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Hi, I build my simulation system in MapleSim, using the module 'Stepper Permanent Magnet' like this


But when I run this system, I get the followiing error:

System is underdetermined


When I delete the part about 'Stepper Permanent Magnet', the system runs well.

How to solve this problem? How to use 'Stepper Permanent Magnet'?

The Help file of MapleSim cannot proviod such help information.

Thank you.





With this application we can meet safety characteristics of a relationship and simple or compound functions. Made with maple 2015.

(in spanish)



I need to model some spherical joints with damping.

Apparently, contrary to revolute joints, it seems that it is not possible to add damping in spherical joints. Indeed, contrary to the revolute joint where there is an option to put spring or damping element, i don't find this option in the spherical joint. 

Do you have a idea to model spherical joints with damping (without changing the spherical joint)?

Thanks a lot for your help


I was trying to find the solution for two theta variables in a couple of simultaneous equations (infact this is an iverse kinematics problem for a two link system pendulum).
The following are the initial inputs/equations to be manipulated:

Then I use the folowing command to rearrange for the theta values which I am after:

which gives me the result:

This is all fine until I give in values for l1, l2, x and y:


I have a RootOf in there with a _Z term poping up here and there. I know that this configuration of the two link mechanism in fact dows have a solution and that these numbers are reasonable. Thus I have three questions:

Why does this happen?
What does the "signum" mean here?
how do I go about getting the nummerical values?

Many thanks,
- pjf

the calculation is like the following command, the result in the picture

SetCoordinates(spherical[r, theta, phi]);
Fv := rho*VectorField(`<,>`(v[r](r, theta, phi), v[theta](r, theta, phi), v[phi](r, theta, phi)));



1) when the Divergence act on the Fv, then it will be expanded, which is lengthy and not like most book's formulation , especially when I want to continue for a Conversation law like in fluid mechanics, this will be too long and a messy for later check.

could there be a way to not expand this result, just as the eq(3) like.

2) when I want to calculate the Divergence of Fv, I must construct a VectorField at first, but this is in components way, is there a quick way for Vector Field Function


I want to make the model which moves along the specific direction (translational), So I used the "prismatic joint" to give the direction that I want to enforce. However, "Prismatic joint" only offers the direction along the "X", "Y", and "Z" axes though I want to give other direction.


Is there any way to give the specific direction (vector) to make my model move in that way ?

Could anyone please hepl me? I have the following system

e1 := exp(F(r)/phi_0)*L*A(r) = (1/2)*(2*(diff(A(r), r, r))*B(r)*A(r)*r*C(r)+2*B(r)*A(r)*(diff(A(r), r))*(diff(C(r), r))*r-(diff(A(r), r))^2*B(r)*r*C(r)-(diff(A(r), r))*(diff(B(r), r))*A(r)*r*C(r)+4*B(r)*A(r)*(diff(A(r), r))*C(r))/(B(r)^2*A(r)*r*C(r));
e2 := alpha*(diff(F(r), r, r))+(alpha^2+omega)*(diff(F(r), r))^2+(1/4)*(4*(diff(C(r), r, r))*B(r)*A(r)^2*C(r)*r+2*(diff(A(r), r, r))*A(r)*B(r)*r*C(r)^2-2*B(r)*A(r)^2*(diff(C(r), r))^2*r-(diff(A(r), r))^2*B(r)*r*C(r)^2-2*A(r)^2*C(r)*(diff(C(r), r))*(diff(B(r), r))*r-(diff(A(r), r))*(diff(B(r), r))*A(r)*r*C(r)^2+8*B(r)*A(r)^2*C(r)*(diff(C(r), r))-4*A(r)^2*C(r)^2*(diff(B(r), r)))/(r*A(r)^2*B(r)*C(r)^2)-(1/4)*(2*(diff(A(r), r, r))*B(r)*A(r)*r*C(r)+2*B(r)*A(r)*(diff(A(r), r))*(diff(C(r), r))*r-(diff(A(r), r))^2*B(r)*r*C(r)-(diff(A(r), r))*(diff(B(r), r))*A(r)*r*C(r)+4*B(r)*A(r)*(diff(A(r), r))*C(r))/(B(r)*A(r)^2*r*C(r)) = 0;
e3 := (1/4)*(-2*(diff(C(r), r, r))*B(r)*A(r)*r^2-B(r)*(diff(A(r), r))*(diff(C(r), r))*r^2+A(r)*(diff(C(r), r))*(diff(B(r), r))*r^2-8*B(r)*A(r)*(diff(C(r), r))*r-2*B(r)*(diff(A(r), r))*C(r)*r+2*A(r)*C(r)*(diff(B(r), r))*r+4*B(r)^2*A(r)-4*B(r)*A(r)*C(r))/(B(r)^2*A(r)) = -(1/4)*(2*(diff(A(r), r, r))*B(r)*A(r)*r*C(r)+2*B(r)*A(r)*(diff(A(r), r))*(diff(C(r), r))*r-(diff(A(r), r))^2*B(r)*r*C(r)-(diff(A(r), r))*(diff(B(r), r))*A(r)*r*C(r)+4*B(r)*A(r)*(diff(A(r), r))*C(r))*r/(B(r)^2*A(r)^2);
e4 := -(alpha^2+2*omega)*(diff(F(r), r))*(-(1/2)*(-(diff(A(r), r))*B(r)*r^4*C(r)^2-A(r)*(diff(B(r), r))*r^4*C(r)^2-4*A(r)*B(r)*r^3*C(r)^2-2*A(r)*B(r)*r^4*C(r)*(diff(C(r), r)))/(A(r)*B(r)*r^4*C(r)^2)-(diff(B(r), r))/B(r)+(diff(F(r), r, r))/(diff(F(r), r))+alpha*(diff(F(r), r)))/B(r) = -exp(F(r)/phi_0)*V_0*(alpha-1/phi_0);

phi_0 := -alpha/(2*alpha^2+2*omega); L := V_0*(1-(alpha-1/phi_0)*alpha/(3*alpha^2+2*omega)); V_0 := -lambda*exp(-fc/phi_0); fc := ln((4*alpha^2+2*omega)/(G_0*(3*alpha^2+2*omega)))/alpha; m := (2/(1+g))^(1/2); n := g*(2/(1+g))^(1/2); P := (G_0*(3*alpha^2+2*omega)/(4*alpha^2+2*omega))^(-2*alpha/(n-m)); eta := 1.4*G_0*Ms*(2/(1+g))^(-1/2)/c^2; g := 1-alpha^2/(2*alpha^2+omega);

omega := -10^5; alpha := 1; G_0 := 6.67*10^(-11); lambda := 10^(-52); c := 2.9*10^8; Ms := 1.9*10^30;
ri := evalf(1000*eta);

ics := A(2.109660445*10^6) = 1, (D(A))(2.109660445*10^6) = 2.370091128*10^(-15)*sqrt(2)*sqrt(99998)*sqrt(199997), B(2.109660445*10^6) = 1, C(2.109660445*10^6) = 1, (D(C))(2.109660445*10^6) = 4.740182256*10^(-15)*(1-(99999/19999300006)*sqrt(2)*sqrt(99998)*sqrt(199997))*(1-1.000017501*10^(-8)*sqrt(2)*sqrt(99998)*sqrt(199997))^(-(99999/19999300006)*sqrt(2)*sqrt(99998)*sqrt(199997))*sqrt(2)*sqrt(99998)*sqrt(199997), f(2.109660445*10^6) = 23.43081116, (D(f))(2.109660445*10^6) = 4.749681180*10^(-15):

eta:=2109.660445: sys:=e1,e2,e3,e4; vars:=[A(r),B(r),C(r),F(r)];

dsn3 := dsolve([sys, ics], numeric, vars, range = 3*eta .. 50*eta);

Results in

Warning, cannot evaluate the solution past the initial point, problem may be complex, initially singular or improperly set up

Setting f(r)=Const,V_0=0 which is a physically relevant case, results in

Error, (in simplify/normal) numeric exception: division by zero

I suugest the problem is that the equation contain sqared derivatives, hence there are several solution branches corresponding to different signs of square root. Maple chooses the singular branch. How can I force it to choose another branch or calculete all of them?

Thanks in advance..

Attached is a photo with the code I am working for.  

On the top is practice code with a simpler ODE to help with trouble shooting, on the bottom is the ODE I am working with.

I was hoping to gain insight about the _z1 symbol in the solution, I haven't been able to find much help on other threads.  I would like to know how I can go about working with it - if it is something on my end or if it is the nature of the equation I am working with.


Thank you for any help,


so if i want to perform a function on elements from a list i do 


X:=[3, 5, 6, 7]


add(x*ln(x), x = X)



but what if i want to have two lists X and Y and perform the function xi*ln(yi) and add that up?



PS: keep in mind that im working with integers aswell as floats


thanks :)

I would like, for an arbitary point in R^2, to calculate the projection onto a curve, such as 1/x or 1/exp(x). Is there a "cheap" way to do so? If it makes it easier, I could mostly be interested in points strictly below the curve.

hi, is there a way to change color of the page in Maple 18? in fact I am preparing my lectures using slideshow option and want to change the color instead of the default white.

I am having issues opening my final year project, it was working an hour ago and now it will not open. When I try to open the file it brings up a box title TEXT FORMAT CHOICE with the options: MAPLE TEXT, PLAIN TEXT, MAPLE INPUT and CANCEL.

This is all my work and I need it to complete my year.

Any help with how I can rectify this would be extremely helpful.

Many thanks

A rigid rotating body is a moving mass, so that kinetic energy can have expressed in terms of the angular speed of the object and a new quantity called moment of inertia, which depends on the mass of the body and how it is such distributed mass. Now we'll see with maple.

(in spanish)


L. Araujo C.

In this section, we will consider several linear dynamical systems in which each mathematical model is a differential equation of second order with constant coefficients with initial conditions specifi ed in a time that we take as t = t0.

All in maple.

(in spanish)




The equations of motion for a rigid body can be obtained from the principles governing the motion of a particle system. Now we will solve with Maple.

(in spanish)


Lenin Araujo Castillo

Corrección ejercico 4


4.- Cada una de las barras mostradas tiene una longitud de 1 m y una masa de 2 kg. Ambas giran en el plano horizontal. La barra AB gira con una velocidad angular constante de 4 rad/s en sentido contrario al de las manecillas del reloj. En el instante mostrado, la barra BC gira a 6 rad/s en sentido contrario al de las manecillas del reloj. ¿Cuál es la aceleración angular de la barra BC?


restart; with(VectorCalculus)



m := 2

L := 1

theta := (1/4)*Pi

a[G] = x*alpha[BC]*r[G/B]-omega[BC]^2*r[G/B]+a[B]NULL


a[B] = x*alpha[AB]*r[B/A]-omega[AB]^2*r[B/A]+a[A]


aA := `<,>`(0, 0, 0)

`&alpha;AB` := `<,>`(0, 0, 0)

rBrA := `<,>`(1, 0, 0)

`&omega;AB` := `<,>`(0, 0, 4)

aB := aA+`&x`(`&alpha;AB`, rBrA)-4^2*rBrA

Vector[column](%id = 4411990810)


`&alpha;BC` := `<,>`(0, 0, `&alpha;bc`)

rGrB := `<,>`(.5*cos((1/4)*Pi), -.5*sin((1/4)*Pi), 0)

aG := evalf(aB+`&x`(`&alpha;BC`, rGrB)-6^2*rGrB, 5)

Vector[column](%id = 4412052178)


usando "(&sum;)M[G]=r[BC] x F[xy]"

rBC := `<,>`(.5*cos((1/4)*Pi), -.5*sin((1/4)*Pi), 0)

Fxy := `<,>`(Fx, -Fy, 0)


`&x`(rBC, Fxy) = (1/12*2)*1^2*`&alpha;bc`

(.2500000000*sqrt(2)*(-.70710*`&alpha;bc`-25.456)+(.2500000000*(57.456-.70710*`&alpha;bc`))*sqrt(2))*e[z] = (1/6)*`&alpha;bc`



"(&sum;)Fx:-Fx=m*ax"           y             "(&sum;)Fy:Fy=m*ay"

ax := -28.728+.35355*`&alpha;bc`



ay := .35355*`&alpha;bc`+12.728



Fx := -2*ax



Fy := 2*ay



`&x`(rBC, Fxy) = (1/12*2)*1^2*`&alpha;bc`

(.2500000000*sqrt(2)*(-.70710*`&alpha;bc`-25.456)+(.2500000000*(57.456-.70710*`&alpha;bc`))*sqrt(2))*e[z] = (1/6)*`&alpha;bc`


.2500000000*sqrt(2)*(-.70710*`&alpha;bc`-25.456)+(.2500000000*(57.456-.70710*`&alpha;bc`))*sqrt(2) = (1/6)*`&alpha;bc`

.2500000000*2^(1/2)*(-.70710*`&alpha;bc`-25.456)+(14.36400000-.1767750000*`&alpha;bc`)*2^(1/2) = (1/6)*`&alpha;bc`



[[`&alpha;bc` = 16.97068481]]





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