Space Wars Challenge

Yes it would be midway between.

As both ships are going at the same velocity they form their own 'inertial frame of reference' thus are stationary relative to each other. e.g. No need to shoot ahead of the other ship (there is no aerodynamic drac on the bullet)...blimey they could even shoot arrows at each other!

It seems to boil down to simple vectors...so I'm probably wrong.

Del

Hi Jorrie - Thanks for naming Me the observer on the moon. It is the closest I will ever get to space.

Drawing a sketch I realised I would not know.

The moon I am standing on is also moving and rotating their relative speeds in relation to each other therefore cannot be the same.

Do we ignore my movement?

The second question - How do I observe? Sound , light or gravitational waves. (ok sound should be out).

Third question - What . . . How do I resign as the observer?

Hi Hendrik.

Del thinks it's too easy while you think it's too difficult...

It is however all about relative movement, but ignore the Moon's rotation - it's sooo slow.

Jorrie

Oh...just re-read it...

chasing each other!

I had assumed they were broadside to eachother!

However they can't be chasing eachother! If one is behind the other, then one is chasing and one is being chased.

Del

Whoo! I didn't catch it on first read either.

However they can't be chasing eachother! If one is behind the other, then one is chasing and one is being chased.

That is unless they are in orbit (implied by "horizontally") at 180º from each other. One fires "forward" and one fires "back". They still have their own frame of reference, and the projectiles would collide midpoint in the orbit section between the two ships.

I think

Ha...

Horizontally at constant speed...relative to you.

Hmmm...what does that mean????

The horizon is curved...does it mean they are on an orbital path? Or that they are on a straight path?

I s'pose a curved path would not be at constant speed relative to you....But why doesn't he just say velocity if they are on a straight path?

I think I'll order another moon beer and think some more.

Hi Del, yea, should have said one chasing the other.

For other lateral thinkers, no they are also not in orbit around the Moon - too darn slow for war-making. For the duration of this short test, they just zip by at constant speed on near straight paths...

Jorrie

Okay, two frames of reference and this is what I would observe. I assume that the two ships are on the same vector, one behind the other?

The lead ship's projectile has a lower relative speed (to me) than the trailing ship's projectile.

An object that has greater velocity will appear shorter to an observer on a "stationary frame of reference than the slower projectile. The trailing ship's projectile will appear shorter because the projectile's velocity is the sum of the muzzle velocity and the ship's velocity. What I don't know is from which end does it shrink? I am going to guess it is the center, so both leading and trailing edges of the projectile shrink toward the projectile's center point. The lead ship's projectile will have an apparent lower velocity to the observer and thus shrink less.

Due to this shrinkage of the trailing ship's projectile the impact would appear to be biased slightly toward the trailing ship.

I will have to think if there are other relativistic effects to consider, but I can't right now because I have to run to work!

Sorry, I don't understand "from which end does it shrink"? The "shrinkage would be a "relativity" over-all shrinkage, where 1 meter rulers would still be 1 meter long to the people inside the ship, but to an outside observer, the meter stick would be less than 1 meter long, depending on the speed the ship is travelling, relative to velocity c. Otherwise, a person in the front part of the ship, in your message, might have shrunk smaller than a person in another part of the ship, which does't make sense. They are all going at the same speed.

Yes, but imagine that meter stick traveling at relativistic speed. To another observer outside the inertial frame of reference for the meter stick taking an instantaneous measurement (comparing his meter stick to the one in motion) finds a discrepancy.

L = L0 (1-v2 /c2)^1/2

This makes sense, but from the outside observer from which end does the meter stick shrink (the leading edge, trailing edge, or both)? This is a difficult question, because the leading edge has a constant velocity (in our thought experiment) and if true, the trailing edge must gain a slight amount of velocity to shrink toward the leading edge. Or does the trailing edge remain at a constant velocity and the leading edge slow down slightly? Or do they shrink towards the middle of the meter stick?

Again, this is only from the outside observer's point of view and the shrinkage is homogenous in the direction of travel.

This effect is not what Jorrie is talking about that drives the challenge question's answer.

If they were coming head to head with each other I would say that collision would be at mid point. If one was being chased by the other they should still occur at mid point. The ship being chased and firing from the rear would reduce the velocity of the projectile by it's own speed. Say the point at which the projectile is fired is at point X. The chase ship would naturally increase the speed of it's projectile relative to the projectile released by the lead ship. Both projectiles could theoretically, depending on speed of ships and muzzle velocity, meet at point X. Mid point between the ships where the lead ship fired it's round.

The more I think about it, the more I question my answer.

I'd say 3/4 of the distance between the two ships, or 1/4 of the distance behind the lead ship. The velocity of the projectile should be considers to exceed the velocity of the ships but the speed of the ships will slow the projectile of the lead ship and would be an adder to projectile of the trailing ship.

if that makes any sense at all??

Wait, isn't the velocities a sum? For instance, if the pursuing ship's velocity is X and the lead ship is Y, then it should be a case of vector math. Let the projectile velocities be denoted by X1 for the pursuing ship's projectile and Y1 for the lead ship. Then:

X = Y <-- Both lead and pursuit ships have the same velocity and the same direction.

X1 = |-Y1| <-- where the absolute value of the lead ship's projectile is equal to the pursuit ship's projectile velocity, but has the opposite sense (opposite vector direction), so its sign is negative.

The pursuit ship's projectile velocity = X + X1.

The lead ship's projectile velocity = Y + (-Y1).

So, the observed projectile velocities for the observer on the Moon, which is taken as a stationary frame of reference, has the Pursuit ship's projectile velocity faster than the lead ship.

Given Jorrie is submitting the challenge we should assume that there is a relativistic ace up his sleeve. :D

Ace up his sleeve

I agree with you otherwise he would have used the example of two horse riders shooting at each other (Western movies) - with up to 40 rounds per load you could calculate the average.

:)

Hi Hero. You are making a valiant effort to calculate this lot!

However, there is just a principle involved here that 'solves' the challenge completely. It is obviously good to put symbols and numbers into formulas (that's the engineer's way of understanding stuff), but sometimes that can get us bogged down severely - as you probably know very well...

Jorrie

Doh!

You did state the problem such that the frame of reference at impact was the position of the two ships, not the observer's.

Not counting the relativistic effects, the point of impact would be precisely between the two ships. at the moment of impact since both ships and projectiles are in the same frame of reference (i.e., the ships reference frame).

Hi Jorrie

If they would be using laser beams, always visible, the beam leaving the chasing craft would be shorter to the point of impact so the point of impact would be closer to the chaser. It has traveled towards the point of contact, were as the chased has traveled away from it. The point of contact was determined at the time of simultaneous projectile release. Chaser would be there first. I think, or at least the visualisation part of my brain does so.

Good morning. Ky.

"If the muzzle velocities were identical, did that collision happen exactly halfway between the two spaceships?"

In a word, NO. The collision would be closer to the chasing ship! How close would depend on velocities of both ships and projectiles.

What if you think of the problem this way. Think of two guns facing each other on either side of a bridge. If both are fired at the exact same point at time and both have the exact same velocities, just in opposite directions, they would meet exactly at the center of the bridge (or at least cross).

Do you agree with the bridge analogy?

But wait! The Earth is moving! It is moving in orbit around the Sun and the Sun around the Milkyway, and so on and so on.

So, what if you think of the two space ships as one long space ship. The two ships were traveling at the same velocity and the same vector direction. Now what's the difference between the two scenarios?

If what you said is true, why wouldn't the two projectiles fired from either side of a bridge meet toward the perusing side of the bridge? Or if you just fire a bullet on Earth, if your prediction is true, the bullet would never fly straight! If you point the rifle into the direction the Earth was moving when you fired it would come out the back of the barrel. We know that doesn't happen, right?

Hi Stan.

Care to give us a rationale behind your brief answer? If you think it will spoil the fun for others, it's fine to delay it a bit...

Jorrie

This is probably wrong and I am probably missing something, but:

The velocity relative to me of the projectile fired from the ship behind is the velocity of that ship plus muzzle velocity. The velocity of the projectile relative to me fired by the ship ahead is muzzle velocity minus the velocity of the ship ahead.

So the closing velocity is biased toward the ship behind and the actual point in space that the collision occurs would be be closer to the location of the ship ahead at the time the projectiles were fired. But at the time of collision the ships have continued to move and so it appears to me standing on the moon that the collision occurs exactly half way in between the ships at the time of impact.

The actual location of impact is not the mid point between the ships when the projectile was fired, but it is the mid point between the ships when the projectiles collided.

This hurts my head in a way that is peculiarly pleasing....

Hi Steve, you wrote: "The actual location of impact is not the mid point between the ships when the projectile was fired, but it is the mid point between the ships when the projectiles collided."

Yep, the latter is the mid-point we are talking about. But will the bullets really collide there?

Jorrie

Not to the observer. The speed of the bullets relative to the observer on the moon would differ based on the velocity of the ships even though muzzle velocities were identical.

If the observer were on one of the ships then the collision would have taken place at the midpoint between the 2 ships.

"The speed of the bullets relative to the observer on the moon would differ based on the velocity of the ships even though muzzle velocities were identical."

I buy into that, but what I don't understand is why it appears different to the Moon observer?

Let's say for simplicity that the muzzle velocity and ship velocities were identical. If the lead ship fires backward the velocity of the projectile, relative to the Moon observer is zero.

The pursuing ship also fires and the net velocity of the projectile is 2X the ships' velocity.

If the two ships are spatially separated by a distance of 10 seconds (T0 is the point where the pursuing ship is now and T1 is 10 seconds apart where the lead ship is now) at their current velocity, then two things can be observed:

1) In 5 seconds the pursing projectile will be at the point where the lead ship is now (T1) and the pursing ship will have reached the midpoint between T0 and T1.

2) in 5 seconds the lead ship's projectile will also be where the lead ship is now (T1) and the lead ship will be at T1 + 1/2 the spatial distance between T0 and T1.

At the point of collision between the two projectiles the projectiles will meet at T1 or where the tail of the lead ship was when both fired.

However, the challenge question asked where would that collision take place relative to the two ships. Quoting the original question; "did that collision happen exactly halfway between the two spaceships?"

The answer would seem to be yes (ignoring any special relativity effects). That is, because the question implies that the time you make the measurement of where they collided is at the time of collision. The two ships and the pursuing projectile have all moved relative to the Moon observer, but the lead ship's projectile did not move relative to the Moon observer.

From the frame of reference of the Moon observer the collision point is at T1, where the lead ship was when they both fired given the example velocities I cited above.

So, what accounts for the outcome to be different than what I described?

If your assumption that the muzzle velocities are equal to the ships velocities is true and the weapons were fired at the exact instant the lead ship was directly over the moon observer then what you state would be correct. The barrel of the lead ship would leave the bullet at muzzle velocity and to the moon observer it would be dropping straight down at some gravitational constant. The collision would then take place directly over the moon observer.

That is probably not the case as it was not stated in the problem.

Doesn't matter what the muzzle velocities are as long as they are the same and greater than zero.

The same applies to the ships' velocities. Both are the same velocities.

The difference would be where the two projectiles collided, but it would be at the best midway between the two ships when they are fired (instantaneous velocities of the projectiles) and at worst they would never leave the barrels. As long as conditions for the two ships remain the same (no velocity or vector change) and the two projectiles had a muzzle velocity greater than zero, the projectiles would intersect at a distance that is midway between the two ships at the moment of impact.

For the observer on the Moon the flash of the impact would appear to be dead center between the two ships (that is, between the tail end of the lead ship and the nose of the pursuit ship) when it happens.

Obviously, it would not be dead center between the two ships original positions because the ships moved between the time of firing and the time of impact.

Hi Hero, you asked: "So, what accounts for the outcome to be different than what I described?"

The relativity of simultaneity.

Jorrie

Ah! Thanks.

I like Anonymous Hero's bridge analogy. If the lead ship fires a shot backwards and the chase ship fires one forward, the bullet from the lead ship will be traveling at a velocity of Vship-Vbullet while the bullet from the chase ship will be traveling at Vship+Vbullet where the absolute values of each velocity component are the same as stated in the initial post, the bullet from the chase ship should gain distance on the lead ship at the same rate with which the bullet from the lead ship will lose distance from the lead ship therefore they should meet at the midpoint between the ships. The bridge analogy makes the same point much more simply by stating that since the two ships are traveling at the same velocity, they are in the same frame of reference with respect to each other. I don't think trying to bring relativity into the question is necessary since Jorie never said how fast the ships were traveling, he just said they were going very fast...if they were approaching c then we could discuss the effects of relativity.

I deposit my 2 cents.

Oops, I forgot to login again!!!

I'd forgotten about gravity...but then the gravitational pull of the moon is pretty weak and the bullets wouldn't be in flight long enough to 'fall' very far.

So....I'll modify my answer and say 3mm below* the centre point !

(Below meaning, 'in the direction of the moon')

Del

I'd forgotten about gravity...but then the gravitational pull of the moon is pretty weak and the bullets wouldn't be in flight long enough to 'fall' very far.

Whether gravity is weak or strong the bullets will drop at the same rate. The distance of drop is a function of time before collision. A 3mm drop would mean the ships were pretty close to each other. Maybe within 100 yards.

I think assuming that gravity is equal on both bullets is a false assumption. The actual "pull" of gravity might be equal but the vectors will be different.

If we include gravity in this challenge then the bullets will not hit exactly midway between the two ships.

since the center between the two ships was perpendicular to the viewer at the time of firing then the bullets Will collide just slightly closer to the trailing of the two very fast spaceships. at the time of firing the bullet from the trailing spaceship will have a small vector forward while the bullet from the leading space ship will have a small vector toward the rear.

As the bullets move the forward gravity vector for the trailing bullet goes away while the rear vector for the bullet from the leading ship gets larger.

If we do not include the moons weak gravity then the bullets will meet exactly half way between the two ships.

Hi silvCrow, you concluded with: "If we do not include the moons weak gravity then the bullets will meet exactly half way between the two ships."

Yes, the Moon's gravity will play a small part, but there's also something else...

Jorrie

Imagine both ships travelling at 1000 f/s. The shots fired both travel at the same speed (1000f/s). The shot from the first ship would appear to stand still, while the second would travel at 2000 f/s. The collision would occur at the location of the first ship at the time of firing. This is not the midpoint which was directly overhead when they fired, but it is the point equidistant from the ships in their relative frame.

The original post did not ask if the point of collision was at the point that the observer saw as being directly overhead...I think this was irrelevant information designed to skew the reader's perception of the question and cloud the issue. The original post asked if the point of impact would happen "exactly halfway between the two spaceships"...this would be the "point equidistant from the ships in their relative frame" as you stated. this would also be the point equidistant from the ships in any other frame of reference at the speeds you used to describe the question. Unless the ships are approaching the speed of light, frame of reference from the moon to overhead probably wouldn't be affected that much, I think that Jorrie worded the question in such a way as to try to make all of us overthink the answer.

Jorrie, Nice problem. Yes mid point of new ship position when shots impact. Not being an engineer I didn't feel the need to over-think the problem.

I enjoy your physics and astonomical comments.

Is the govt going to do the measuring? Because I always love the statement "Close enough for Govt work".

We should try this same problem except use beams of light instead of bullets. I would love to see the theories on this. Would the beams ever meet?

Yes, the beams of light would meet since the speed of light in a given medium is constant in all frames of reference according to the theory of special relativity (as I recall from my college physics...if I'm wrong here, I'm sure Jorrie will correct me).

Imagine both ships are travelling at the speed of light. Now you fire beams of light. Is the answer still the same, because as Jim correctly stated, "the speed of light in a given medium is constant in all frames of reference".

Hi Poison, you wrote: "Now you fire beams of light. Is the answer still the same,..."

I assure you, the answer will be the same - all that's required is to find that answer!

Jorrie

If your traveling at the speed of light, in space, driving a 68 camaro, and turn on the headlamps, will they work?

If your traveling at the speed of light, in space, driving a 68 camaro, and turn on the headlamps, will they work?

Yes they will work but they will look like taillights.

Yes, but try the same thing at any speed with a British-Leyland car and they won't.

If I had a 68 Camaro and a red-headed woman, who would care about simultaneity?

Very true Jorrie but the red and blue shift would be a function of the speed of the ships.

2 cents

Brad

Gallilean relativity seems to apply.

The observed POI is midway between the two, but not directly overhead.

An observer on the moon will see the collision closer to the forward ship because of the difference in speed of the bullets relative to him.

An observer on any of the ships will see the collision at midpoint.

I think Pragmatist has it right--the collision happened halfway between the two ships because the bullets were fired within the ships' inertial framework.

The observer would see the point of impact appear to move in the direction the ships motion.

Jorrie--thanks for the cite on Relativity of Simultaneity--interesting reading and I learned something new.

Not to overthink things, but as the original problem was stated the ships were "chasing each other". One way to read that is that the ships were in a horizontal circular pattern above the observer. In this case, the midpoint of the two ships does not change with respect to the ships' angular velocity. A bullet (laser, tomahawk, etc...) fired toward the other ship (with due respect for the challenge involved) would always meet at the center of the circle--directly over the observer.

Hi habib, you wrote: "One way to read that is that the ships were in a horizontal circular pattern above the observer. In this case, the midpoint of the two ships does not change with respect to the ships' angular velocity."

Sorry, it was bad wording in my original post, which I 'red-faced' in reply #13 above: "Hi Del, yea, should have said one chasing the other."

I have also clarified later that I meant the ships to be flying more or less on a straight path. (Since this is not a Blog post, I could not go back and edit the original post.)

Your horizontal circular pattern is interesting. I'll have to think about the relativistic effects, which are not necessarily as simple as the straight line scenario...

Jorrie

Simple down-to-earth vectors indicate (to my non-relativistic mind) that the projectiles would collide at a point closer to the pursued spaceship. If the bullets are released simulatenously then the time taken by each to reach the collision point is equal.

Constant distance between the spaceships is 1.

Distance of the collision point from the pursued is p.

Ship velocity is v.

Bullet velocity is w.

Then c/(w-v) = (1-c)/(w+v) => c=1/2 - v/(2w)

For midpoint collision we should have had c=1/2 but we see that c is smaller than 1/2.

What is 'c' ?

Sorry, sorry, sorry.

I must have been hypnotised by Jorrie's relativity proneness and the speed of light crept in unnoticed.

Delete c, put p instead. The distance of the bullet collision from the pursued ship.

Hi dovy, sorry for the 'hypnosis', but with your c replaced by p, things still don't look right!

"Constant distance between the spaceships is 1.

Distance of the collision point from the pursued is p.

Ship velocity is v.

Bullet velocity is w.

Then p/(w-v) = (1-p)/(w+v) => p=1/2 - v/(2w)"

You forgot that the ships have moved while the bullets traveled and so has the midpoint. But, you can just as well subtract the ship velocity v from everything and let the ships be stationary in their inertial frame...

Jorrie

I stand corrected!

Therefore we now have (keeping the same definitions and without 'c' intrusions...):

p/w = (1-p)/(2v+w) or p = 1/2 * w/(v+w)

In other words, the collision point is still closer to the pursued ship but closer by a small multiplicative factor, not a small additive one.

Or have I missed something again?

Hi dovy, nope, you are still erring, because you are making it too complicated.

With the assumption that you originally made, the bullets must collide exactly equidistant from the two ships (at the 'moving' midpoint). Here's why: you do not need the complication of v in the calcs - assume that the moon is moving backwards at v and that the two ships are stationary and then compute it again in the ships' frame.

However, this doesn't give the correct answer either, because there's a problem with the assumption...

Jorrie

Hello Jorrie,

The way I read your statement of the challange and your 2 corrections, I'm forced to conclude that the bullets should collide midway between the ships in their moving frame of reference.

But this point will no more be directly above the observer. It will be v/(2*w) away from the midpoint in the direction the ships are travelling, where v is the ships' velocity and w is the bullets' velocity.

In my defence speech I apologise for knowing nothing about relativistic vector calculus.

Jorrie

Nice reading on the relativity of simultaneity. If the ships appear to fire at the same time and the rules of simultaneity and physics are in affect, then according to the "observer" as seeing them fire at the same time would be wrong. Fact is they did not fire at the same. But you observed it as such. If this were the case then they projectiles would not collide at mid point. As to which direction the observer viewed it from, that would determine which one fired first and where they would meet.

That point in time-space was specified as at the midpoint directly above the observer (check Post #1 again). So what would be your prediction?

Hello Jorrie,

If those spaceships were travelling away from you at slightly less than the speed of light, relative to you the observer, and the bullet speed if added to the ship speed exceeded the speed of light, relative to you, the observer, there would be an interesting observation:

You would not actually see the bullet fired from the rearmost spaceship, and once that bullet exceeded the speed of light relative to you, going away from you, it cannot be detected, or seen, so for you it would not exist.

The Theory of relativity is quite often misunderstood, the so-called "shrinkage in the axis of travel" for extremely fast objects, is only an apparent "shrinkage", not a real "shrinkage".

The other thing to remember, is that any observation or measurement, actually interferes with what is observed or measured, so an accurate observation or measurement is never obtainable - just a near approximation.

This can be easily proved, because each observer is entitled to assume that he/she has THE fixed frame of reference, understood by Einstein, but many clever people, Physics Professors and others, fall into the wee trap ......

Hi Sparkstation, you wrote: "You would not actually see the bullet fired from the rearmost spaceship, and once that bullet exceeded the speed of light relative to you, going away from you, it cannot be detected, or seen, so for you it would not exist."

Huh? No bullet, no nothing, can pass from 'this side of the speed barrier' to 'that side of the speed barrier'!

Velocities add up us follows: v = (v₁+v₂)/(1+v₁v₂/c²), so even if the ship could fly at v₁= c and it could fire a bullet at muzzle velocity of v₂= c, the bullet would still move at v = c relative to me. In that sense I would not be able to see the bullet - it would not have left the muzzle of the gun! But if the velocities were both 0.999c, the result would still be less than c and I would be able to observe the bullet.

Lastly, Heisenberg uncertainty does not quite apply to macroscopic things like bullets. The effect is there, but immeasurably small, unless the 'bullet' is a subatomic particle...

Jorrie

Hello Jorrie,

Because the fastest thing we know of, is the speed of light relative to an observer, it is impossible for that observer to observe anything travelling faster than the speed of light, in a direction away from him.

This is easily proved, using basic reasoning.

Let us suppose we have an observer (A), who assumes his/her position is fixed relative to the rest of the entire Universe.

A spaceship travels away from that observer at 99.99999% the speed of light, relative to observer (A).

On that spaceship is observer (B) who is also entitled to assume that he/she has the "fixed frame of reference".

A bullet is fired at a speed to exceed the speed of light, relative to observer (A) out of the front of that spaceship.

So Observer (A) cannot perceive the bullet, (for him it travels faster than light), once it leave the muzzle of the cannon, and has no way of knowing it exists, therefore for him it does not actually exist, because "Nothing can exceed the speed of light relative to an observer", as formulated by Einstein, who reasoned it out in the same manner.

However observer (B), who sits in the nose cone of that spaceship, looking out the viewing port, he sees the bullet travel away from him, at the muzzle velocity when fired from the ship's cannon. He has no problem seeing that bullet, which will eventually vanish into the depths of space.

As I said earlier, it is all about assumptions, and the "fixed frame of reference" which each observer is entitled to assume he/she has - If we do not allow each observer to have that fixed frame of reference, then we all go crazy, because we are not designed to cope with the infinite.....

Hello Sparkstation.

I see you totally ignore the well proven relativistic addition of velocities rule?

Why?

Because of this, your reasoning is completely contra-Einstein.

Jorrie

In what I know will be an over simplification, I will state that the answer is no. My reasoning is that equal muzzle velocities requires unequal projectile velocities (relative to the observer on the moon). Collision should appear to observer to be closer to the fleeting (not chasing) ship.

This ignores the fact that 'firing' a projectile will reduce 1 ship's velocity, while increasing the other ship's velocity. I may be wrong but I am assuming our combatants have developed cruise control and the thrusters are automatically adjusted to compensate.

I had not considered the ship reaction to firing its own projectile. I really don't know how to quantify the trailing ships just slight slowing from firing its projectile vs the leading ships just slight increase in speed. My first thought "was" that each ship will be effected equally in speed change for the purposes of this challenge. Each ship is also effected by the moons gravity. With the trailing ship being just slightly slower it will be effected by the moons gravity for a slightly longer time than the bullet, but with similar gravity vectors, changing the center point as compared to the other ship just a little toward the trailing ship. What I don't know, is how to quantify, with the details given, this as compared to the bullets collision point.

Jorrie,

I look forward to reading your answer on Monday. I hope I have time to get to it on Monday. Monday is the first day of finals week here. I have a lot of student projects to go over. By Monday evening I may prefer getting a tooth pulled.

Hi silvCrow.

The recoil actions actually cancel out in this challenge, provided the ships, guns and bullets are identical. Even if they are not, the effect will be tiny. The slight asymmetry of the Moon's gravity on the two ships are also very small compared to the 'gross' effect that I'm after.

Good luck with the final week!

Jorrie

"The recoil actions actually cancel out in this challenge"

I must be picturing something very wrong. If 1 gun is firing out of the front of the trailing ship, and the other gun is firing out of the rear of the leading ship, the velocity of the trailing ship will be decreased, while the velocity of the leading ship is increased. This hardly constitutes canceling, since the result is that the ships are now traveling at 2 different velocities. I realize this may have nothing to do with the final solution, but I disagree with your statement. I have tested this theory, firing my 2 civil war cannons at each other, when spaced 15 feet apart. After the firing the distance between the cannons was 20 ft, and both my neighbors have new windows.

Also I believe there is confusion in the term 'muzzle velocity', but will wait to Monday to read your final answer.

Go back to Post #13 and/or forward to post #77. Where the correction is restated.

"ONE SHIP IS CHASING THE OTHER."

Capiche?

Keeping things simple, I never mis-understood the original post, and have always been picturing and describing "ONE SHIP CHASING ANOTHER"

Please read post 73--- He get's it,

CAPICHE?

Mevel123 Pass the 'crow' and I'll have the 'humble pie' later.

The wires must have been crossed and/or short circuited in the central office!

I regret even bringing up the effects of 'firing' it had no bearing on my answer of "no." It was just a separate thought that occured while typing. Jorrie, has served my slice of pie already, by pointing out how the effect is cancelled.

I believe the 'projectile-muzzle velocity' is the key, thus I still answer "no."

I expect to have additional humble pie on Monday when; regardless of Jorrie's answer, I will be wrong.

Hi again Mevel. In order to serve it "slice-wise" and in "layers", your 'no' answer is correct, but the direction and reasoning (post #78, I think) around it is not!

Think about your two cannons again. Does the fact that mother Earth is hurtling around the Sun at 30 km/s make any difference as to where the two cannon balls crashed, as you measure it? Obviously not, because you shot them simultaneously. Even the 'man-on-the-moon' will agree that they crashed at a point equidistant from the cannons. But will that man agree that you shot them simultaneously?

Jorrie

"...but the direction and reasoning (post #78, I think) around it is not!"

It was your #72, not #78!

The 'official solution' has been posted in the CR4 Relativity/Cosmology Blog.

Jorrie

Hi Mevel, you said: "I have tested this theory, firing my 2 civil war cannons at each other, when spaced 15 feet apart. After the firing the distance between the cannons was 20 ft, and both my neighbors have new windows. "

Have you checked if the mid-point between the two cannons changed position?

If you didn't, I'm afraid you'll have to go and shoot again!

Jorrie

Thanks for that Jorrie, I checked, and you are right!

However I am still sticking with my answer of "no" and it is still based on 'muzzle velocity.' If Monday your answer is "Yes" then I will have more problems.

Hello again Jorrie,

Your Question....Copy/Pasted:

"During the space wars, you were standing on the Moon, watching two fast spaceships passing horizontally overhead, chasing each other at identical constant speeds relative to you. When the midpoint between them was directly overhead, you observed them to simultaneously fire a single shot at one another, but the two bullets spectacularly collided and exploded. If the muzzle velocities were identical, did that collision happen exactly halfway between the two spaceships?"

I just re-read your question, and it appears the spaceships are travelling in a closed circle, or an ellipsoid, since they are "chasing each other."

Therefore the spaceships, although they were capable of "fast" motion, would have been travelling quite slowly, or the centripetal effect would have caused them to veer off the absolute circle or ellipsoid "track".

Also your question appears to be "loaded" with the assumption that the spaceships are exactly opposite one another across the circle/ellipsoid.

It is possible the ships are actually touching, nose of the rear touching the rear of the one in front, because your question allows that situation.

Each spaceship could be in any position along the circular/ellipsoidal track, providing it did not occupy the space of the other spaceship.....

So.........what I am saying, is that it is impossible to correctly answer your question, as it is given.

is out there....

If this is NOT so, please advise further, thank you.....

Hi SS, yep it was bad semantics on my part, as pointed out by Del in #6 above and corrected in #13, where it was changed to "one chasing the other".

One cannot go back and edit a post later. Sorry.

Jorrie

Jorrie,

Good question as always. As they near c, would the observer see them as horizontal?

Hmmm.

I don't think 'fast' puts us into relativistic speeds, certainly not if 'bullets' are being shot.

If relativity comes into the answer I shall go scratch the furniture in disgust .

Del