# Help me solve this math equation



## MisterPerfect (Nov 20, 2015)

Barry is going at (MPH)2301.81 in a striaght path. There is a object going at an angle which will run directly in front of Barry. This object is going 595 MPH. At what angle would these two objects meat and at what piont. 

Part Two 

Evan is going at 3836.35(MPH) in a striaght path. There is an object going at an angle which will run directly in front of Evan. This object is going 595 MPH. At what angel would these two objects meet and at which piont. 

I forgot the exact formula to solve this but I have the numbers.


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## Psychophant (Nov 29, 2013)

You need to know where the objects start from.


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## ae1905 (Jun 7, 2014)

it's a bit of a trick question cuz a collision can happen at any point on the straight path and at any angle...you just have to put the object in the right starting position for each solution


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## MisterPerfect (Nov 20, 2015)

ae1905 said:


> it's a bit of a trick question cuz a collision can happen at any point on the straight path and at any angle...you just have to put the object in the right starting position for each solution


Well I trying to figure out at which angle going that speed would the Object have to be to hit the person traveling at that speed. Since the person is going much faster than the object but for the object to hit you have to propel it at a particular angle. I not sure how to figure out the angle though.


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## ae1905 (Jun 7, 2014)

MisterPerfect said:


> Well I trying to figure out at which angle going that speed would the Object have to be to hit the person traveling at that speed. Since the person is going much faster than the object but for the object to hit you have to propel it at a particular angle. I not sure how to figure out the angle though.


take any point on the straight path--call it x1...in the time it takes either guy to arrive at x1 the object will travel a distance equal to the elapsed time multiplied by its speed--call this distance x2...the object can come from _any angle _as long as it travels that distance, x2--ie, the object can start from any point on a circle centered on x1 with radius x2...and as x1 increases, the radius of this circle increases in direct proportion


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## ae1905 (Jun 7, 2014)

Psychophantic said:


> You need to know where the objects start from.



actually, the starting point of the object doesn't uniquely determine the angle as there are _two _possible angles corresponding to the two circles (that I described above) that intersect at that point...so starting from the same point, the object can either travel _towards _the guys or _away _from the guys and still collide--two angles for any starting point--_except _the points that lie on the envelope enclosing all points; these have only one angle (equal to inverse sine[speed obj/speed guy] for the techies


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## Psychophant (Nov 29, 2013)

ae1905 said:


> actually, the starting point of the object doesn't uniquely determine the angle as there are _two _possible angles corresponding to the two circles (that I described above) that intersect at that point...so starting from the same point, the object can either travel _towards _the guys or _away _from the guys and still collide--two angles for any starting point--_except _the points that lie on the envelope enclosing all points; these points have only one angle


Well sure, but my point was just that we're missing necessary information.


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## MisterPerfect (Nov 20, 2015)

Psychophantic said:


> Well sure, but my point was just that we're missing necessary information.


The equation is trying to figure out at which angle you would have to be in order to hit the moving target. Since its probobly going to find anything that really goes at these speeds(Persons speed) we have to use the equation to figure it out. Its not impossible its just more hard constructing it with just the speeds. If it was slower objects I could propel the objects to hit the moving object based on skill and prediction but at these speeds eyeing it is not really possible.

For this equation the object is going left to right. So the object being propelled would either have to be to the Left or towards the middle in order to hit the moving object.

A simplified example 

John is runing left to right at a speed of 100 MPH 

I want to hit john with an arrow 

Now I have to do this by shooting where he will be not where he currently is or else ill miss 

If we are doing this on a real life scale figuring this out is pretty simple but that is at much slower speeds. 

If we have two trains on a track that go at the same time but different speed we have to do 

Speed of trains(MPH) 
Length of trains 
Angel at which they intersect 
Time at which they each departure 
Wieght of trains(but that can be taken out if we know MPH)


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## HAL (May 10, 2014)

Not enough information.

Your spelling is appalling.


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## Psychophant (Nov 29, 2013)

MisterPerfect said:


> The equation is trying to figure out at which angle you would have to be in order to hit the moving target. Since its probobly going to find anything that really goes at these speeds(Persons speed) we have to use the equation to figure it out. Its not impossible its just more hard constructing it with just the speeds. If it was slower objects I could propel the objects to hit the moving object based on skill and prediction but at these speeds eyeing it is not really possible.
> 
> For this equation the object is going left to right. So the object being propelled would either have to be to the Left or towards the middle in order to hit the moving object.
> 
> ...


That still doesn't provide enough information to find a single angle since I don't know where you're standing or when you shoot the arrow.


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## ae1905 (Jun 7, 2014)

MisterPerfect said:


> The equation is trying to figure out at which angle you would have to be in order to hit the moving target. Since its probobly going to find anything that really goes at these speeds(Persons speed) we have to use the equation to figure it out. Its not impossible its just more hard constructing it with just the speeds. If it was slower objects I could propel the objects to hit the moving object based on skill and prediction but at these speeds eyeing it is not really possible.
> 
> For this equation the object is going left to right. So the object being propelled would either have to be to the Left or towards the middle in order to hit the moving object.
> 
> ...




the equations are 

angle moving _towards _the guys = inverse tangent [y/[x-x1)] 
angle moving _away _from the guys = inverse tangent [y/(x1-x)]

where x1 is the point of collision, (x,y) is the starting point of the object, and (0,0) is the starting point of the guys

note that x, y, and x1 are not all independent variables...you can either choose an x1 in which case (x,y) is determined, or you can choose (x,y) in which case x1 is determined...the equation relating x, y, and x1 is a simple quadratic algegraic equation involving the two speeds...I'll let you figure that out yourself


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## Simpson17866 (Dec 3, 2014)

@MisterPerfect Let's say you have an object starting at xy-coordinate (0,-1) going up 1 unit per time, while another object is moving at 2 units per time. Even if we know that the objects will collide at (0,0), that still doesn't tell us if the faster object is moving

left from an initial position of (2,0) for a 90º angle
right from (-2,0) for another 90º angle
down from (0,2) for a 180º angle
up from (0,-2) for a 0º angle
down-right from (√2,√2) for a 45º angle...

do you see the problem we're having?


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## g_w (Apr 16, 2013)

ae1905 said:


> it's a bit of a trick question cuz a collision can happen at any point on the straight path and at any angle...you just have to put the object in the right starting position for each solution


It's even trickier than that...they might miss altogether.


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## sprinkles (Feb 7, 2010)

The angle shouldn't change the impact point if the two vectors are intersecting since the velocity is not changing. 

Imagine that two objects are going from the edge of a circle to the center on straight lines at the same speed. They will always meet in the center no matter where on the edge they start. The angle is not defined.


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## MisterPerfect (Nov 20, 2015)

Simpson17866 said:


> @MisterPerfect Let's say you have an object starting at xy-coordinate (0,-1) going up 1 unit per time, while another object is moving at 2 units per time. Even if we know that the objects will collide at (0,0), that still doesn't tell us if the faster object is moving
> 
> left from an initial position of (2,0) for a 90º angle
> right from (-2,0) for another 90º angle
> ...


Both Objects are moving at different speed but the goal is for the Object to hit the much faster moving object which is moving in a straight line. Since the object at an angle is slower than the object it can only attempt to hit an object based on will it will BE vs trying to hitting it striaght like a still object. 

I think 45 degree angle would work for the slow moving object.


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## sprinkles (Feb 7, 2010)

I think it would be something like _target speed_ divided by _projectile speed_ multiplied by the distance of the projectile from the target vector at an angle of 90 degrees off the target vector. 

So if your projectile is 1 mph and the target is 1000 mph, you would have to aim at a point that is 1000 miles in front of the target for every mile of distance you are away from the target vector intersection.


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## MisterPerfect (Nov 20, 2015)

sprinkles said:


> I think it would be something like _target speed_ divided by _projectile speed_ multiplied by the distance of the projectile from the target vector at an angle of 90 degrees off the target vector.
> 
> So if your projectile is 1 mph and the target is 1000 mph, you would have to aim at a point that is 1000 miles in front of the target for every mile of distance you are away from the target vector intersection.


*MPH)2301.81(barry) Diveded by (Projectile)595 MPH X Distance136957.95 at Degree(?)*

Like that?


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## Psychophant (Nov 29, 2013)

You just don't have enough information for any practical purpose, which makes this a frustrating problem to solve since you'll need a bunch of parameters. This depends on _where_ you're standing and the object starts, and _when_ you throw. In theory, if you say your moving target is translating along the x-axis at some velocity dx/dt (starting at some big negative value at t = 0) and you're standing on the y-axis at some point, _when_ you throw changes everything. In most cases, you could simply toss it straight ahead if you timed correctly and the two would collide.

I assumed you threw at t = 0 and used x0 as the starting parameter (where the target starts), and I got this mess of an equation. Give us the problem you're actually trying to solve and this will be cake, but solving it generally just makes it a pain for no good reason.










Vp is the projectile velocity vector, s is its speed (or |Vp|), Vt is the speed of the target (it only moves along x, so I'm making it a scalar), theta is angle from that dotted horizontal line and the angle you shoot the arrow or whatever from, tc is the time it collides at, and the final expression relates your choice of y0 (where you stand) and x0 (where the moving object starts) to theta, so if you plug those in you _could_ solve easily with a calculator. But what on earth are you using this for? I think you're failing to grasp the complexity of the problem when you phrase it as you do.


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## sprinkles (Feb 7, 2010)

MisterPerfect said:


> *MPH)2301.81(barry) Diveded by (Projectile)595 MPH X Distance136957.95 at Degree(?)*
> 
> Like that?


Something like that, but what you're looking for is a vector which is a speed and a direction. It can't be done with speed alone. If you have vectors you can do some trig to figure things out because it would work like the sides of a triangle.


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## HAL (May 10, 2014)

@MisterPerfect Here's a general model with every possible variable. As you can see, there's so much stuff missing from your description. The whole premise is incomplete. I wonder if you're only here to get as many people as possible on this wild goose chase.


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## ae1905 (Jun 7, 2014)

Metalize said:


> this, with the other series of posts in this vein, honestly smacks of someone feeling bitter about their own age and projecting it outwards.
> 
> *Why else would you randomly insult someone else's presumed age and intelligence *in a discussion about a somewhat *ambiguous math problem*? And getting angry over a better response isn't going to automatically make yours correct, it'll just make it more apparent that you don't actually understand the problem.


first, the problem is not ambiguous...that is the original point of contention

second, I explained why I question his age and knowledge of physics when I wrote in answer to his complaint that I ignored his solution:



> your "solution" was so inept, you were left with an equation you couldn't solve except iteratively on a calculator and only for individual cases
> 
> why would you not simplify the solution and put one of the objects at the origin?...that's the first thing I noticed and told me immediately that you are a klutz and your solution was not worth my time
> 
> ...


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## Psychophant (Nov 29, 2013)

ae1905 said:


> really!...what school?


A relatively prestigious one, actually. Not that it matters.


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## ae1905 (Jun 7, 2014)

Psychophantic said:


> Grab a calculator or find one on the internet? Do I need to explain to him that his calculator's cosine function uses a series expansion to give him a precise estimate of the function's value too? The guy wanted a method and I gave him one.


you still don't understand what I'm asking you...you can't directly calculate theta from your solution...that's why you admitted it was "not easy to solve"....so what exactly does the calculator's "numeric solver" do when it solves for theta?

hint: it's more than just approximating the cosine by a truncated series expansion


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## ae1905 (Jun 7, 2014)

Psychophantic said:


> A relatively prestigious one, actually. Not that it matters.



why don't you let us judge how "relatively" prestigious it is?


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## Psychophant (Nov 29, 2013)

ae1905 said:


> you still don't understand what I'm asking you...you can't directly calculate theta from your solution...that's why you admitted it was "not easy to solve"....so what exactly does the calculator's "numeric solver" do when it solves for theta?
> 
> hint: it's more than just approximating the cosine by a truncated series expansion


Depends, actually. My calculator often gives exact answers if possible, and if not, it gives a very high precision approximation, much like it does for just about everything else. If it is an approximation, it's almost always good enough for all practical purposes.

You insist this isn't a dick measuring contest but want to know about my education?


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## sprinkles (Feb 7, 2010)

Like I said before


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## ae1905 (Jun 7, 2014)

sprinkles said:


> Like I said before



no, that's not the problem posed by the op...in your diagram, the angle is known and the only unknown is the time the object is fired...in the op's problem, it is the opposite: the angle is not known, but the time of release is


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## ae1905 (Jun 7, 2014)

Psychophantic said:


> *Depends, actually.* My calculator often gives exact answers if possible, and if not, it gives a very high precision approximation, much like it does for just about everything else. If it is an approximation, it's almost always good enough for all practical purposes.
> 
> You insist this isn't a dick measuring contest but want to know about my education?



what does it "depend" on?


and this isn't a contest...I just wanted to confirm my intuition...you've answered that question to my satisfaction


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## sprinkles (Feb 7, 2010)

ae1905 said:


> no, that's not the problem posed by the op...in your diagram, the angle is known and the only unknown is the time the object is fired...in the op's problem, it is the opposite: the angle is not known, but the time of release is


The angle is chosen because we don't need to calculate it. The time of release is what you want. If the target is traveling on a vector in front of you, you can choose to fire at any angle which intersects that vector. 90 degrees is the most optimal to actually hit the target because it's the least distance for the projectile to travel. The only reason you'd want another angle is if something is in the way.

This is the right answer for the wrong question.


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## Psychophant (Nov 29, 2013)

ae1905 said:


> what does it "depend" on?


Whether or not algebraic rules can be applied to give an exact expression. I think I'm done here. If you want to unveil your brilliant technique that somehow uses only the information he gave to provide viable, numerical solutions (absent other undefined variables), go ahead.. otherwise, stop. I don't know the details of most computer algebra systems and it's not relevant anyway.


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## sprinkles (Feb 7, 2010)

ae1905 said:


> no, that's not the problem posed by the op...in your diagram, the angle is known and the only unknown is the time the object is fired...in the op's problem, it is the opposite: the angle is not known, but the time of release is


Also you missed the same important point the OP did which is how far apart the objects are. You can't calculate when they will intersect because of this, but you can calculate where they will intersect and when you need to fire.

It doesn't help to know that Barry is going 2301.81 mph if you _don't know how far away Barry is. _ Barry could be starting a million miles away. However, choosing when to shoot instead of where is still a working solution as long as Barry starts far enough away that it's not impossible for the projectile to intersect before Barry passes by.


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## sprinkles (Feb 7, 2010)

Also if you shoot at anything other than 90 degrees you will need to calculate an even longer lead time, because the distance to intersect is longer, which means the projectile takes longer to get there which means your lead distance must be proportionally longer.


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## ae1905 (Jun 7, 2014)

sprinkles said:


> Also you missed the same important point the OP did which is how far apart the objects are. You can't calculate when they will intersect because of this, but you can calculate where they will intersect and when you need to fire.
> 
> It doesn't help to know that Barry is going 2301.81 mph if you _don't know how far away Barry is. _ Barry could be starting a million miles away. However, choosing when to shoot instead of where is still a working solution as long as Barry starts far enough away that it's not impossible for the projectile to intersect before Barry passes by.


that's one way to deal with the unknown but it doesn't work if the guy is too close




Psychophantic said:


> Whether or not algebraic rules can be applied to give an exact expression. I think I'm done here. If you want to unveil your brilliant technique that somehow uses only the information he gave to provide viable, numerical solutions (absent other undefined variables), go ahead.. otherwise, stop. I don't know the details of most computer algebra systems and it's not relevant anyway.


I gave the equations in a post above...you need a relation between x, y, x1,and x2...you have

x2=x1*u2/u1

and the equation for the circle of radius x2 centered on x1

(x-x1)^2 +y^2 = x2^2

if (x,y) is given, then simply solve these equations for x1...it's a quadratic algebraic equation that has an explicit solution...then plug x1, x, y into the equations for the angle

when I thought about this problem and realized that any point (x,y) on a circle of radius x2 centered on x1 could serve as a starting point for the object, this picture came into my mind:










it depicts the sounds waves (the blue circles) emanating from a jet traveling at supersonic speeds...sound propagates within the triangle bound by the jet at the apex and the trails of shock waves on its two sides (actually, in 3D it's a cone, not a triangle, but our problem is in 2D, as is this image, so I call it a triangle)...simply put the guy at the apex and reverse the direction of travel from right to left...the circles become all the points from which the object can be fired and intersect the guy at their centers...obviously, there are an infinite number of such circles and points...but only those points _within _the triangle can hit the object...points outside start too far from the guy to ever intersect...notice too that each point inside the triangle consists of the_ intersection of two circles_...so an object shot from that point can be fired in _two different directions_ corresponding to the radii of these two circles...the exceptions are objects shot from the outer bounds (sides) of the triangle...these objects can be fired at a single angle only, namely, inverse-cosine(u2/u1)


moral of the story: it pays to think about the problem before you start jotting down equations


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## Psychophant (Nov 29, 2013)

ae1905 said:


> I gave the equations in a post above...you need a relation between x, y, x1,and x2...you have
> 
> x2=x1*u2/u1
> 
> ...


This doesn't address the complaint... So you assumed the object starts at the origin, fine, but now you need the archer's starting position. He still didn't provide enough information to find some finite set of acceptable solutions. _That_ was the complaint. And, ironically, this is a lot harder to work through for someone with limited algebra experience than plugging a single expression into a calculator and getting a result. Your comments were rude, unnecessary, and you _completely_ misunderstood (probably deliberately) what was being said. Had you simply posted this before, we could have avoided pages of useless debate, but clearly it was more important to jack yourself off (with very stupid complaints, honestly) before posting a different method for doing the exact same thing that didn't actually rectify any of the other concerns. Well done.


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## Jamaia (Dec 17, 2014)

This thread may be the strangest thing I've seen on PerC roud:



MisterPerfect said:


> Barry is going at (MPH)2301.81 in a striaght path. There is a object going at an angle which will run directly in front of Barry. This object is going 595 MPH. At what angle would these two objects meat and at what piont.
> 
> Part Two
> 
> ...





MisterPerfect said:


> Would it be better if I specified what the exact purpose for the Equation is? Instead of just breaking it into Objects, names and Numbers?
> 
> Reasoning
> 
> Its trying to figure out if a object moves at these speeds can be shot with a gun or would the gun miss. You can shoot moving objects but in order to do so you have to shoot where they will be vs where they are. In order to figure it out you must figure out the speeds, and than at which angle the gun is shot. To do that we must figure out the Angle and intersecting piont. Someone sugested 45 which would probobly be reasonable to start at as it would be further in front of the object. Than we have to figure out if there is a piont of collison and which angles would work for this equation.


So the question is, can you shoot Barry and later on Evan with a very sluggish bullet, with a hundred year old pistol perhaps? Barry is in an aircraft traveling at the speed of Mach 3 and Evan is in a space shuttle traveling at the speed of Mach 5. Can you hit them, sure you can, theoretically, assuming the dudes aren't moving directly away from you faster than your bullet travels. You could even throw a dead goose at them and have it hit, since theoretically there can always be a point of collision. But of course in IRL you'd miss. 

In real life your biggest issues would be that the target is too fast, your reactions are too slow and your gun is rubbish. IRL your only chance of hitting the plane would be to stand directly in front of it and shoot directly at it just as it's about to hit you. Does it count as a hit, if you're holding the gun with the bullet still in it when the plane hits you?


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## ae1905 (Jun 7, 2014)

Psychophantic said:


> This doesn't address the complaint... So you assumed the object starts at the origin, fine, but now you need the archer's starting position. He still didn't provide enough information to find some finite set of acceptable solutions. _That_ was the complaint. And, ironically, this is a lot harder to work through for someone with limited algebra experience than plugging a single expression into a calculator and getting a result. Your comments were rude, unnecessary, and you _completely_ misunderstood (probably deliberately) what was being said. Had you simply posted this before, we could have avoided pages of useless debate, but clearly it was more important to jack yourself off (with very stupid complaints, honestly) before posting a different method for doing the exact same thing that didn't actually rectify any of the other concerns. Well done.


your Fi is showing, infp

my solution does address the complaint and offers a direct and easy way to calculate any angle given the position of the object...it also yields insights into the nature and number of solutions

your "solution", otoh, did none of that...it was garbage--like your brain

and go back and _read _and _think _about my earlier posts...I _did _explain this solution...it was _you _who didn't understand what I was talking about


tbh, unintelligent infps like you should never study the hard sciences--it overtaxes your *inferior *Te


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## Psychophant (Nov 29, 2013)

ae1905 said:


> your Fi is showing, infp
> 
> my solution does address the complaint and offers a direct and easy way to calculate any angle given the position of the object...it also yields insights into the nature and number of solutions
> 
> ...


Lol, no, it does not. The information it requires was never specified by the OP thus it fails (yes, I understand the method perfectly and that's not the point). That is the final time I repeat that. 

And this is seriously pathetic. You're old enough to sport the Gen-X tag and _this_ is still how you settle disagreements? Take it elsewhere, old man.


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## ae1905 (Jun 7, 2014)

Psychophantic said:


> Lol, no, it does not. The information it requires was never specified by the OP thus it fails (yes, I understand the method perfectly and that's not the point). That is the final time I repeat that.


yeah, it does...you're just making excuses for your pathetic display of "skillz"



> And this is seriously pathetic. You're old enough to sport the Gen-X tag and _this_ is still how you settle disagreements? Take it elsewhere, old man.


I suppose this was "age-appropriate"?:



psychopathetic said:


> Your comments were rude, unnecessary, and you _completely_ misunderstood (probably deliberately) what was being said. Had you simply posted this before, we could have avoided pages of useless debate, but clearly it was more important to *jack yourself off *(with very stupid complaints, honestly) before posting a different method for doing the exact same thing that didn't actually rectify any of the other concerns. Well done.


now fuck off, little man...you've wasted enough of my time


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## sprinkles (Feb 7, 2010)

ae1905 said:


> that's one way to deal with the unknown but it doesn't work if the guy is too close
> 
> 
> 
> ...


I know this. This is why you don't need to/can't calculate an angle and this has been my point the whole time. 

As I said before, the projectile can intersect at any angle, and in fact an infinite number of angles. This is why it cannot be defined by speed and you need to calculate the release point instead, because that is the ONLY thing you can.










As you can see here, the projectile can travel at any angle which is lower than parallel to the target vector. It's an infinite number of angles therefore you cannot calculate a specific angle that you need to hit. The only thing that changes is the lead distance. Angles further off perpendicular have increasingly long lead times increasing to infinity.


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## sprinkles (Feb 7, 2010)

Also my method gives the worst case scenario. If the target and projectile can collide at all, it will at least be at 90 degrees. If the target is too fast to hit at 90 then it is too fast to hit period because that's the closest you can get to the target.


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## MisterPerfect (Nov 20, 2015)

Jamaia said:


> This thread may be the strangest thing I've seen on PerC roud:
> 
> 
> 
> ...


Seriously? Of all the things that I or anyone on this forum have posted. This is the strange one? Yes, that was the idea.


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## sprinkles (Feb 7, 2010)

Jamaia said:


> This thread may be the strangest thing I've seen on PerC roud:
> 
> So the question is, can you shoot Barry and later on Evan with a very sluggish bullet, with a hundred year old pistol perhaps? Barry is in an aircraft traveling at the speed of Mach 3 and Evan is in a space shuttle traveling at the speed of Mach 5. Can you hit them, sure you can, theoretically, assuming the dudes aren't moving directly away from you faster than your bullet travels. You could even throw a dead goose at them and have it hit, since theoretically there can always be a point of collision. But of course in IRL you'd miss.
> 
> In real life your biggest issues would be that the target is too fast, your reactions are too slow and your gun is rubbish. IRL your only chance of hitting the plane would be to stand directly in front of it and shoot directly at it just as it's about to hit you. Does it count as a hit, if you're holding the gun with the bullet still in it when the plane hits you?


What Barry is and what the projectile is practically speaking are irrelevant because you could easily replace them with missiles or anything that goes fast.

Also if we can land a probe on Mars we can hit a plane with a bullet. You'd just have to try hard enough.

Edit:
Also to put that in perspective, Mars is 249 million miles away and is moving at almost 54,000 mph. But we still hit it.


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## Jamaia (Dec 17, 2014)

sprinkles said:


> What Barry is and what the projectile is practically speaking are irrelevant because you could easily replace them with missiles or anything that goes fast.
> 
> Also if we can land a probe on Mars we can hit a plane with a bullet. You'd just have to try hard enough.
> 
> ...


Lols . But I got it right. He was asking a real life question, to which there is a real life answer with the limited information provided.

@MisterPerfect the question was a bit strange, but this thread was stranger ! 

If you know all the variables, if you know exactly what route the aircraft is going to go and at what time, if you've done measurements of how the gun works, the delays, the range, the flight patterns, etc. and the weather or wind or such things are not an issue, you could calculate where you need to position your gun, where to aim it and exactly at what time it needs to be fired to get the plane hit the bullet. That's what the others have argued about, you could really pick any angle you like and just adjust the time when your gun will go off, the limiting factor is the range of your gun and the accuracy of your measurements. The range determines how far the bullet can travel before you need to have it's trajectory intersect with the air crafts trajectory, if you want any chance of a hit. But there's no one angle and one point of collision.

But without the precalculations and having set up the gun in advance etc. your chance of hitting the plane as it goes past are up to luck. The aircraft goes so fast, you can't aim or estimate it's speed before it's gone. Basically you'd have to fire blindly towards the direction where you think the plane will be before the plane comes. 

If want to try to vision it, let's say your pistol has a point-blank range of about the distance the bullet travels in one second, that's accurate enough for this I think. With the assumed 595 mph speed and negligible air resistance etc, in one second your bullet travels 270 meters or about 300 yards (that's about right for MPBR). So you'd want the bullet to leave the muzzle one second or less before the collision, and you could stand at most 300 yards away from future collision point. Your best chance of hitting the plane would be if you stood directly in it's path and fired at it, that way you can maybe see the far away plane as a dot and aim at it without having to move your hand with the target. If you want to be alive after, you could be slightly off the plane's path to any side. But because everything is so vague in this example, I'm not going to take one yard off the range because of that angle.

At most the Mach 5 plane could be about 2 km or 1,2 miles away at the moment when the bullet has to be in the air, that is if the plane is moving directly towards you and the bullet. You would have to consider the delay in your reaction time and the delay of the gun. Maybe the combined delay is 0.5 seconds. Every second more you need puts the plane 1700 meters further. You have positioned yourself close to where you know the plane will go, you see the plane as a dot in the distance, you fire your gun at it, the bullet leaves the muzzle after the delay, the plane has within one second come a mile closer, another second or half, the plane sweeps off your airborne bullet (collision), in 0.2 seconds the plane passes you and perhaps doesn't hit you too. One more second and the plane is a mile away again.

For the Mach 3 plane you'd have a better chance since it moves half the speed. But still it's about knowing the planes trajectory, aiming at it when it's still very far away and getting the timing right with maybe one second room to choose your time. If you intend to be standing at 45 degree angle, you'd have a correct time frame of "launching" the bullet of maybe milliseconds, which is impossibly small window considering human reaction times.


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## sprinkles (Feb 7, 2010)

@Jamaia

That's right.

Though if you miss the first time you have more chances to hit.

For example if the lowest angle you can hit at is 45 degrees, you can also hit at every angle between 45 and 135 degrees, so you could launch a spray of projectiles at many intersecting angles. 

135 degrees works the same as 45 degrees because they both have a hypotenuse of the same length. The angle is equal, it is just flipped across your axis and the lead time is the same but the aim point is inverted. And of course you can aim at anything in between.


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## Psychophant (Nov 29, 2013)

I think you'd need to automate this if you wanted any chance of it actually working in most cases heh.


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## Jamaia (Dec 17, 2014)

sprinkles said:


> @Jamaia
> 
> That's right.
> 
> ...


There's no time to even determine if it was a hit or miss. The plane is within the guns reach for half a second and then it's gone. His gun will probably just blow up when it's fired anyways. But if he had a gun that would shoot a spray then sure, his best bet would be to sweep the general area. 

One thing I forgot was that at super- and hypersonic speeds I don't know if the bullet would actually reach the plane (come in contact with it). I don't have the time for that now but the bullet might just burn (if it's like a musket ball) or be swept aside by the shock layer.


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## RobynC (Jun 10, 2011)

You need the starting point of both objects, and the angle relative to each other. You also need to know if they are at the same altitude


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## sprinkles (Feb 7, 2010)

Jamaia said:


> There's no time to even determine if it was a hit or miss. The plane is within the guns reach for half a second and then it's gone. His gun will probably just blow up when it's fired anyways. But if he had a gun that would shoot a spray then sure, his best bet would be to sweep the general area.
> 
> One thing I forgot was that at super- and hypersonic speeds I don't know if the bullet would actually reach the plane (come in contact with it). I don't have the time for that now but the bullet might just burn (if it's like a musket ball) or be swept aside by the shock layer.


It depends on the inertia of the projectile. Many projectiles could probably penetrate the shockwave before being diverted, especially because the shockwave is coming so fast. 

It's like if you throw a bullet at a soda can, it'll just knock the can over, but if you fire the bullet out of a gun it will penetrate.

Edit:
This also relates to inertial dampening. Forces that act quickly have less time to overcome inertia as illustrated by the inertial ball experiment, where you have a lead ball hanging from a string, and another string hanging from the ball. If you pull gently on the bottom string, the top string will break first, but if you yank very fast on the bottom string, the bottom string will break first because the force passing through the string is dampened by the inertia of the ball, which protects the top string from breaking.


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## sprinkles (Feb 7, 2010)

Also we would be looking at the sum of the two velocities.

From a relativistic perspective, a mach 5 plane would be hitting a mach 5 bullet from the plane's perspective.


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## Jamaia (Dec 17, 2014)

**


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## Jamaia (Dec 17, 2014)

sprinkles said:


> Also we would be looking at the sum of the two velocities.
> 
> From a relativistic perspective, a mach 5 plane would be hitting a mach 5 bullet from the plane's perspective.


Or Mach 6 if combined, but still regardless of perspective, the shock waves are propagating with the plane against the direction of the bullet (if shot head on). So while there would be an impact of equal force of shooting a plane at standstill with a "Mach 6 bullet", the impact in this case might just be the end of the bullet and not touch the plane. The load of inertia is on the plane and very little is on the bullet even if the impact itself is large. Is it the plane or the air condensed around the plane that takes the hit, I don't know and didn't quickly find the answer online, so I'm hoping @MisterPerfect will try this and report back to us =).

Edit: Also, do try to hit the hornet with a dead goose too. Take a video


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## HAL (May 10, 2014)

ae1905 said:


> honestly, if you gave that answer on an exam, you would be lucky to get 1 point...I would give you 0





ae1905 said:


> your effort befits a high school student


Aside from that fact that this is not the first time you've tried to be 'the one with the most knowledge' in a thread where it's totally irrelevant, and aside from the fact that you became a total dickhead toward all the rational responses in this thread in particular, I'd like to say that if you're going to start attacking people for their intellectual powers in any way, you need to first address the shocking quality of writing in your posts, wherein the efforts befit that of a high school student (or worse) and for that alone you'd be lucky to get 1 point. In fact I would give you 0.

By the way your use of circular waves emitting from a faster-than-sound object is absurd. You make so many assumptions that are simply not given in @MisterPerfect 's ridiculously vague description.

Everyone in this thread, apart from you, has made a good effort at giving the most general solution such that any variable can be plugged in to form an answer, _as soon as the starting variables are known_.

Now STFU.


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## ae1905 (Jun 7, 2014)

HAL said:


> Aside from that fact that this is not the first time you've tried to be 'the one with the most knowledge' in a thread where it's totally irrelevant, and aside from the fact that you became a total dickhead toward all the rational responses in this thread in particular, I'd like to say that if you're going to start attacking people for their intellectual powers in any way, first address the shocking quality of the writing in your posts, wherein the efforts befit that of a high school student (or worse) and for that alone you'd be lucky to get 1 point. In fact I would give you 0.


lol

you and I have answered the same questions in a number of threads, and each time I've schooled you...easily...not that I was out to embarass you...I was simply answering the questions...it was _your ignorance _that made you look bad

so _you _deserve all the credit for your impotent performance, hal...that's how brilliant you are


btw, you didn't get your physics degree from trump U, did u?...if you did, there's still time to get your money back


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## HAL (May 10, 2014)

ae1905 said:


> lol
> 
> you and I have answered the same questions in a number o threads, and each time I've schooled you...easily...not that I was out to embarass you...I was simply answering the questions...it was your ignorance that made you look bad
> 
> ...


Dude you came across as a bit of a dick in all those threads, clearly trying to prove a point where there was nothing to prove. Weird that you think it's about 'schooling' other respondents, too. Personally, I just gave answers, then you gave weird cryptic counter-answers that didn't really make a point here or there. 

What did/do you study by the way? And where?


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## ae1905 (Jun 7, 2014)

HAL said:


> Dude you came across as a bit of a dick in all those threads, clearly trying to prove a point where there was nothing to prove. Weird that you think it's about 'schooling' other respondents, too. Personally, I just gave answers, then you gave weird cryptic counter-answers that didn't really make a point here or there.
> 
> What did/do you study by the way? And where?



dude, don't blame me for answering the questions correctly

blame yourself for giving the wrong answers


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## Theories (Mar 24, 2016)

Are you referring to, Barry Allen? As in, The Flash? If so, he can phase dodge that and it will have no point of impact.

If not then oh well. Maybe this might help: https://www.mathsisfun.com/algebra/trig-solving-ssa-triangles.html

And give OP some slack about his grammar. English may not be his native language, so he may actually be doing better than you would with his native language.


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