Dog and Rabbit Problem

I would look at it as progress towards the rabbit is equal to the difference in speed, plus the starting distance between them, and make a triangle that intersects at the meeting point...that should give you all the information to solve the distance...

http://sections.maa.org/okar/papers/2006/lloyd.pdf

That's on the assumption that the dog anticipates where the rabbit will get to and runs in a straight line to that point. That's quite a simple problem and I worked it out to give a bit of a check on the result from iteration.

But thanks for the link, it's very useful. Took a bit of fiddling about to get it to suit the problem, but then I plotted the curve from the formula and we agree spot-on.

A minor thing - it says at one point k = fox's speed/rabbit's speed, but later if k > 1, which corresponds to the rabbit running faster than the fox. I used k = 2

Thanks for that. Your simulation looks identical to mine. Interesting that there is an analytical solution, but it's beyond my calculus skills to get the result in SolarEagle's link. I'd be surprised if the original question setter expected people to do it that way, but it was a Mensa magazine, so maybe!

I'd like to attach my graph, but I can never work out how to do it on CR4. I saved it as a pdf but no joy. Can you give me any hints? I can copy and paste a photo (JPG) but it doesn't work with other docs.

Just do a screen capture with your graph on display and paste it into paint as a jpg....save it as a file and use file path to post...

I save my graphs as jpg pictures out of Octave and then use the camera widget in this box to post it. If you can't save your graph as a gif or jpg, you might be able to use "Snipping Tool" (in Windows) to save a picture.

I typed this but got error message below when I tried to send it.

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Thanks for that. Managed to add graph below, but I had to print it, then scan it to get a jpg, which I can copy and paste. Seems a bit of a rigmarole, I must be missing something. I can't see any other way to get a jpg, and copy/paste doesn't work with other formats for some reason.

Anyway, my curve is identical to yours, the diagonal line is if the dog intercepts. The curve from the formula is also there, but it coincides with the simulation except near the end.

Thanks to everybody who commented.

With the simulation, I've come up with the formula for the time it takes the dog to catch the rabbit. It works only for when the dog's speed is twice the rabbit's speed.

Let the rabbit's speed be Vr (m/s), and the dog's speed Vd = 2 x Vr (m/s).

The rabbit's distance at the start is D meters from the dog's location, 100 meters in the example.

Time=(2*D/(3*vr)

In the example D=100 m, vr=5 m/s, Time=13.333 seconds

Nice! My simulation works differently, so can't calculate time, apart from seeing what time corresponds to dog's 100m easterly travel.

I'd guess you could come up with a formula for time with other values of D and vr, if those values were used in the simulation.

I don't know how your simulation works, but what I did is to advance time (Δt) one increment at a time, adjusting the canine velocity vector, Vd, at each iteration, the same magnitude re-aimed at the updated position of the rabbit. I got the travel time by multiplying the loop counter by Δt and the eastward distance measurement by summing the horizontal component of Vd * Δt. I set a flag so that when it reached 100 meters, it records the time once only. (The dog is fixated on that rabbit and once it reaches it, continues orbiting.)

In my simulation, I have time interval 0.1s, and at the start of each interval calculate the angle between the line from dog to rabbit and the x-axis, θ = atan((rabbit's y-value minus previous)/(100m minus dog's previous x-value)).

Hypotenuse is constant, h = dog speed*time interval, so new x = h*cos θ + previous, new y = h*sin θ + previous.

Clearly both our approaches work.

In the link, I just about understand the calc in Special Case of Pursuing a Straight Line Target, but I'd never have got there from first principles. Interesting though.

I am unable to post a link, or I would do so.

Search for ‘Radiodrome’.

A radiodrome is the specific type of pursuit curve we are considering here.

Wiki link below seems to work. There's also stuff on pursuit curves on the web.

Radiodrome

Thanks for the link.

I’m having an issue posting things, I believe this to be related more to my antique seven year old device than to CR4.

I thought I'd sent this already, but either I forgot to hit send or there's another problem with the site.

So am I with posting issues. When I try Insert Image I can browse for a file, but when I click OK I get "There was an error uploading the image."

I attached a photo (jpg) some years ago, can't remember whether it's ever worked with other types. But trying it now, with my simulation or a random photo as a test, get message in #13.

Edit - seems to have gone OK this time.

It is a vector quantity,like crossing a river that is running 5mph in a boat that is running 10mph...also similar to a hunter leading a flying bird.

But the analogy with a river crossing is if the dog anticipates where the rabbit will get to and runs in a straight line to that point. A much simpler problem.

The dog has no grasp of plane geometry, time, or distance. His only actionable information is where the rabbit is right now.

This results in the pursuit curve shown in Rixter’s comment.

OK,I will take a SWAG at it:,no formulas,no graphs,just Kentucky Windage: 11.180339 seconds after the start,approximately.

I know it was a joke, but it must be more than 11.18 seconds because it takes 11.55 seconds on a straight run interception basis. From the data in SolarEagle's link it comes to 13.3333 seconds.

Anything after 11 seconds is beyond "approximately." But you were teasing us I assume.

A SWAG I said,son,A SWAG! (Foghorn Leghorn)

Here is a good explanation of the solution to the dog and rabbit problem:

https://www.researchgate.net/publication/1899909_The_dog-and-rabbit_chase_problem_as_an_exercise_in_introductory_kinematics

You can compute the intercept time for any values of distance, rabbit-speed and dog-speed, provided that dog-speed > rabbit-speed.

The formula is: Time = (Distance x Vdog) / (Vdog² - Vrab²)

The table below is the intercept time for values of Vdog (horizontal) vs Vrab (vertical) for a distance = 100 meters.

Thanks for that, very interesting. So it's easy to calculate the time, and with that, the distances of each at point of contact. Harder to calculate the dog's trajectory.

The original question just asked for the time, if I remember right, and may have assumed people would know the formula, or work it out.