de cuong cad-cam

8/3/2019 De Cuong Cad-cam

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Cu 1: Trnh by khi nim v CAD/CAM?

CAD/CAM l lnh vc nghin cu nhm to ra cc h thngt ng thit k v ch to trong my tnh in t c s dng thc hin mt s chc nng nht nh.

CAD/CAM to ra mi quan h mt thit gia hai dng hot ng:Thit k v Ch to.

-T ng ho thit k l dng cc h thng v phng tin tnh tongip ngi k s thit k, m phng, phn tch v ti u ho cc gii

php thit k.

-T ng ho ch to l dng my tnh in t k hoch ho,

iu khin v kim tra cc nguyn cng gia cng.- CAD: (Computer Aided Deisgn ) l cng ngh lin quan n

vic s dng h thng my tnh gip vic to, sa i, phn tchv ti u ha thit k.

+ Vai tr trong thi k:- i vi pha khi nim ha thit k cc cng c CAD cn c l cc

k thut m hnh ha hnh hc, cc h tr ha v cc thao tc ha.- Pha m hnh ha v m phng thit k cn cc cng c k trn,

cng c m phng chuyn ng, lp rp v mt s gi m hnh hakhc.

- Pha phn tch thit k cn cc gi v phn tch, cc gi v ccchng trnh ty bin.

- Pha thit k ti u cn cc ng dng ty bin v ti u ha ktcu.

- Pha nh gi thit k cn cc cng c v dung sai, kch thc vbng cc vt liu.

- Pha to ti liu v truyn t thng tin thit k cn cng c tobn v k thut v cng c to nh t bng.- CAM: CAM ( Computer Aided Manufacturing ) l cng ngh linquan vi vic s dng h thng my tnh lp k hoch, qun l viu khin cc qu trnh ch to.

Cu 2: Trnh by cc chc nng ca h thng CAD/CAM?

Khc bit c bn vi qui trnh thit k theo cng nghtruyn thng, CAD cho php qun l i tng thit k di dng

m hnh hnh hc s trong c s d liu trung tm, do vy CAD ckhnng h tr cc chc nng k thut ngay t giai on pht trin sn


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phm cho n giai on cui ca qu trnh sn xut, tc l h triukhin cc thit b sn xut bng iu khin s.

H thng CAD c nh gi c kh nng thc hin chc

nng yu cu hay khng, ph thuc ch yu vo chc nng x lca cc phn mm thit k.

Ngy nay nhng b phn mm CAD/CAM chuyn nghip phcv thit k v gia cng khun mu c kh nng thc hin c cc chcnng c bn sau:

- Thit k m phng hnh hc 3 chiu (3D) nhng hnh dng phctp.

- Giao tip vi cc thit b o, qut to 3D thc hin nhanhchng cc chc nng m phng hnh hc t d liu s.

- Phn tch v lin kt d liu: to mt phn khun, tch khun,qun l kt cu lp ghp...

- To bn v v ghi kch thc t ng: c kh nng lin kt ccbn v 2D vi m hnh 3D v ngc li.

- Lin kt vi cc chng trnh tnh ton thc hin cc chc nng

phn tch k thut: tnh bin dng khun, m phng dng chy vt liu,

trng p sut, trng nhit , co rt vt liu,...

- Ni suy hnh hc, bin dch cc kiu ng chy dao chnh xccho cng ngh gia cng iu khin s.

- Giao tip d liu theo cc nh dng ho chun.

- Xut d liu ho 3D di dng tp tin STL giao tip vi

cc thit b to mu nhanh theo cng ngh to hnh lp th.

Cu 3: Trnh by ng dng ca h thng CAD/CAM?

- To mu nhanh thng qua giao tip d liu vi thit b to

mu nhanh theo cng ngh to hnh lp th (o qut to ).

- Gim ng k thi gian m phng hnh hc bng cch to m

hnh hnh hc theo cu trc mt cong t d liu s.

- Chc nng m phng hnh hc mnh, c kh nng m t


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nhng hnh dng phc tp nht.

- Kh nng m hnh ho cao cho cc phng php phn tch,

cho php la chn gii php k thut ti u.

Cu 4: Trnh by qu trnh thit k v gia cng to hnh theo cngngh truyn thng?

Trong cng ngh truyn thng, cc mt cong 3D phc tp c gia cngtrn my vn nng theo phng php chp hnh s dng mu hoc dng.Do vy qui trnh thit k v gia cng bao gm c 4 giai oan phn bit:

1. To mu sn phm,

2. Lp bn v k thut,3. To mu chp hnh,

4. Gia cng chp hnh.

Qui trnh ny c nhng hn ch:

- Kh t c chnh xc gia cng, ch yu do qu trnh chp hnh,


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- D dng lm sai do nhm ln hay hiu sai v phi x l mt s ln d

liu,

- Nng sut thp do mu c thit k theo phng php th cng v

qui trnh c thc hin tun t: to mu sn phm - lp bn v chi tit

- to mu chp hnh - phay chp hnh.

Cu 5:Trnh by qu trnh thit k v gia cng to hnh theo cngngh CAD/CAM?

S pht trin ca phng php m hnh ho hnh hc cng vi thanhtu ca cng ngh thng tin, cng ngh in t, k thut iu khin s c

nhng nh hng trc tip n cng ngh thit k v gia cng to hnh:

- Bn v k thut c to t h thng v v to bn v vi s tr gip

ca my vi tnh.

- To mu th cng c thay th bng m hnh ho hnh hc trc tip

t gi tr ly mu 3D.

- Mu chp hnh c thay th bng m hnh ton hc - m hnh hnh

hc lu tr trong b nh my vi tnh v nh x trn mn hnh di


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dng m hnh khung li.

- Gia cng chp hnh c thay th bng gia cng iu khin s (CAM).

V cng ngh, khc bit c bn gia gia cng to hnh theo cng ngh

truyn thng v cng ngh CAD/CAM l thay th to hnh theo mu bngm hnh ho hnh hc.

Kt qu l mu chp hnh v cng ngh gia cng chp hnh c thay thbng m hnh hnh hc s (Computational Geometric Model - CGM) vgia cng iu khin s. Mt khc kh nng kim tra kch thc trc tip vkh nng la chn ch gia cng thch hp (gia cng th, bn tinh vtinh).

Theo cng ngh CAD/CAM phn ln cc kh khn ca qu trnh thitk v gia cng to hnh theo cng ngh truyn thng c khc phc vrng:

B mt gia cng t c chnh xc v tinh xo hn.

Kh nng nhm ln do ch quan b hn ch ng k.

Gim c nhiu tng thi gian thc hin qui trnh thit k v gia

cng to hnh.

Cu 6: Trnh by cu to chung ca h thng phn cng ca h thngCAD/CAM?

Mt h CAD ni chung thng c cc thit b phn cng sau y:

1. Mt hoc mt s trm thit k vi mt u cui ho v cc thit bvo ca ngi thit k.

2. Mt hoc mt s my v v cc thit b ra khc.

3. Mt my tnh.

4. Cc b lu tr ngoi (b nh ngoi).


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CPU (Central Processing Unit), thit b nhp dliu, thit b xut d liu, thit b nh d liu l ccphn cng c bn trong cc h CAD/CAM. Mt my tnhc nhn PC (personal computer) thng thng (nh hnh1.2) vi cc thit b ngoi vi ti thiu bao gm bn phmchut, mn hnh v mt a mn c th mt s phn mn CAD/CAM lm vic, nhng rt kh thchin c vic thit k v ch to cho cc d n ln, phctp. tng nng sut v cht lng thit k ngi ta tora nhiu phn cng m rng cho CAD/CAM. V d nh:

Thit b nhp liu ph bin nht l con chut

v bn phm, nhng cc thit b ny khng kh nng nhp thng tin ca cc i tng thit k phc tp nhm hnh ca mt pho tng. nhp thng tin cho cc itng phc tp nh vy phi s dng cc thit b kt ni vimy tnh nh my o to 3D, my qut lade ( scanrlaser) 4D.

s dng cho cc nh thit k chuyn nghip,

ngi ta ch to ra cc bng s ho trn. l ton b


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cc chc nng ca phn mn v vng ho ghp ni viphn mn CAD/CAM gi l tablet.

Cng cn lu rng trong phn cng ca h

CAD/CAM khng bao gm cc my cng c. V kt thcca giai on CAM l kt sut ra cc chng trnh giacng NC.

H CAD c loi c lp trong ch c mt my tnh c nhn iu khin,c loi ni mng cuc b vi nhiu trm thit k do mt my trung tm iukhin.

Cu 7: Trnh by c im ca h thng trn c s my tnh ln, my tnhnh,my vi tnh v trm cng tc?

Cu 8: Nu cc nguyn tc c bn cho vic thit k phn mm ha?

Cc nguyn tc c bn cho vic thit k phn mm ho nh sau :

1. Tnh n gin : Phn mm ho phi d s dng.


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2. Tnh nht qun : Go phn mm phi em n cho ngi s dng mtphng php vn hnh nht qun v c th d on c cc thao tc tip theo.

3. Tnh ng b : Khng b st mt chc nng no ng k trong b chcnng ho ca phn mm.

4. Tnh bn vng : Chu ng c nhng sai phm nh trong vn hnh cangi s dng.

5. Tnh hiu qu : Vi nhng hn ch ca phn cng, phn mm cn khai thcc n mc ti a kh nng ca n.

6. Tnh kinh t : Phn mm khng qu ln hoc qu t tin n mc khchhng khng chp nhn c.

Cu 9: Trnh by chc nng ca phn mm h thng?

Cu 10: Trnh by cc chc nng ca gi phn mm ha?

1/ To sinh cc yu t ho

Trong ha, thut ng Entity dng ch mt yu t ha hai chiunh im, on thng, ng trn, v.v... Bn cnh cn phi k n b ch

ci, b ch s cng cc k hiu c bit khc, to nn b yu t ha ctrong h thng. Thng c mt b phn phn cng c bit (mt chip c cythm vo card mn hnh) nhm tng tc ca qutrnh to sinh cc yu t ha v cho php ngi s dng kin to ra m hnh ng dng t nhng yu tkhng thuc vo cc yu t c sn trong h thng.

Thut ng Primitive (nguyn th) thng c dng ch mt yu t ho ba chiu nh hnh cu, hnh tr hay lp phng. Trong m hnh khung dy

ba chiu v m hnh c, cc nguyn th c s dng vi t cch l nhng

khi c s xy dng nn mt m hnh ba chiu v i tng c th mngi ta s dng ang quan tm.

2/ Cc php chuyn i:

Cc php chuyn i c dng thay i hnh nh trn mn hnh v nh v li cc yu t trong c s d liu. l cng c gip cho ngi sdng xy dn m hnh ng dng theo mun. Cc php chuyn i ni y

bao gm php dch chuyn (ti nh v), php quay, php t l (phng to hocthu nh)...

3/ iu khin hin th v cc chc nng ca s


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Nhm chc nng ny gip ngi s dng quan st hnh nh t mt gc mong mun v vi mt t l yu cu. lm c nh vy, cc phpchuyn i c vn dng hin th m hnh ng dng theo cch mngi s dng mong mun. Cng vic ny i khi c gi l php Window

v mn hnh ha c quan nim nh mt khung ca s qua ngi sdng nhn vo m hnh. Ch rng ca s c th t bt c u nhn voi tng ang xt.

Bn cnh , kh b nt khut cng thuc nhm chc nng ny. Trong as h ha, hnh nh c to nn bi nhng ng nt th hin mt itng c th. Kh b nt khut l mt th tc m nh hnh nh c chia rathnh nhng nt thy c v khng thy c (hay nhng nt khut). mts h ho, ngi thit k phi ch ra nt no (hay phn no ca nt)

khngthy c kh b khin cho hnh v d hiu hn. mt s h khc,gi phn mm mnh t ng kh b nt khut.

4/ Chc nng phn on : Cc chc nng thuc nhm ny cho php ngi sdng thay th, xo c la chn hoc sa cha nhng phn hnh nh mongmun. Thut ng on dng ch mt phn c th ca bn v cn chn ra sa cha. Php phn on l php chia mt i tng phc tp thnh nhng

b phn n gin hn c th qun l c.

Mi b phn nh th c gi l mt on. on c th l mt phn tn nh on thng, ng trn, hoc cng c th l mt nhm phn t nnhng v mt logic th vn c xem nh mt phn t n m ta c th thaotc trn .

5/ Cc chc nng Input ca ngi s dng: y l nhm cc k quan trngtrong gi phn mm ho v chng cho php ngi s dng a lnh v dliu vo cho h thng. Nhng cng vic ny c thc hin nh cc thit bInput. Cc chc nng Input tt nhin phi c vit sao cho ph hp vi ccthit b vo c th c trong h thng, ng thi phi d nh, d lm quen v

tin li cho ngi s dng.

Cc chc nng Input cn c vit sao cho c th khai thc n mc ti ali th tng tc ca h ICG. Mt bn thit k phn mm tt va c michc nng p ng mi tnh hung vo d liu, li va khng c qu nhiulnh khin ngi s dng ngp chm trong . Tuy nhin iu ny rt kh tc i vi nhng gi phn mm ln v phc tp.

Mt trong nhng hng gii quyt ch yu l n gin ho s giao din

ngi - my n mc mt ngi s dng khng hiu bit nhiu v my tnhcng c th khai thc phn mm mt cch hiu qu cho mc ch chuyn mn


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ca mnh. Cc phn mm ha chy trong DOS gn y v c bit l chytrong Windows l nhng v d thnh cng theo hng y.

Cu17: Trnh by c s i s vect trong khng gian 3D?

Cu 18: Trnh by cc biu din v tch cht ca m hnh ha hnh hcng cong?

1 Biu din ng cong.

V ton hc, ng cong c th dc biu din di cc dng:

- Phng trnh n.

- Phng trnh tng minh.

- Phng trnh tham s.

Xt ng trn n v trn mt phng (x - y), c tm trng vi gc h to trn hnh 2.1. Mi quan h gia cc to x v y c m t bi phng trnh:

( ) 01, 22 =+= yxyxf : Phng trnh n (2.1)

Nu ch xt phn na trn ca ng trn, phng trnh biu din l:

( ) ( )2

1

1xxgy

== : Phng trnh tng minh (2.2)

Nu t gc gia on thng PO v trc x l tham s ca ng trn, ta c:

cos)( ==xx ( ) sin==yy : Phng trnh tham s (2.3)

Trng hp t gc to bi PQ v trc x l tham s, th ( )1+==

x

y

tagt

Kt hp vi phng trnh (2.1) ta c:


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y cng l phng trnh tham s ca ng trn v c gi l phng

trnhtham s a thc hu t. Qu trnh thit lp phng trnh tham s hu tca ng cong v mt cong t phng trnh a thc n c gi l tham sho.

Nn biu din ng cong 3D thch hp di dng phng trnh tham s:

hay di dng vect:

Theo dng phng trnh tham s, ng cong c nh ngha mt cch d

dng bng cch xc nh min gii hn ca tham s. Khng th xc nh ngcong 3D bi phng trnh n hay tng minh, bi v phng trnh ng(x,y,z)=0 biu din mt cong, do cn hai phng trnh xc nh ngcong 3D. Trong trng hp ny, ng cong c nh ngha nh giao tuyngia hai mt cong.

2 c tnh ca ng cong.

Trong phn ny biu din ng cong, ta s dng phng trnh tham s

chun tc:

c tnh c bn ca ng cong, bao gm:

a. chy ca ng cong.

b. Vect tip tuyn n v.

c. Vect php tuyn chnh.

d. cong v bn knh cong.

a. chy

ln ca vect o hm( )tr

.

c gi l chy ca ng cong:


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Hy tng tng ng cong l con ng v tham s t tng trng cho thi

gian. Nh vy, chy ca ng cong tng ng vi tc chy xe. ilng ny c s dng trong thut ton ni suy hnh hc theo phng phpqut hnh.

Nu t qung ng i c l tham s s, phng trnh ng cong dng r(s)tr thnh phng trnh tham s t nhin vi chy bng 1. chy cang cong khng phi l c tnh ring ca ng cong, l kt qu ca

php tham s ho.

b. Vect tip tuyn n v:

Cho s l tham s t nhin ca ng cong r(t), sao cho:

Vect tip tuyn n v ca ng cong r(t) c nh ngha nh sau:

(2.6)

hay di dng vi phn: (2.7)

c. Vect php tuyn chnh

Ly o hm vect tip tuyn n v T theo t v chun ho gi tr, chng ta cvect n v N, c gi l vect php tuyn chnh ca ng cong:

(2.8)

V T l vect n v (T.T=1), do

vect N vung gc vi vect T (Hnh 2.2).

Mt phng nh ngha bi vect T

v N c gi l mt phng mt tip. Vect

B vung gc vi vect N v T c gi l

vect php tuyn i xc nh bi quan h:


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B = TxN

d. cong v bn knh cong:

Hy cho s l tham s t hin v T l vect tip tuyn n v ca ng congr(t). cong c nh ngha nh sau:

(2.9)

hay di dng vi phn:

(2.10)

trong : .i vi ng cong 2D dng phngtrnh tng minh y = y(x), phng trnh trn c dng:

trong :

Hy xt ng trn trn mt phng mt tip (Hnh ve), i qua im hin thir(t) v cong ca n bng chnh cong ca ng cong ti im ny.ng trn ny c gi l ng trn mt tip, bn knh ca ng trn mttip c gi l bn knh cong v c xc nh bi:

(2.11)

e. xon ca ng cong:

xon ca ng cong 3D c nh ngha nh sau:

trong N l vect php tuyn chnh; B l vect php tuyn i. Phng trnhc bn m t c tnh ca ng cong 3D c gi l phng trnh Serret-Frenet:

(2.12)


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Cu 19: Trnh by cc biu din v tch cht ca m hnh ha hnh hcmt?

1 Phng php biu din mt cong:

a. M hnh mt cong cong dng phng trnh n.

Hy xt mt cu n v vi tm ti gc to cc. Cc im pha trong mtcu tho bt ng thc:

v phng trnh: 01222 =++ zyx (1)

biu din cc im thuc mt cu.

Xt mt cch tng qut, phng trnh n g(x,y,z) = 0 biu din mt cong giihn bi hai na khng gian g(x,y,z) > 0 v g(x,y,z) < 0.

b. M hnh mt cong dng phng trnh tham s.

Theo hnh hc vi phn, mt cong c nh ngha nh l nh ca php nh xchnh qui tp hp im trong khng gian 2D vo khng gian 3D v c biudin bi phng trnh:

(2)

trong : u v v l tham s ca mt cong.

i vi hnh cu n v, ta c th d dng tham s ho phng trnh (1) bngcch t tham s u l v tuyn v tham s v l kinh tuyn ca mt cu:

(3)

Tng t nh ng trn n v c th tham s ho phng trnh mt cu dihnh thc khc, bng cch s dng a thc hu t.

c. M hnh mt cong dng phng trnh phi tham s.

Khi min xc nh ca mt cong l mt phng x-y ca h to Descarte

(u x, vy ) m hnh tham s (2) tr thnh phi tham s:

(4)


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Nu ch xt bn cu trn ca mt cu n v th phng trnh (1) c biudin di dng tng minh:

(5)

Hnh hc mt cong c minh ho trn hnh 2.3. Ta thng gi phn

mt cong trong min tham s gii hn l mt li. Cc mt li lin kt theoiu kin kt ni lin tc to thnh mtcong phc hp.

2. Tnh cht.

a. cong.

Ma trn c s th hai.

Xt ng cong r(t) trn mt cong r(u,v) o hm bc hai ca r(t) theo t c gi

tr nh sau:

(6)

Thc hin php nhn v hng vi vect php tuyn n v n ca mt cong vich rng 0.. == nrnr vu , ta c:

(7a)

trong :


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cong php tuyn.

T phng trnh (2.12), o hm bc hai ca r(t) c tnh nh sau:

Thc hin php nhn v hng mt ln na vi vect n v ch ya rng:T.n = 0:

(7b)

Gi tr kN.n biu thc trn c gi l cong php tuyn kn. T cc cng

thc (7) v (3), ch rng..

rs= , cong php tuyn c xc dnh bi cng

thc sau:

(8)

ngha vt l ca cong php tuyn nh sau:

Ti im hin thi r(u(t),v(t)) trn mt cong r(u,v), dng mt phng i quavect tip tuyn n v T v vect php tuyn n v n ca mt cong. congca ng cong vi mt phng l cong php tuyn ca mt cong ti im

r(t) theo phng vect.

q

cong chnh.

cong php tuyn (8) l hm ca.

q :

Do c th tnh gi tr cc i ca cong php tuyn t biu thc:

(9)


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Gi tr cc i ca cong php tuyn c gi l cong chnh v c xcnh t (8) nh sau:

(10)

trong :

Vi eddhgg ,,,,, 2121 l cc s hng tng ng ca ma trn c s G vD

Tch gi tr hai cong chnh c gi l cong Gauss c s dng biudin trn lng ca mt cong.

Cu 20: Trnh by cc biu din v tch cht ca m hnh ha hnh khi?

Cu 21: Trnh by v phn tch cc dng m hnh ton hc ng congphng2D?

ng cong 2D c s dng nh cc i tng hnh hc c s trncc bn v k thut truyn thng m t hnh th 3D.

1. M hnh ng cong di dng phng trnh a thc n.

Phng trnh n g(x,y) = 0 biu din ng cong trn mt phng x-y, v dnh ng trn v ng thng c biu din bi phng trnh:

M hnh ny c u im:

- D dng xc nh vect tip tuyn v php tuyn,

- D dng xc nh v tr tng i gia im vi ng cong.

Phng trnh a thc bc 2 g(x,y) = 0 biu din h ng cong conic lgiao tuyn gia mt ct phng v mt nn tr. Tu theo v tr tng i giamt phng ct v mt nn, ng cong conic c th l:


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Nhc im chnh ca m hnh ng cong di dng phng trnh n l kh

thc hin hnh tun t, y l chc nng quan trng trong ho in ton.Do vy :

trong m hnh ho hnh hc, ng cong conic di dng phng trnh tham sc s dng ph bin hn c. Thc t m hnh dng phng trnh a thc nc bc cao hn 2 rt t c s dng.

2. M hnh ng cong di dng phng trnh a thc tng minh.

Phng trnh tng minh dng :

m t ng cong trn mt phng x-y. Nu f(x) l a thc bc 2, ng congl Parabol.

c tnh tiu biu ca a thc tng minh l c th chuyn i thnhphng trnh n hoc phng trnh tham s. Nu y = f(x), trong f(x) l athc ca x, tc l:

(1)

Do vy phng trnh a thc tng minh c u im ca phng trnh nv phng trnh tham s, l:

- D dng xc nh vect tip tuyn v php tuyn.

- D dng xc nh v tr tng quan gia im vi ng cong.


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- D dng thc hin hnh tun t.

Nhc im chnh ca dng phng trnh tng minh l khng th iukhin ng cong khp kn hoc ng thng ng. Dng phng trnh (3.1)cn c gi l dng phi tham s.

Cu 22: Trnh by v phn tch m hnh ton hc ng cong a thc ?

Kho st vic thit lp ng cong vi iu kin bin cho trc bao gm to

v tip tuyn ti 2 im u v cui: P0, P1, t0, t1. V rng ng cong cnh ngha bi 2 vect v tr v 2 vect tip tuyn c th biu din chng didng phng trnh a thc vect bc 3. a thc bc 3 c s dng rt ph

bin, bi v l bc ti thiu, dng cc loi ng cong trong khnggian 3D.

1. M hnh ng cong di dng phng trnh a thc chun tc.

c tnh ca m hnh a thc chun tc l d dng xc nh. Xt phngtrnh a thc vect bc 3:

C th biu din phng trnh a thc ny di dng ma trn theo vect c sU v vect h s A nh sau:

(1)

Phng trnh a thc bc 3 (1) khng th hin c ngha hnh hc, nhngcth c s dng thit lp ng cong trn lng i qua 4 im d liu { Pi: i=1,...,4} theo phng php sau:

t di l chiu di ct tuyn gia im Pi v 1+iP :


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T gi tr tham s ui ti cc im Pi c xc nh nh sau:

ng cong bc 3 (1) i qua cc im d liu phi tho iu kin:

Tng qut, ng cong a thc bc n i qua (n+1) im d liu c biu din

bi phng trnh a thc:

2. ng cong Ferguson.

Ferguson gii thiu mt

phng php khc s dng phng

trnh (3.2). Theo ng cong

c thit lp bi (Hnh 3.1):

a. Hai im u cui P0 v P1.

b. Tip tuyn u cui t0 v t1.

ng cong bc 3 (1) tho iu kin bin P0, P1, t0, t1 chng phi m bo:

(2)

Sau cc php bin i, h s PT a thc c xc nh theo biu thc:


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(3)

Kt hp biu thc (1) v (3), ng cong Ferguson r(u) theo iu kin bin

nh trn c biu din bi ma trn h s Ferguson C v vect iu kin binFerguson S nh sau:

(4)

Thc t d dng xc nh c ln ca vect tip tuyn, do ln ca

vect c chn bng chiu di ct tuyn 0110 PPtt == . S la chn nytho yu cu v hnh dng.

Phng trnh (1) v (4) u c biu din di dng ma trn c s. C th

biu din (4) di dng khc:

(5)

Trong :


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D dng xc nhn rng phng trnh (5) tho iu kin bin (2).

Phng trnh (5) l nh ngha chun v ng cong kt ni Hermite.

3. ng cong Bezier

ng cong Bezier c nh ngha bng nhiu phng php.

Hy xt phng php xy dng ng cong Bezier bc 3 t phng trnh

ng cong Ferguson (4).

Bn nh iu khin Bezier V0, V1, V2, V3 (hnh 3.2a) tho iu kin:

V0 l im u ca ng cong,

V1 l v tr 1/3 chiu di trn vect tip tuyn u,

V2 l v tr 2/3 chiu di trn vect tip tuyn cui,

V3 l im cui ca ng cong.

Cui cng ta thay th kt qu (3.7) vo phng trnh ng cong Ferguson


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(4) t c phng trnh ng cong Bezier bc 3 biu din bi ma trn hs Bezier M v vect nh iu khin R:

(8)

trong :

c tnh tiu biu ca ng cong Bezier l hnh dng ca ng cong phthuc vo a tuyn li gii hn bi cc nh iu khin ( Hnh 3.2) . Tng tnh ng cong Ferguson c th biu din ng cong Bezier (3.8) di dng

phng trnh a thc:

(9)

trong :

a thc Bernstein bc n c dng :

(10a)

a thc Bernstein c gi l hm c s Bezier s dng nh ngha ng

cong Bezier bc n bng cch kt ni (n+1) nh iu khin:


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(10b)

ng cong Bezier bc n tho iu kin bin sau:

(11)

nh ngha chun v ng cong Bezier theo hm c s Bezier (3.10b) thhin tnh cht hnh hc ca ng cong tt hn so vi biu din di dng matrn (3.8), v d nh c th chia nh hoc tng bc cho ng cong. Ngc midng ma trn c u im l d dng x l d liu.

4. ng cong B-spline u

M hnh ton hc ca ng cong B-spline l phng trnh i s. Ta snghin cu php dng hnh hiu r tnh cht hnh hc ca dng m hnhny.

Xt 4 nh iu khin V0,...,V3 v cc im M0, M1, P0, P1 vi tnh chtnh sau: (Hnh 3.3).

Nh vy ta c th biu din im bin P0, P1 v tip tuyn t0, t1 theo nh iukhin nh sau:


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hay di dng ma trn:

Thay kt qu trn vo phng trnh ng cong Ferguson (3.5) t cphng trnh ng cong B-spline u bc 3 biu din bi ma trn h s B-spline u N v vect nh iu khin R:

Cu 23: Trnh by v phn tch m hnh ton hc ng cong hu t ?


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Hm hu t c nh ngha nh l t s ca 2 hm a thc. ng cong hut c linh hot v hnh dng cao hn so vi cc dng ng cong a thcchun tc khc. ng cong hu t s c dng a thc chun tc nu nh c

biu din theo h to ng nht. Ta s kho st dng hu t ca m hnh

ng cong Bezier.

1. To ng nht.

Ta bit phng trnh tham s ca ng trn n v nh sau:

V mi tyhnh phn ca vect 3D trn c cng mu s, nn ta c th chuyn

chng thnh vect ng nht 4 thnh phn R(u) vi 3 thnh phn u tin ngvi t s v thnh phn th 4 ng vi mu s chung:

Vect R(u) c gi l vect ng nht v thnh phn ca chng tr thnh to ng nht ca im 3D (r(u)). Ta c th chuyn i (X, Y, Z, h) thnh (X/h,

Y/h, Z/h, 1) tc l thnh (x, y, z, 1). S chuyn i ny gi l s chun ho. ngha hnh hc ca s chun ho l vect 4D c chiu ln mt phng h = 1trong khng gian 4 chiu.

Nh vy vect ng nht (x, y, z, 1) v (hx. hy, hz, h) biu din cng mt im3D (x, y, z) nu h 0. Theo m hnh hu t mi dnh iu khin Vi(xi, yi, zi)c nh ngha nh nh iu khin ng nht:

trng s Wi lm tng tnh linh hot v hnh dng.

Biu din to cc di dng ng nht c s dng rng ri trongcc php bin i to ng dng trong ho cng nh Robot hc.

2. ng cong hu t bc 2.

Nu vect nh iu khin Bezier chun tc c thay th bi vect ngnht tong ng ta s t c ng cong Bezier hu t. ng cong Bezier

bc 2 c dng:


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Ta chuyn i ng cong Bezier bc 2 ny thnh dng hu t.S dng Hi biu din nh iu khin ng nht ca Vi, sao cho:

To cc ca nh iu khin ng nht Hi trn biu thc (3.25) tngng vi nh iu khin Vi, khng ph thuc vo trng s wi. Do vy ngcong Bezier hu t bc 2 c biu din di dng:

Ta s kho st hnh dnh ng cong ng nht ny. iu kin bin ca ngcong hu t c xc nh bng cch tnh biu thc (3.26) v o hm cachng ti u = 0 v u = 1.

Ly o hm phng trnh trn theo u ta c:

C th i vi ng cong Bezier bc 2 ta c:

Kt qu trn chng t rng tip tuyn ca ng cong Bezier ng nht (3.26)v ng cong Bezier chun tc (3.24) ti cc im bin c cng phng vinhau nhng ln ca chng thay i theo t l w1/w0 v w1/w2 (Hnh 3.6).


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Phng trnh Bezier ng nht (3.26) c biu din di dng thnh phnng nht nh sau:

V phng trnh Bezier c biu din ngn gn hn di dng hu t:

trong : Vi = (Xi, Yi, Zi) : nh iu khin Bezier, Wi : trng s.

Nh vy ta ch ra rng c th biu din ng cong Bezier hu t hoc didng ng nht (3.26) hoc di dng hu t (3.30) v ng cong Bezier hut bc 2 c chuyn i thnh ng cong chun tc khi Wi = 1 vi mi i.M hnh ng cong hu t bc 2 c s dng rt ph bin trong php thams ho ng cong mt ct cnic.

3. ng cong hu t bc 3.


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Ta c th d dng xc nh m hnh hu t cho ng cong Bezier v B-splinebc cao hn. ng cong Bezier bc 3 hu t c dng ng nht tng t nhng cong Bezier chun tc (3.7):

M hnh ng cong hu t c bc t do cao hn dng nh ngha

hnh dng. S dng cc gi tr trng s khc nhau c th iu khin hnh dngng cong hu t trong min gii hn bi a tuyn c tnh. Nhng qunhiu bc t do thng khng phi l tt, thc t rt t khi s dng bc cao hn2.

Cu 24: Trnh by v phn tch m hnh ton hc kh rn v mt

Cu 25: Trnh by v phn tch cc gii thut c bn trong thit k v giacng?

Cu 26: Trnh by khi nim v nguyn tc v phng php trnh bin dchchng trnh gia cng.

Cu 27: Trnh by v ng dng phng php tnh ton qu o dng c mts trng hp c bn?

Cu 28: Phn tch cng ngh gia cng in hnh ?

Cu 29: phn tch cng ngh gia cng nhm? ngha cng ngh nhm tronggia cng CAD/CAM?

Cu 30: Phn tch cng ngh gia cng linh hot?


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