 第五讲：求首差一，尾合十的两个两位数之积。方法：用较大数的十位数的平方减去 1，在差的后面添上较大的个位数的平方对于100的补数，所得的数就是积。 例1：42×38(42－1) ×100(100－22)=1596 例2：57×63=(62－1) ×100(100－32)=3591

Mass Balance Calculation

Mass Balance Calculation 1. Measured Items: ●Crude Ore ●Waste ?Wet End Waste ?Dry End Waste ?Unpacked Filler ?Leaking ●Products ?Filter Aid ?Filler ?Naturals 2. Methods: ●Crude Ore Ask load truck operators to count the number of loads during his shift. Multiply this number and the average dry weight of each load that we measured in the past. Comments: there are a lot of uncertainties going on. Operators could forget putting down the number or to make the number of loads higher, namely better, he could lift less material in each load. The measured average dry weight of each load varies from shift to shift and depends on what combination of crude ores is being used to make product. Improvement: we need to make the operators accountable for these numbers. Award and punishment system can be implemented. (Short-term). There is no metrics of mass flow in the system. To eliminate human errors and have a better monitoring system, a weight belt is necessary and doable at this point. The data can be retrieved and treated due to the presence of PLC that is used on the spot. Similarly, we are tracking diesel usage every day by looking at the data history. (Long-term) ●Waste ?Wet end waste, dry end waste, unpacked filler These components are tracked every day on production report. Basically, they use the number of loads or bags multiplied by the average dry weight per unit. Similar problems could occur at any time. But because the variance is a small amount, we just need to make the production leader accountable for these numbers. ?Leaking It is happening all the time. Pretty hard to measure and quantify, we may be able to calculate it if we have good measurement on crude ore.

工程量快速计算的基本方法75434

工程量快速计算的基本方法 工程量快速计算的基本方法包练好“三个基本功”；合理安排工程量计算顺序；灵活运用“统筹法”计算原理；充分利用“工程量计算手册”等四项内容。在实际工作中，只要能够熟练掌握，充分利用以上“基本方法”，就可以快速提高工程量计算业务水平。 第一节：练好“三个基本功” 练好“三个基本功”包括：提高看图技能；熟悉常用标准图做法；熟悉工程量计算规则，等三个方面。 一、提高看图技能工程量计算前的看图，要先从头到尾浏览整套图纸，待对其设计意图大概了解后，再选择重点详细看图。在看图过程中要着重弄清以下几个问题： （一）建筑图部分 1、了解建筑物的层数和高度（包括层高和总高）、室内外高差、结构形式、纵向总长及跨度等。 2、了解工程的用料及作法，包括楼地面、屋面、门窗、墙柱面装饰的用料及法。 3、了解建筑物的墙厚、楼地面面层、门窗、天棚、内墙饰面等在不同的楼层上有无变化（包括材料做法、尺寸、数量等变化），以便采用不同的计算方法。 （二）结构图部分 1、了解基础形式、深度、土壤类别、开挖方式（按施工方案确定）以及基础、墙体的材料及做法。 2、了解结构设计说明中涉及工程量计算的相关内容，包括砌筑砂浆类别、强度等级，现浇和预制构件的混凝土强度等级、钢筋的锚固和搭接规定等，以便全面领会图纸的设计意图，避免重算或漏算。 3、了解构件的平面布置及节点图的索引位置，以免在计算时乱翻图纸查找，浪费时间。 4、砖混结构要弄清圈梁有几种截面高度，具体分布在墙体的那些部位，圈梁在阳台及门窗洞口处截面有何变化，内外墙圈梁宽度是否一致，以便在计算圈梁体积时，按不同宽度进行分段计算。 5、带有挑檐、阳台、雨篷的建筑物，要弄清悬挑构件与相交的连梁或圈梁的连结关系，以便在计算时做到心中有数。 目前施工图预算和工程量清单的编制主要是围绕工程招投标进行的，工程发标后按照惯例，建设单位一般在三天以内要组织有关方面对图纸进行答凝，因此，预算（或清单）编制人员在此

梁的剪力方程和弯矩方程 常用弯矩图

5-7．试列出下列梁的剪力方程和弯矩方程，并画出剪力图和弯矩图。 解：首先求出支座反力。考虑梁的整体平衡 由 0,0=+?=∑e RA B M l F M 得 l M F e RA - = 由 0,0=-?=∑e RB A M l F M 得 l M F e RB = 则距左端为x 的任一横截面上的剪力和 剪力图 弯矩表达式为： ()l M F x F e RA S - == ()x l M x F x M e RA ?- =?= 剪力方程为常数，表明剪图应是一条平行梁轴线的直线；弯矩方程是x 的一次函数，表明弯矩图是一条斜直线。( 如图) 解：首先求出支座反力。考虑梁的平衡 由 04 5 2,0=??-?=∑l l q l F M RB c 得 ql F RB 8 5= 由 02 1 ,02=+?=∑ql l F M RC B 得 ql F RC 2 1 -= 则相应的剪力方程和弯矩方程为： AB 段：（2 01l x ≤≤） ()()21 11 12 1qx x M qx x F S -=-= BC 段：（ 2 322l x l ≤≤） 剪力图 弯矩图

()()? ?? ?? -?+??? ??-??-==-= 285428 21852222l x ql l x l q x M ql ql ql x F S AB 段剪力方程为x 1的一次函数，弯矩方程为x 1的二次函数，因此AB 段的剪力图 为斜直线，弯矩图为二次抛物线；BC 段剪力方程为常数，弯矩方程为x 2的一次函数，所以BC 段剪力图为平行梁轴线的水平线段，弯矩图为斜直线。（如图） 5-9 用简便方法画下列各梁的剪力图和弯矩图。 解：由梁的平衡求出支座反力： KN F KN F RB RA 12,8== AB 段作用有均布荷载，所以 AB 段的剪力图为下倾直线，弯矩图为下凹二次抛物线；BC 段没有荷载作用，所以BC 段的剪力图为平行梁轴线的水平线段，弯矩图为直线。 在B 支座处，剪力图有突变，突变值大小等于集中力（支座反力F RB ）的大小；弯矩图有转折，转折方向与集中力方向一致。（如图） （5） 解：由梁的平衡求出支座反力： KN F KN F RB RA 5.6,5.3== AB 与BC 段没有外载作用，所以AB 、BC 段的剪力图为平行梁轴线的水平线段，弯矩图为直线；CD 段作用均布荷载，所以CD 段的剪力图为下倾直线，弯矩图为下凹二次抛物线。

UFI统计标准和定义 ufi_calculation_standards_definitions

UFI CALCULATION STANDARDS and DEFINITIONS The figures requested for an UFI approved event, as mentioned in article 3 of the UFI Internal Rules, will be counted and audited according to the following definitions and rules. A. Calculation Standard for the Surface Area of an Exhibition For an Organizer, the figure to be certified and provided is the "total net exhibition space", defined as follows: total floor space - indoors and outdoors - occupied by exhibitors. This is also called “contracted space”, and may include both paid and unpaid space. It also includes space allocated to special shows having a direct relation to the theme of the exhibition. For an Exhibition Centre operator, the figure to be provided is the "total gross exhibition space". This is the total space provided by the venue operator for use by the organizers or, the total space used by the fair, including circulation. Catering areas, offices, storage, etc. are excluded. When exhibition space figures are communicated, they must always be specified as “total net” or “total gross”. B. Calculation Standard for the Number of Exhibitors B.1. Exhibitors (“direct” exhibitors) Only the exhibitors (“direct” exhibitors) will be counted. Considered as such are both the main exhibitors and the co-exhibitors. The main exhibitors are those bodies contracting directly with the organizer. The co-exhibitors are those organizations/companies present on a main exhibitor's stand, with their own staff and their own products and/or services. They must be clearly identified via several means, e.g. mentioned on the application form of the main exhibitor or declared by an official co-ordinating body, or in the exhibition catalogue forms. In the case of a collective participation, the space must be rented and paid for by the exhibitor organising the collective participation. The area is shared by several participants who are considered to be co-exhibitors if they occupy their own area, appear under their own name and present their own products/services by their own staff. If each of these conditions is not fulfilled, they are considered as “represented companies” (“indirect” exhibitors), and may not be counted in the exhibitor tally. In any communication with reference to the UFI standard, or to the UFI approval of an event, only the figures related to direct exhibitors may be used. B.2. Represented companies (“indirect” exhibitors) Represented companies are those organizations/companies not present with their own staff, and whose products or services are present on a main exhibitor's or co-exhibitor's stand. These represented companies are excluded from the calculation of the total number of exhibitors. B.3 To avoid any confusion, it must be clearly mentioned which category of exhibitors were counted.

Seal or Reseal Design Calculation Sheet

RTA Form 395K September 2006 Roads and Traffic Authority, NSW SEAL OR RESEAL DESIGN CALCULATION SHEET (FOR CONVENTIONAL, EMULSION, POLYMER BINDER and GEOTEXTILE TREATMENTS) T y p e o f T r e a t m e n t Seal (S) or Reseal (RS) Single / Single (S/S) Single / Double (S/D) Double / Double (D/D) Geotextile Reinforced Seal (GRS)High Strength Seal (HSS) Strain Alleviating Membrane (SAM) or Strain Alleviating Membrane Interlayer (SAMI)Primersealed (PS), Sealed (S) Existing Aggregate Size (nominal) Primed (P) Asphalt (AC), Slurry Surfacing (SS) Cement Concrete (CC)Timber Bridge Deck (TD) Unit Shoulder Lane 1 Lane 2 Lane 3 Lane 4 Surface Texture mm Ball Penetration Depth of Prime or Primerseal (required for seals only)mm E x i s t i n g S u r f a c e C o n d i t i o n s Age of Surface Years Aggregate Design Unit 1st layer 2nd layer Nominal Size -mm Crushed (C), Partly Crushed (PC) or Rounded (R)--Compatible with existing seal (size checked)-yes/no Average Least Dimension ALD ALD mm ALD 1 = ALD 2 = Basic Aggregate Spread Rate (Table 1A/1B)F m 2/m 3 Factors for Spread Rate (Table 2)I -Design Aggregate Spread Rate = F x I H m 2/m 3 Geotextile Type - eg Polyester or Polypropylene M a t e r i a l s f o r S e a l o r R e s e a l Binder Type/Class Road Number/Name:Location: J o b D e t a i l s Length Width Area Number of Lanes Roadloc: to Date: Job/Order Number: Office: Segment Number: km to km from towards m m m 2 mm File

Different Methods of Depreciation Calculation

Different Methods of Depreciation Calculation Depreciation Calculation Methods Various depreciation calculation methods are mentioned below: i. Base Method ii. Declining Balance Method iii. Maximum Amount Method iv. Multi Level Method v. Period Control Method i. Base Method Base Method- SPRO> IMG> Financial Accounting (New)> Asset Accounting>Depreciation> Valuation Methods> Depreciation Key> Calculation Methods>Define Base Methods Base method primarily specifies: ?The Type of depreciation (Ordinary/ Special Depreciation) ?Depreciation Method used (Straight Line/ Written Down value Method) ?Treatment of the depreciation at the end of Planned useful life of asset or when the Net Book value of asset is zero (Explained in detail later in other related transactions ). Straight Line Method (SLM) ?This is the simple method of depreciation. ?It charges equal amount of depreciation each year over useful life of asset. ?It first add up all the costs incurred to bring the asset in use and then it divides that by the useful life of asset in years to calculate the depreciation expense. ? E.g.: Say a Computer costs Rs. 30,000 and Rs. 11,000 (as additional set-up/installation/maintenance expenses) = Rs 41,000 and it is anticipated that its scrap value will be Rs. 1,000 at the end of its useful life, of say, 5 yrs. Total Cost = Cost of Computer + Installation Exp. + Other Direct Costs Depreciable Amount over No. of years = Total Cost - Salvage Value (At end of useful life) 30,000 +11,000 =41,000 (Total cost) 41,000 – 1,000 = 40,000 as the Depreciable Amount Depreciable Amount = Rs. 40,000, Spread out over 5 years = Rs. 40,000/5(Yrs) = Rs. 8000/- depreciation per annum. Written Down Value Method (WDV) ?This method involves applying the depreciation rate on the Net Book Value (NBV) of asset. In this method, depreciation of the asset is done at a constant rate. ?In this method depreciation charges reduces each successive period.