Earl

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13 years, 41 days

MaplePrimes Activity


These are replies submitted by Earl

@Rouben Rostamian   I made an attempt to adapt your solution to my original surface and start and end points, however the following command failed:

dsol := dsolve({de, bc}, numeric, method=bvp[middefer], maxmesh=500, abserr=1e-1);

The error message said that dsolve could not meet the abserr criteria and suggested increasing maxmesh and/or abserr. I tried both but dsolve continued to fail.

I suspected that the values of Pi in z(x,y) were causing the problems with dsolve and changed these values to 3.14 and your solution then displayed what appeared to be the correct fastest path, however the following failure occurred when attempting to calculate the time for this path:

dsol := dsolve({sys}, numeric, method=bvp[middefer],
  maxmesh=500, abserr=1e-1);
Error, (in fproc) unable to store 'HFloat(722917.2413392215)*I' when datatype=float[8]

Comments?

@vv I also isolated u^2 but attempts with commands minimize and Optimization[Minimize] failed, apparently due to the presence of tan(alpha) and cos(alpha). Your substitution of polynomials for these has enlightened me.

@Rouben Rostamian  Trial and error with your solution shows that a value on the range 0.65059998845 < omega < 0.6506  has the combined disks rolling until theta = PI/2 and then stopping in this position. Does this seem reasonable?

Why is your value for g = 1 rather than 9.8?

JohnS's question has anticipated my next question. When the combined disks rise from the x axis do they rotate about their combined centre of gravity?

@Rouben Rostamian  Sorry for the delayed response, I was out of town. I will spend some time absorbing your solution and reply again with any supplementary questions.

@Preben Alsholm Thank you for an enlightening solution.

Can your method also find the value of DthetaZero which yields the simultaneous values theta(t)=Pi and diff(theta(t),t)=0?

@rlopez Your remarks help clarify this powerful coding.

@rlopez Your deconstruction very much helps me to understand Kitonum's code.

However I remain puzzled by the statement:

Hence, y prime = yk and Q*yk is fdy

In the above statement is y prime diff(y(x),x) or is it diff(y(t),t) ?
 

Please expand on your explanation of this statement.

Also, is it true that a series of line segments must be listed in counterclockwise order for their integrated values to have the correct signs when they are summed toward the total integral over the region?

In Kitonum's second example it appears that the integral of an upward convex curve e.g. sin(Pi*t/3) has a negative value where the integral of an upward concave curve e.g. -sin(Pi*t/3) has a positive value. These values add correctly toward the total integral over the region.

Would this be true for more complex curves such as one which has both upward convex and concave sections within the region?

@Rouben Rostamian  vv:- your suggestion alone permits the boat to cross the same river (same velocity profile, same destination across from the departure) with the boat's speed reduced from 1.0 to 0.82. The time of crossing increases from 1.15 to 1.52 seconds.

  Rouben Rostamian:- your suggestion, while also using vv's suggestion, permits the boat to cross from [0,0] to [1,-0.3] provided its speed is increased from 1.0 to 1.1.

Both of your suggestions allow me to test the limits of the Maple solution. I greatly enjoy such exploration! I thank you both. 

@Rouben Rostamian  Thank you for reading and commenting on my worksheet.

After examining your solution to the least time crossing, here are a few observations:

I animated the boat (as a plot arrow) as it crossed the river following the least time path. It always headed upstream (against the current) with remarkably little variation in its bearing (the value of alpha).

I tried reducing the boat's speed, but any value below 0.935 produced this error message from dsolve;

   Error, (in dsolve/numeric/bvp) initial Newton iteration is not converging

This same error occurred if I moved the destination upstream from the starting point, however a destination downstream from the starting point produced a correct least time path.

 

@Rouben Rostamian  Thank you for your wonderful analysis and solution to my question. I will take lots of time to analyze it, including substituting my worksheet's river velocity function for yours.

@Carl Love Please see my latest reply to Rouben Rostamian. I hope you too can now access Rivercrossing.mw

@Rouben Rostamian  Following Carl Love's advice I renamed my worksheet as one word and it appears to have uploaded successfully. Please let me know if you can now access it.

@Rouben Rostamian  Sorry, I have tried the upload of the worksheet several times and it fails each time, so I deleted the non-working link. Is there another way I can make the worksheet available to you?

In addition I have altered the question's wording to try to clarify that the worksheet animates one river crossing path, namely the boat always heads towards its destination. I would also like to animate a least time path for crossing from the starting point to the destination, which I assume requires a functional definition of the boat's constantly changing heading.

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