## simultaneous equation...

Hi,

I have attached a Maple file. My problem is that the solve for the simultaneous equation does not give me understandable results. I even simplified my equations by saying some parameters are zero although my final goal is to find an expression for a and varphi. Any idea how to solve this analytically? I know how to do it numerically. I need an analytical expression.

Thanks,

Baharm31

## Roots of the characteristic polynomial ...

`how i can calculate roots of the characteristic polynomial equations {dsys and dsys2}and dsolve them with arbitrary initial condition for differennt amont of m and n?thanksKr.mw  `

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` `

` `

## Difficulty in Solving a Problem...

I have been working on a general solution to motion analysis and seem to be going backwards.  I have an numerical solution in Octave I use for comparison.  I have reduced the problem to a small example that exhibits the problem.

I posted a question similar to this, but, without a set of known values.

I am doing something wrong, but, what?

Tom Dean

## bearing.mpl, solve the target motion problem with bearings only.
##
## Consider a sensor platform moving through points (x,y) at times
## t[1..4] with the target bearings, Brg[1..4] taken at times t[1..4]
## with the target proceeding along a constant course and speed.
##
## time t, bearing line slope m, sensor position (x,y) are known
## values.
##
## Since this is a generated problem the target position at time t is
## provided to compare with the results.
##
#########################################################################
##
restart;
##
genKnownValues := proc()
description "set the known values",
"t - relative time",
"x - sensor x location at time t[i]",
"y - sensor y location at time t[i]",
"m - slope of the bearing lines at time t[i]",
"tgtPosit - target position at time t[i]";
global t, m, x, y, tgtPosit;
local dt, Cse, Spd, Brg, A, B, C, R, X;
local tgtX, tgtY, tgtRange, tgtCse, tgtSpd;
## relative and delta time
t := [0, 1+1/2, 3, 3+1/2];
dt := [0, seq(t[idx]-t[idx-1],idx=2..4)];
## sensor motion
Cse := [90, 90, 90, 50] *~ Pi/180; ## true heading
Spd := [15, 15, 15, 22];  ## knots
## bearings to the target at time t
Brg := [10, 358, 340, 330] *~ (Pi/180);
## slope of the bearing lines
m:=map(tan,Brg);
## calculate the sensor position vs time
x := ListTools[PartialSums](dt *~ Spd *~ map(cos, Cse));
y := ListTools[PartialSums](dt *~ Spd *~ map(sin, Cse));
## target values  start the target at a known (x,y) position at a
## constant course and speed
tgtRange := 95+25/32; ## miles at t1, match octave value...
tgtCse := 170 * Pi/180; ## course
tgtSpd := 10; ## knots
tgtX := tgtRange*cos(Brg[1]);
tgtX := tgtX +~ ListTools[PartialSums](dt *~ tgtSpd *~ cos(tgtCse));
tgtY := tgtRange*sin(Brg[1]);
tgtY := tgtY +~ ListTools[PartialSums](dt *~ tgtSpd *~ sin(tgtCse));
## return target position vs time as a matrix
tgtPosit:=Matrix(4,2,[seq([tgtX[idx],tgtY[idx]],idx=1..4)]);
end proc:
##
#########################################################################
## t[], m[], x[], and y[] are known values
##
## equation of the bearing lines
eq1 := tgtY[1] - y[1]    = m[1]*(tgtX[1]-x[1]):
eq2 := tgtY[2] - y[2]    = m[2]*(tgtX[2]-x[2]):
eq3 := tgtY[3] - y[3]    = m[3]*(tgtX[3]-x[3]):
eq4 := tgtY[4] - y[4]    = m[4]*(tgtX[4]-x[4]):
## target X motion along the target line
eq5 := tgtX[2] - tgtX[1] = tgtVx*(t[2]-t[1]):
eq6 := tgtX[3] - tgtX[2] = tgtVx*(t[3]-t[2]):
eq7 := tgtX[4] - tgtX[3] = tgtVx*(t[4]-t[3]):
## target Y motion along the target line
eq8 := tgtY[2] - tgtY[1] = tgtVy*(t[2]-t[1]):
eq9 := tgtY[3] - tgtY[2] = tgtVy*(t[3]-t[2]):
eq10:= tgtY[4] - tgtY[3] = tgtVy*(t[4]-t[3]):
##
#########################################################################
##
## solve the equations
eqs  := {eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9,eq10}:

Sol:= solve(eqs, {tgtVx, tgtVy, seq([tgtX[k], tgtY[k]][], k= 1..4)}):
##

genKnownValues():
## these values are very close to Octave
evalf(t);evalf(m);evalf(x);evalf(y);evalf(tgtPosit);
## The value of tgtX[] and tgtY[] should equal the respective tgtPosit values
seq(evalf(eval([tgtX[idx],tgtY[idx]], Sol)),idx=1..4);

## Integer solution of system...

Dear all,

I wold like to find the solution of the next system of two equations with three unknowns but we assume that the unknows are positive integers. How the following code can work. Many thanks

> restart;
> assume(J, integer, J >= 0);
> assume(A, integer, A >= 0);
> assume(T, integer, T >= 0);
> eq1 := J+10*A+50*T=500;
eq2 := J+A+T = 100;
solve( {eq1,eq2},{J,A,T});

## Equidistant surface

by: Maple 15

Example of the equidistant surface at a distance of 0.25 to the surface
x3
-0.1 * (sin (4 * x1) + sin (3 * x2 + x3) + sin (2 * x2)) = 0
Constructed on the basis of universal parameterization of surfaces.

equidistant_surface.mw

## Generate matrix form ...

Hello,

I cant find solution how to create matrix form from equations of motion. Equations looks like this:

My equations ar much more complicated and one of them looks something like this:

http://i63.tinypic.com/21c5ctk.png

and I want form like this:

I tried to do it using the Generate Matrix but it does not work as I expected. How can you get this form?

## RootFinding[Isolate] vs solve...

For solving polynomial systems I used RootFinding[Isolate]. But after discussing the question http://www.mapleprimes.com/questions/211774-Roots-Of--Expz--1
I decided to compare Isolate and evalf(solve ([...], [...])). It seemed to me that solve some convenient. The only if in the equation there are integers as a real, they should be recorded with a decimal point. (For real solutions of this procedure should be used with (RealDomain).)  Examples:

SOLVE_ISOLATE.mw

I wonder why then the need Root Finding [Isolate]?

## How to rearrange equations...

Hello,

I have a complex set of non linear diff eqns in the form :

y1'' = f(y1',y1,y2'',y2',y2,y3'',y3',y3,y4'',....,y6'',y6',y6,u1,u2,u3,u4) ;

y2'' = f(y1'',y1',y1,y2',y2,y3'',y3',y3,y4'',....,y6'',y6',y6,u1,u2,u3,u4)

and so on ... y6''=(...)

As I want to resolve this coupled systeme in matlab using @ODE45... I wanted the equations in the form : y1''=f(y1',y1,y2',y2,....) and so on ... => X'[] = f(X[],U[])

How can I force maple to rearrange a system of coupled eqns with only the variables i want ?

I know this is possible beacause it is a nonlinear state space model but maple do not work with nonlinear state space model... It give me error when I tried to create statespace model with my non linear diff eqns.

Thanks a lot !

## Eliminating redundant equations ...

I have a system of 16 polynomial equations in 15 variables. Independently I know there is at least a one parameter familiy of solutions to this system, so there is reason to think at least two of the equations are redundent. I would like to use Maple to decipher which of the equations are redundent, but I am unsure how to proceed.

So far I have looked at the Groebner package, and it seems like the Reduce and InterReduce commands will be useful. Say I call the set of 16 polynomials X and define a lexicographical order T on the variables. I then ask maple to compute

Reduce(X,X,T)

and receive a list with 7 zeroes and 9 polynomials. What exactly is this telling me? Does this mean that maple has used polynomial division and found that 7 of the equations are redundent?

## How to solve algebraic equations?...

solve or fsolve dos not any answer?

i think use Newton  Raphson, because solve or fsolve not work.

thankscomparision.mw

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## How to solve this ODE system?...

Here is a code

Maple Worksheet - Error

Failed to load the worksheet /maplenet/convert/EQ.mw .

A fragment of code

for b in extra_bcs do try print(b = 10^(-2)); res[b] := dsolve(dsys4 union {b = 10^(-2)}, numeric, initmesh = 2024, output = listprocedure, approxsoln = [omega2 = 0.1e-2, s(x) = cosh(upsilon*x)-cos(upsilon*x)-(cosh(upsilon)+cos(upsilon))*(sinh(upsilon*x)-sin(upsilon*x))/(sinh(upsilon)+sin(upsilon)), g(x) = sin(((2*n+1)*(1/2))*Pi)], abserr = 0.1e-1) catch: print(lasterror) end try end do; indx := indices(res, nolist); nops([indx]); res[indx[i]]; seq(subs(res[indx[i]](1), omega2(1)), i = 1 .. nops([indx]))

## Solving system with constraints ...

SYSTEMATIC APPROACH TO LIFTING EYE DESIGN

Moses

 Nomenclature Ab  = Required bearing area, sq in. (mm2) As  = Required shear area at hole, sq in. (mm2) Aw = Required cheek plate weld area, sq in. (mm2) b     = Distance from center of eye to the cross section, in. (mm) C    = Percentage distance of element from neutral axis D    = Diameter of lifting pin, in. (mm) e     = Distance between edge of cheek plate and edge of main plate, in. (mm) Fa   = Allowable normal stress, ksi (kN/mm2) Fv   = Allowable shear stress, ksi (kN/mm2) Fw  = Allowable shear stress for weld electrodes, ksi (kN/mm2) Fy   = Yield stress, ksi (kN/mm2) fa   =  Computed axial stress, ksi (kN/mm2) fb   =  Computed bending stress, ksi (kN/mm2) fmax = Maximum principal stress, ksi(kN/mm2) fv    = Computed shear stress, ksi (kN/mm2) g     = Distance between edge of cheek plate and main structure, in. (mm) h    = Length of lifting eye at any cross section between A-A and C-C, in. (mm) n   = Total number of lifting eyes used during the lift P  = Design load per lifting eye, kips (kN) R  = Radius to edge of lifting eye, in. (mm) Rh  = Radius of hole, in. (mm) r    = Radius of cheek plate, in. (mm) S   = Safety factor with respect to allowable stresses s  = Cheek plate weld size, in. (mm) T  = Total plate thickness, in. (mm) Tp  = Main plate thickness, in (mm) t    = Thickness of each cheek plate, in (mm) te  = Cheek plate weld throat, in. (mm) W  = Total lift weight of structure, kips (kN) α  =  Angle of taper, deg. β  =  Angle between vertical and lifting sling, deg. θ  =  Angle between attaching weldment and lifting sling, deg.

The design load for each lifting eye is given by:

 >
 >

In the above equation, n refers to number of lifting eyes to used for the lift, S is the safety factor with respect to allowable stresses, W is the total weight to be lifted, and β is the angle between the vertical direction and the lifting sling.This analysis applies only to lifting eyes shaped like the one in Fig. 1. For other shapes, the designer should re-evaluate the equations.

Radius of liftimg eye hole will depend upon the diameter of the pin, D, used in the lifting shackle. It is recommeded that the hole diameter not greater than 1 / 16 in. (2 mm) larger than tha pin diameter. The required bearing area for the pin is

 >

where Fy is the yield stress. This equation is based on allowable stresses as definde in Ref. 1, which considers stress concentrations in the vicinity of the hole. The designer may choose to use a technique which determines the stresses at the hole and should appropriately adjust the allowable stresses. The total plate thickness is then given by

 >

At this point, if the thickness, T, is too large to be economically feasible, it may be desirable to use cheek plates (Fig.2) around the hole in order to sustain the bearing stresses. In this case, the above thickness, T, is divided into a main plate of thickness Tp and two cheek plates each of thickness t:

 >

It is recommended that t be less than Tp to avoid excessive welding. The radius to the edge of lifting eye plate and the radii of the cheek plates, if they are used, are governed by the condition that the pin cannot shear through these plates. The required area for shear is

 >

It is possible to compute the required radii by equating the shaering area of the cheek plates plus the shearing area of the main plate to the total shear area. Theis a degree of uncertianty in choosing the appropriate shearing area. Minimum areas are used in the following equation, therefore, leading to conservative values for the radius of the main plate, R, and the radius of the cheek plate, r,

 >
 >

where Rh is the radius of the hole and e is the distance between the edge of the cheek plate and the edge of the main plate (Fig. 2). This difference should be large enough to allow space for welding the cheet plate to the main plate. A reasonable value for e is 1.5*t. It should be noted that the above equations assume there are two cheek plates. If cheek plates are not used, then simply let t equal zero and use Eq. 6 to determine R.

It is not necessary to check tension on this net section, since the allowable stress for shear is 0.4*Fy (Eq. 5); whereas the allowable stress for tension on a net section at a pin hole is given as 0.45*Fy (Ref. 1) which is greater than for shear. Size of weld between the cheek plates and the main plate can be determined as follows. The necessary weld area per cheek plate is

 >

where Fw is the allowable shear stress for the welding electrodes. The weld thickness, te is given by

 >

For a manual weld the size, s is given by

 >

To assure that this weld size is large enough to insure fusion and minimize distortion, it should be greater than the AISC suggested Minimum Fillet Weld Sizes (Ref. 1).

The axial stress due to uniform tension along a section is

 >

where h is the length of the section. The elemental bending stress which is distributed linearly along the section may be expressed as

 >

where C represents the distance of an element from the neutral axis and b is the distance from the center of the eye to the cross section. The shearing stress varies parabolically for section between A-A and B-B and is given as

 >

It is felt that Eq. 13 (i.e., parabolic shear stress distribution) is applicable to the cross sections between A-A and B-B and does not apply to the cross sections between B-B and C-C in the area of the taper. The taper creates discontinuities on the shear plane, which result in significantly large shear stress concentratons along the edge of the taper coincident to point of maximum bending stress. This problem will be addressed ina subsequent section of this article.

The maximum principal stress that exists on an element is given by

 >

or after dividing by the maximum allowable normal (i.e., tension) stress, Fa, gives a ratio that must be less than unity, where Fa has been taken as 0.6*Fy. A similar analysis for the maximum shear stress on the element yields

 >
 >
 >
 >
 >

The designer should now select several critical elements throughout the plate and apply the restrains of Eq. 16 and 17 to obtain a required minimum length for the selected cross section. Eq 18 through 21 apply for an element at the neutral axis of the section. C will be zero and Eq. 11, 13 and 16 reduce to

 >
 >

An element at the end of the section will be subjected to bending stresses but not shearing stresses.

For this case C = 0.5 and Eq. 16  becomes

 >

Using the quadratic formular to solve for h yields

 >

The largest value of h predicted by Eqs. 19, 21 and 23 can be used as a first estimate for the length of the cross section; however, intermediate elements, that is, between the edge and the center of the cross section, should also be checked to determine the appropriate length, h, of the section under consideration.

Cross sections A-A and B-B should be analyzed using the above approach. The designer should use his own discretion to select other cross sections for analysis.At cross section A-A, the lifting eye is assumed to be welded with complete penetration to the support structure. Once length, h, is determined, the angle of taper, α, should be investigated. It can be shown that normal stress and shear stress are related by

 >

Minimum required length, h, for cross sections between B-B and C-C can be computed by calculating the maximum shear stress for the most critical element of the cross section, which occurs at the tapered surface. It can be shown that the maximum principal stress would not control the required length, h. Using Eqs. 11 and 12 in conjunction with Eq 24, the maximum shear stress yields the following:

 >

If the above inequality is not satisfied, the angle of the taper, α, must be adjusted.

The adequacy of the structure to which the lifting eye is to be attached should be checked to verify that it is capable of sustaining the loads from the lifting eye.

In some instances, it may be justifiable to use a more sophisticated technique for analyzing the lifting eye as well as the supporting structure.

Input Variables

 >
 >
 >
 >
 >
 >
 >
 >
 >
 >
 >
 >
 >
 >

 Output

Good Evening Everybody,

Could any one help me with the attached file. I'm trying to solve 9 equations with 9 unknowns with many constraints, I'm getting no output from Maple. Please help.

Regards,

Moses

## Rearrange 7 equations ...

Hello,

I tried to rearrange below set of equations to have the equations in term of P[0], P[1], P[2], P[3], P[4], P[5] and P[6]. I used the symbol := for function definition for all Ps except one of them. Thus maple will rearrange that excepted one. However, I got error massage stating "Error, (in P[3]) too many levels of recursion" when I tried to rearranged equations for P[0].

Can I get help to rearranged them.

P[0](s) = (P[1](s)*mu[1]+P[2](s)*mu[2]+1)/(s+3*lambda+3*sigma)

P[1](s) = (3*P[0](s)*lambda+3*P[3](s)*mu[1]+P[4](s)*mu[2])/(s+mu[1]+2*lambda+2*sigma)

P[2](s) = (3*P[0](s)*sigma+P[4](s)*mu[2]+P[3](s)*mu[1])/(s+mu[2]+2*lambda+2*sigma)

P[3](s) = 2*lambda(P[1](s)+P[2](s))/(s+2*mu[1]+lambda)

P[4](s) = 2*sigma(P[1](s)+P[2](s))/(s+2*mu[2]+sigma)

P[5](s) = lambda(P[3](s)+P[4](s))/s

P[6](s) = sigma(P[3](s)+P[4](s))/s

## Writing equations in procedure...

 Pile spacing of each pile from a refrence pile: Angle of each pile from a reference pile in radian: Angle of each pile from a reference pile in degrees: The total number of piles in a pile group: The length of each pile in the pile group in (meters):       Diameter of pile (meter): The radius of each pile (meter):

Good Evening Maple Community,

Can anybody please help in the attached file. Highlighted in yellow are the three equations I'm trying to program in Maple procedure, but could not get any output due to an error message. Please Help.

Boyer

## Solving function to vo/vs ...

Hi

I have 3 equations

eq1 := vs = (Rs+Z)*i1+Z*i3

eq2 := A*vi = Z*i1+(Z+Rf+ro)*i3

eq3:= vo = (Rf+Z)*i3+Z*i1

and I want to solve for vo/vs . How to do that ?

the expected solution is

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