Couple of quick maths questions

[P3nT4gR4m]

Prolly easy if you know sums.

Object A is travelling in a straight line, at a constant speed of 60 Mph toward object B.

Object B rapidly accelerates to 6 mph on a heading perpendicular to object A's trajectory.

How close can object A be to object B if object B wants to leave it til the last second before paddling like fuck into the bow wave of an approaching oiltanker

Hypothetically speaking?

part 2 is the exact same question but object A is travelling at 40mph and is only the size of a fishing boat

I'd appreciate if you could tell me how you worked it out. Now that I've formulated the question I'm kinda interested in how maths would be applied to solving it.

[LMNO]

I think more information is needed. 

Do you want Object B to go in front of the boat, hit the boat, or go behind the boat?

What is the rate of acceleration? For example, gravity accelerates at 32 m/s².

If you want to cut in front of it, how wide is object A?

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Wait, I'm trying to visualize this.  "How close can object A be to object B if object B wants to leave it til the last second before paddling like fuck into the bow wave of an approaching oiltanker."

You want to paddle (along an X axis) at an oil tanker (travelling on a Y axis) and hit the bow wave, right?  So the answer depends on two variables.  How far away on the X axis you are, and how far away on the Y axis you are.  One would determine the other.

[Faust]

Prolly easy if you know sums.

Object A is travelling in a straight line, at a constant speed of 60 Mph toward object B.

Object B rapidly accelerates to 6 mph on a heading perpendicular to object A's trajectory.

How close can object A be to object B if object B wants to leave it til the last second before paddling like fuck into the bow wave of an approaching oiltanker

Hypothetically speaking?

part 2 is the exact same question but object A is travelling at 40mph and is only the size of a fishing boat

I'd appreciate if you could tell me how you worked it out. Now that I've formulated the question I'm kinda interested in how maths would be applied to solving it.

Well I don't have the separation or size of A and B, so as long as I want?

[LMNO]

Yeah, that's an easier way to answer it.  I had to take the long way around.

[P3nT4gR4m]

To be honest a ballpark would do it.  I'll be hitting 6mph within a couple of secs. I'm wondering if its 20 feet or two hundred meters kind of thing?

[Mesozoic Mister Nigel]

You have to provide at least an approximation of the other distance, or it's unsolvable.

[Faust]

And if not unsolvable, multivariate, allowing for nigh on infinite answers.

It's like saying what are x and y in the equation x+y =10, sure there are only certain combinations that are true, but there are many many values that fit.

[P3nT4gR4m]

Either I'm not explaining the problem or I don't get this "other distance" thing 

So I'm going to spot the tanker from about a mile away and put myself directly in it's path then just wait patiently while he comes barrelling down the shipping lane toward me.

I can take off at a perpendicular angle at any time but there's a photo op from someone in front of me to get a shot of me piling down the wave with the tanker in shot (the closer it looks the better) average tanker or cargo freighter width is 200 feet so (if I can line up dead center, I'll have max 100 feet to cover, plus another 50 for good measure. Is this enough?

[LMNO]

OK, let's say the event point is Zero on an XY axis.  The Tanker is heading down the Y axis.  At the zero point, you want to be 150' out on the X axis.

If you start a X=0, and you travel 6mph, how long will that take, and how far away will the tanker be when you start paddling?

Ok, that sounds more manageable.  A two-parter.

One:  How long does it take to go 150' at 6 MPH?

6 MPH = 31,680 feet/hr = 528 feet/minute = 8.8 feet/second
150/8.8 = 17 seconds.

Two:
How far will a tanker go in 17 seconds at 60 MPH?

60 MPH = 316,800 feet/hr = 5280 feet/min = 88 feet/second
88x17 = 1,496 feet

So, if you really can go from 0 to 6MPH almost instantaneously, start paddling like hell when the tanker is 1,500 feet away.

But you still have an enormous chance of dying.  So, you know, don't.  Die, that is.  I know you're going to hypothetically do this anyway.

[LMNO]

Keep in mind that both your paddling speed, and the speed of the tanker are relatively unknown, so distance and time could fluctuate wildly.

[P3nT4gR4m]

Jesus. Where the fuck did I get 60mph from? Must have been a brainfart - tanker will be travelling around 15mph - I'll be tracking it's speed and heading on gps. I also know that my lenseman can do all sorts of f-stop jiggery pokery to make it look like I'm closer than I look to the leviathan. Don't worry about me dying - this is purely hypothetical 

Another thing it's possible to do is cheat a bit and not start dead center on the tanker but a hundred feet or so to the side. This is surprisingly easy to do when you know the rules.

[LMNO]

15 MPH = 79200 FPH = 1320 FPM = 22 f/s

22x17 = 374 feet

Start hypothetically paddling at about 400 feet.

[P3nT4gR4m]

Brilliant, cheers! Seeing it written down like that, it's pretty straightforward, too. I've learned me some math. Now I can tighten up that 17 secs estimate with a bit of field testing...

while we're on the subject, say I travelled at an angle of 45degrees to the Y axis, in the direction the tanker is moving, so I'll be better placed to drop into the wave, would this by any chance double the time taken to reach 150 feet clear on X?

[LMNO]

Ok, we're moving on to geometry.  Tricky.


You've got a right triangle with a single side of 150', which is AB, below

Before, you were simply going from A to B, where B is the zero point (meeting the ship).

Now, you're starting at A, and moving to C, which is the new zero point.

You're in luck, because a 45^o right triangle has a special property in that both AB and BC are the same length.
You're also in luck, because the hypotenuse can be calculated with the handy dandy formula

AB² + BC² = AC²

22500 + 22500 = AC²

45000 = AC²

The square root of 45000 is 212 (roughly)

So, now you're going 212 feet instead of 150.  Back to the formulas!

6 MPH = 31,680 feet/hr = 528 feet/minute = 8.8 feet/second
212/8.8= 24 seconds.

15 MPH = 79200 FPH = 1320 FPM = 22 f/s
22x24 = 528 feet

If you go on a 45 degree angle, start paddling about 530 feet away.  Make it 550, because who knows how exact your angle will be.

[Telarus]

This whole conversation gave me hypothetical shivers down my spine.