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en-us2019 Maplesoft, A Division of Waterloo Maple Inc.Maplesoft Document SystemMon, 21 Jan 2019 03:05:10 GMTMon, 21 Jan 2019 03:05:10 GMTQuestions asked on MaplePrimes that have not yet received an answerhttp://www.mapleprimes.com/images/mapleprimeswhite.jpgMaplePrimes - Unanswered Questions
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How to Lock Links of a Robot ?
https://www.mapleprimes.com/questions/226336-How-To-Lock-Links-Of-A-Robot-?ref=Feed:MaplePrimes:Unanswered Questions
<p><a href="/view.aspx?sf=226336_question/ABB_IRB_120_Model_IK_OK_6_RealParamsLinearization.msim">ABB_IRB_120_Model_IK_OK_6_RealParamsLinearization.msim</a></p>
<p> </p>
<p>I did the model of the ABB IRB 120 in MapleSim, but I would like to lock several links but I do not know how to do it.</p>
<p> </p>
<p>Also I need the differential equations of the robot, which must be six differential equations but in MapleSim equation extraction page there are lots of equations.</p>
<p> </p>
<p>Please can you help me ?</p>
<p><a href="/view.aspx?sf=226336_question/ABB_IRB_120_Model_IK_OK_6_RealParamsLinearization.msim">ABB_IRB_120_Model_IK_OK_6_RealParamsLinearization.msim</a></p>
<p> </p>
<p>I did the model of the ABB IRB 120 in MapleSim, but I would like to lock several links but I do not know how to do it.</p>
<p> </p>
<p>Also I need the differential equations of the robot, which must be six differential equations but in MapleSim equation extraction page there are lots of equations.</p>
<p> </p>
<p>Please can you help me ?</p>
226336Sun, 20 Jan 2019 20:38:30 Zaaeerraaeerrhelp me to solve the eqution by ritz methode codes
https://www.mapleprimes.com/questions/226325-Help-Me-To-Solve-The-Eqution-By-Ritz?ref=Feed:MaplePrimes:Unanswered Questions
<p>how i can solve the eqution by ritz methode codes?</p>
<p>how i can solve the eqution by ritz methode codes?</p>
226325Fri, 18 Jan 2019 17:32:25 Zmohammad12313mohammad12313Visualising 3d subspaces of 6d space
https://www.mapleprimes.com/questions/226315-Visualising-3d-Subspaces-Of-6d-Space?ref=Feed:MaplePrimes:Unanswered Questions
<p><br>
I have an object in 6d I'd like to visualise. The region of 6d space I am interested in is described by these equations:<br>
<br>
{f[10] = -(.2000000000*(5.*f[21]*f[20]*f[22]-5.*f[20]*f[22]^2+20.*f[20]*f[21]-20.*f[20]*f[22]+135.*f[20]+46.*f[21]))/(f[21]*(f[21]-1.*f[22])),<br>
f[11] = -1.*f[22]-4.,<br>
f[12] = -(1.*(f[22]^2+4.*f[22]-27.))/f[21],<br>
f[20] = f[20],<br>
f[21] = f[21],<br>
f[22] = f[22]}<br>
<br>
clearly the first three variables are dependant, and the latter three are independant. I'd like to graph the first three as the latter three vary between bounds and then colour the points on the output based on where they came from in the input, so i can get some intuition about what these equations mean.<br>
</p>
<p><br />
I have an object in 6d I'd like to visualise. The region of 6d space I am interested in is described by these equations:<br />
<br />
{f[10] = -(.2000000000*(5.*f[21]*f[20]*f[22]-5.*f[20]*f[22]^2+20.*f[20]*f[21]-20.*f[20]*f[22]+135.*f[20]+46.*f[21]))/(f[21]*(f[21]-1.*f[22])),<br />
f[11] = -1.*f[22]-4.,<br />
f[12] = -(1.*(f[22]^2+4.*f[22]-27.))/f[21],<br />
f[20] = f[20],<br />
f[21] = f[21],<br />
f[22] = f[22]}<br />
<br />
clearly the first three variables are dependant, and the latter three are independant. I'd like to graph the first three as the latter three vary between bounds and then colour the points on the output based on where they came from in the input, so i can get some intuition about what these equations mean.<br />
</p>
226315Wed, 16 Jan 2019 22:23:59 ZAnnonymouseAnnonymouseHow do I obtain Taylors series expansion of fractional differential equation?
https://www.mapleprimes.com/questions/226301-How-Do-I-Obtain-Taylors-Series-Expansion?ref=Feed:MaplePrimes:Unanswered Questions
<p>Good day everyone.</p>
<p>Please can help me with this code on the taylor series expansion involving Fractional Differential Equation (FDE)? Particularlly, the lines highlighted blue and green respectively in relation to FDE.</p>
<p>Thank you and kind regards</p>
<pre class="prettyprint">
#k=2
restart:q:=n*h:
P:=sum((a[k]*x^(k))/GAMMA(k+1-alpha), k=0..3):
assume(alpha>0,alpha < 1):
Q:=fracdiff(P,x,alpha):
e1:=simplify(eval(P, x=q+h))=y[n+1]:
e2:=simplify(eval(Q,x=q))=f[n]:
e3:=simplify(eval(Q,x=q+h))=f[n+1]:
e4:=simplify(eval(Q,x=q+2*h))=f[n+2]:
var:=seq(a[i], i=0..3):
M:=e||(1..4):
Cc:=eval(<var>, solve(eval({M}),{var}) ):
for i from 1 to 4 do
a[i-1]:=Cc[i]:
end do:
Cf:=P:
E:=collect(Cf, [y[n+1], f[n], f[n+1],f[n+2]], recursive):
print():
s2:=y[n+2]=simplify(eval(Cf, x=q+2*h)):
collect(%, [y[n+1], f[n], f[n+1],f[n+2]], recursive):
s1:=y[n]=simplify(eval(Cf, x=q)):
collect(%, [y[n+1], f[n], f[n+1],f[n+2]], recursive):
<span style="background-color:#0000CD;">Y[n+1]:=convert(taylor(y(x+h),h=0,12),polynom):
F[n]:=convert(taylor((D(y))(x), h = 0,12), polynom):
F[n+1]:=convert(taylor((D(y))(x+h), h = 0,12), polynom):
F[n+2]:=convert(taylor((D(y))(x+2*h), h = 0,12), polynom):
</span>
<span style="background-color:#008000;">W:=asympt(expand(eval(rhs(s2),[y[n+1]=Y[n+1],f[n]=F[n],f[n+1]=F[n+1],f[n+2]=F[n+2]])),h,6);
X:=convert(taylor(y(x+2*h),h=0,12),polynom)-W;
lte:=convert(asympt(X,h,8),polynom);</span></pre>
<p> </p>
<p> </p>
<p>Good day everyone.</p>
<p>Please can help me with this code on the taylor series expansion involving Fractional Differential Equation (FDE)? Particularlly, the lines highlighted blue and green respectively in relation to FDE.</p>
<p>Thank you and kind regards</p>
<pre class="prettyprint">
#k=2
restart:q:=n*h:
P:=sum((a[k]*x^(k))/GAMMA(k+1-alpha), k=0..3):
assume(alpha>0,alpha < 1):
Q:=fracdiff(P,x,alpha):
e1:=simplify(eval(P, x=q+h))=y[n+1]:
e2:=simplify(eval(Q,x=q))=f[n]:
e3:=simplify(eval(Q,x=q+h))=f[n+1]:
e4:=simplify(eval(Q,x=q+2*h))=f[n+2]:
var:=seq(a[i], i=0..3):
M:=e||(1..4):
Cc:=eval(<var>, solve(eval({M}),{var}) ):
for i from 1 to 4 do
a[i-1]:=Cc[i]:
end do:
Cf:=P:
E:=collect(Cf, [y[n+1], f[n], f[n+1],f[n+2]], recursive):
print():
s2:=y[n+2]=simplify(eval(Cf, x=q+2*h)):
collect(%, [y[n+1], f[n], f[n+1],f[n+2]], recursive):
s1:=y[n]=simplify(eval(Cf, x=q)):
collect(%, [y[n+1], f[n], f[n+1],f[n+2]], recursive):
<span style="background-color:#0000CD;">Y[n+1]:=convert(taylor(y(x+h),h=0,12),polynom):
F[n]:=convert(taylor((D(y))(x), h = 0,12), polynom):
F[n+1]:=convert(taylor((D(y))(x+h), h = 0,12), polynom):
F[n+2]:=convert(taylor((D(y))(x+2*h), h = 0,12), polynom):
</span>
<span style="background-color:#008000;">W:=asympt(expand(eval(rhs(s2),[y[n+1]=Y[n+1],f[n]=F[n],f[n+1]=F[n+1],f[n+2]=F[n+2]])),h,6);
X:=convert(taylor(y(x+2*h),h=0,12),polynom)-W;
lte:=convert(asympt(X,h,8),polynom);</span></pre>
<p> </p>
<p> </p>
226301Tue, 15 Jan 2019 09:39:00 Zabdulganiyabdulganiy Optimal control problem
https://www.mapleprimes.com/questions/226295--Optimal-Control-Problem-?ref=Feed:MaplePrimes:Unanswered Questions
<p>How do I plot the <strong>optimal control functions</strong> in an optimal control problem ?</p>
<p>How do I plot the <strong>optimal control functions</strong> in an optimal control problem ?</p>
226295Mon, 14 Jan 2019 12:38:34 ZHaileHaileMultiple Execution of Maple Code with a Varied Parameter
https://www.mapleprimes.com/questions/226273-Multiple-Execution-Of-Maple-Code-With?ref=Feed:MaplePrimes:Unanswered Questions
<p>Hi!</p>
<p>I have a rather long Maple code and want it to be executed multiple times with a parameter changed each time.</p>
<p>Surely this can be done with the loop structure, but it seems the whole loop structure must be contained into one single execution group, which makes it to be a little inconvenient, since the code is too long.</p>
<p> </p>
<p>So is there any alternative way to realize this utility?</p>
<p> </p>
<p>Best regard and thanks!</p>
<p>Hi!</p>
<p>I have a rather long Maple code and want it to be executed multiple times with a parameter changed each time.</p>
<p>Surely this can be done with the loop structure, but it seems the whole loop structure must be contained into one single execution group, which makes it to be a little inconvenient, since the code is too long.</p>
<p> </p>
<p>So is there any alternative way to realize this utility?</p>
<p> </p>
<p>Best regard and thanks!</p>
226273Wed, 09 Jan 2019 08:30:01 ZGlowingGlowing