I am just having some trouble solving this. If somebody knowledgeable

>>8073477
>BallA is traveling at 10 degrees

10 degrees with respect to what?

>>8073481
We can say that the positive X direction is 0 degrees or 360 degrees

File: l.jpg (80KB, 1171x659px) Image search: [Google]

80KB, 1171x659px

I drew a picture to represent it.
I'm trying to think of a way to some how make a right triangle out of this

>>8073477
You know at every time t the position of ball A. You then know the distance of A from the initial position of B. The collision will occur when the distance of A's current position to B's initial position is 6/10 the distance from A's starting position

>>8073633
You can do the calculation yourself, it seemed like you needed guidance to see how you could use the information to solve the problem.

>>8073635
I am not >>8073633

>>8073633
That makes no sense

bumping. I'm gonna try to work on this problem, but it seems like the first thing we need to do is find the portion of time where it is impossible for them to collide. So we need to extend a circle with radius (6t) extending out from (14,2), and see where it intersects with the circle of radius 10t. Then we have to somehow calibrate t so that the intersection point/points will be at 10 degrees from the x-axis. It's an interesting problem and I'm sure there are many ways to go about it. I'll let you know what I come across.

>>8073673
At time t ball A is at a(t)=(2+ 10*cos(10),7+ sin(10)).
Calculate distance between a(t) and (14,2) using sqrt((x1-x2)^2+(y1-y2)^2)).
At time t ball A has travelled 10t.
If the balls collide, whoever the collision takes place, it must be that ball A and ball B arrived at that place at the same time.

File: l.jpg (77KB, 944x531px) Image search: [Google]

77KB, 944x531px

>>8073702
What do I do with t? Here's what I've drawn out

>>8073702
...continuation of >>8073707

I figured I would use the two equations down at the bottom of the picture I just posted and try and deduce either t or theta by using substitution. We'll see...

>>8073707
Don't solve for the angle yet. Work out the collision time using the following observation. If a collision occurs at time t, A will have travelled 10t and B will ave travelled 6t, so the collision is when 1/6 times the distance of moving A at time t from (14,2) is equal to 10t.

>>8073714
If you do need the angle, once you work out collision time as described, you just check A's position at that time to get the collision position and can then work out the angle too.

>>8073714
You lost me at
"so the collision is when 1/6 times the distance of moving A at time t from (14,2) is equal to 10t."

Where did 1/6 come from? How is it possible to work out the time from this?

>>8073732
The knowns are:
Va = 10km/h
Vb = 6km/h
Da = 10 degrees
(Xa,Ya) = (2,7)
(Xb,Yb) = (14,2)
The unknowns are the direction of BallB, time, and the point of collision. aka Db, t, and (Cx, Cy). If I can find any of the three of those things, the other two can be found too.
We can infer that:
timeA=timeB
distanceA=(10km/h) * t
distanceB=(6km/h) * t

I already explained that you can write, as a function of t, the distance from ball at at time t to the position B left from, and how to do so. If 1/6 of that distance equals 10t then (assuming the ball B left in the correct direction) then both balls would collide.
I really can't explain this concept further, it's just realising that if they collide at some time, the distance from the two starting points will be the ratio if their speeds. In most directions there will be no collision

Homework threads go in >>>/hm/. Now get out.

>>8073771
This isn't for homework you ass hat

>>8073770
Where did you get 1/6 though? It seems very arbitrary without some kind of explanation.

Additionally, how would you know if it equals 10t if you don't know what t is, without having to brute force the answer and try a number of inputs into the function.

>>8073775
Because ball B moves at 6 km/h...

>>8073786
I'm sorry but that makes no sense. You didn't explain with any concepts or equations or coherent explanations. Why one sixth? And additionally, how would you check if 1/6 that distance equals 10t anyways without brute forcing and trying repeatedly? It's not a thorough explanation... Nobody has been able to help me with my presented problem thus far, in a few different places I've looked for help

File: DSC_0658.jpg (1MB, 3840x2160px) Image search: [Google]

1MB, 3840x2160px

Is this any better?

>>8073808
Sorry for upside down. On my phone...

>>8073808
>>8073817
Thanks for taking the time to write that out.
I'm not that well versed in math so I wasn't sure about the quadratic formula and using it to solve. I'll look into how to do that myself. Thanks

>>8073825
The problem solving is more important than applying the quadratic formula when you get the equation out.
I hope you can understand why my explanation solves the problem. We found two different descriptions for the distance from b at the collision time (6t and the more complicated one)

>>8073834
Here's what I've deduced

In terms of BallA, the collision coordinates are:
x = 2 + (10km/h) * t * cos(10 degrees)
y = 7 + (10km/h) * t * sin(10 degrees)

The distance between BallB and the collision point can be defined as both:
(6km/h) * t
and
sqrt((14 - x)^2 + (2 - y)^2) where x and y are the coordinates of the collision point.

So x and y can be substituted by the equations in terms of BallA.
After that, the right side is set to 0 and the quadratic equation is used to solve for t.

>>8073863
That's right. Then put the equality in to quadratic form and solve it.

File: DSC_0659.jpg (2MB, 3840x2160px) Image search: [Google]

2MB, 3840x2160px

>>8073825
Here's another way you could solve it, using a side-angle-side approach. Note that you would still solve for t, then work out the angle by getting the position of by seeing where ball A was.

>>8073901
Wow. True that. I didn't even consider that approach. That's much more efficient than what I was working on right now, plus it's more easily implementable in code which is what I'm shooting for.

Thanks so much.

>>8073901
Is this still supposed to turn into a quadratic problem?

>>8073929
I didn't actually solve it and I don't recall the trig rule for side angle side, but probably. Also, I would say the first method probably involves less calculation overall, especially if you want time, position and angle.

>>8073936
Thank you for your help, I'd count this problem as solved. I was able to solve it with both approaches. The work I have to put in now is going to be trying to implement this programmatically, some how.

I will need to figure out how to solve for t in this template equation in javascript somehow:
0 = (14 - 2 + 10t * cos(10))^2 + (2 - 7 + 10t * sin(10))^2 - 6t^2

I got two angles:
41.33
93.43
I basically set up a vertical line from the point of collision to the c axis and it made two right triangles similar to >>8073707. I then put everything in terms of the angle and couldn't be bothered to find the angle using trig so I plugged everything into Wolphram. My method was correct I just don't know how to solve an equation like this:
1.2sin(x)-.5cos(x)=1.1679
I don't even think it's possible without some crazy trig identity that I don't know.

>>8073947
http://www.wolframalpha.com/input/?i=solve+0%3D(xb-xa%2Bva+t+cos(da))%5E2%2B(yb-ya%2Bva+t+sin(da))%5E2-vb+t%5E2+for+t

I ended up going with the first approach posted in this thread. I take the x and y offsets from point a to the collision point to find the collision point in terms of t, and then finding the distance between point b and the collision point in terms of t using the distance formula/pythagorean theorem, and then substituting x and y in the distance formula for x and y of the collision point. After that, t is the only variable left and it is a quadratic problem.

>>8073947
>>8073989
Oops, this link works:
http://www.wolframalpha.com/input/?i=solve+0%3D(a-b%2Bc+t+cos(d))%5E2%2B(e-f%2Bc+t+sin(d))%5E2-g+t%5E2+for+t