# Thread: vector and trajectory calculation

1. Who wants to calculate trajectories and vectors with me? I will start off with an easy one. the target is 5 miles away, ur artillery shell weighs 450 pounds and the angle at which the battery can fire is 20-63 degrees. what angle and speed would u need to hit said target? there is a 500 ft. grace radius from the target in any direction that u have to hit within. if u need hints let me know and have fun!!  2.

3. I have noticed that no has guessed or can figure out the answer. Here is some help: the azimuth of fire is 15 degrees west with true north azimuth at 5 degrees. Hope that helps and please try to answer this question.  4. If air drag is to be ignored the problem is trivial and can be solved with a high-school physics textbook and some work.

Hint: assume the elevation angle is 45 degrees, which gives you the greatest range (still assuming no air drag), then compute the minimum speed. With any more speed, you need to change the angle. When the angle gets beyond the limits of your gun, that speed won't do.

If air drag is to be taken into account, I don't know how to solve it. I would google for some empirical formulae.

I cannot see why you give us the azimuth - it is obvious that the azimuth should be towards the target (unless you need to think about wind), and the whole problem only concerns the elevation and speed.

Good luck.  5. I don't know Leszek. Over 5 miles you might need to account for the Coriolis effect.  6. I know I just figured azimuth always does help as there are methods using azimuth to do this calculation. Assume there is no drag otherwise this problem would be immensely difficult. Do u have the answer Leszek? I know what it is and I would like to know if u know what. I will tell u the angle is not 45 degrees. A low parabolic trajectory like that makes the shell go off target by about 800 feet. U dont need to factor in the coriolis effect for this problem because that makes it very difficult. I will do one later that takes the coriolis effect into account.  7. Well you didn't give the muzzle velocity, so there will be a whole lot of solutions at least one elevation and maybe two, for each velocity, won't there?  8. yep there is more than one right speed and angle of course. with a grace radius of 500 feet it leaves room for different answers. I challenge u to find an answer better than I did. which was a 57 degree shot at 462 miles per hour. My calculations put that at within 125 feet of the target.  9. Originally Posted by the man of science
yep there is more than one right speed and angle of course. with a grace radius of 500 feet it leaves room for different answers. I challenge u to find an answer better than I did. which was a 57 degree shot at 462 miles per hour. My calculations put that at within 125 feet of the target.
Hmm. According to this online calculator, a velocity of 462 mph (206.53248 m/sec) and angle of 57 degrees gives you a range of 3976.31523 meters = 2.47076774 miles and you need 5 miles.  10. well thats weird my calculations were as follows: 1.0266 miles per hour per pound at a 57 degree angle. the vertex was 2000 ft. high and 3 miles to target. the following drop saw an increase in speed to at least 540 mph before hitting the target 125 ft. from target. I dont know where I went wrong especially considering my azimuth of shot was 15 degrees.  11. Torque?? What equations are you using? Let's see your work.  12. well torque is just acceleration per pound correct?  13. No. Torque is force multiplied by the length of a lever arm. There is no way you would use torque in a trajectory calculation, unless you were using a catapult of some sort.  14. nvrmind then harold. take the torque out of the equation then.  15. Originally Posted by the man of science
well torque is just acceleration per pound correct?  16. I see you've take the torque out of your answer, but you still have pounds. If you know the velocity and the angle, and you are ignoring air resistance, the weight of the projectile will not be a factor. Take a look at that web page with the online calculator that I linked to earlier. It has the equations you need.  17. Maybe it will help if we go through the steps (BTW, I got the same answer as Harold14370 did.

First we convert to the FPS system:

5 miles = 26400 ft

462 mph = 677.6 ft/sec.

at 57° the vertical component will be 677.6 *sin(57)= 568 ft/sec.

a=at, so t=v/a

IOW, the amount of time it will take for the acceleration due to gravity to stop the projectile on its upward journey is:

17.75 sec.

It will take an equal amount of time to fall back to the Earth, so the total time is

35.5 sec.

The horizontal component of the velocity will be 677.6 * cos(57) =369 ft/sec

and the projectile will travel horizontally 369 * 35.5 = 13101 ft in this time, which is 13101/5280 = 2.48 miles.  18. You can also solve for a more general equation:

Vertical component of initial velocity: Time of vertical rise and fall: Horizontal component of initial velocity: Distance traveled in time of rise and fall: solving for t and equating to time of rise and fall: substitute for vh and vv: solving for v: Finally, reduce the trig identity: Giving you an equation that can be re-arranged to find any one of the variables (distance, angle or initial velocity) given the other two.  19. Thank you Janus. That was the exact same equation I used, but you didn't take into account the acceleration of the shell to 540 mph at 3 seconds after the vertex of the trajectory is reached. with the added speed the shell goes an additional 11,215 feet beyond ur calculations making the shell fall within 125 ft. of the target.  20. Originally Posted by the man of science
Thank you Janus. That was the exact same equation I used, but you didn't take into account the acceleration of the shell to 540 degrees at 3 seconds after the vertex of the trajectory is reached. with the added speed the shell goes an additional 11,215 feet beyond ur calculations making the shell fall within 125 ft. of the target.
Where did that acceleration come from? The nature of the question in your original post makes no mention of anything that would account for it. As stated it is a simple ballistic trajectory, in which case the only acceleration after leaving the gun would be that due to gravity, which is accounted for by the equation I gave.

Besides that, what does "acceleration to 540 degrees" even mean?  21. I meant to say 540 mph srry    22. Originally Posted by the man of science
I meant to say 540 mph srry  You still haven't explained what the source of this mysterious acceleration is.  23. our friend gravity.  24. Originally Posted by the man of science
our friend gravity.
I had a feeling that is what you were going to say.

Gravity only effects the rise and fall of the projectile, and the instantaneous vertical velocity of the projectile. This, in turn, determines the time of flight. The horizontal velocity is not effected by gravity at all and remains the same throughout the flight.

Ergo, the distance traveled is equal to the time of flight multiplied by the constant horizontal velocity. All of which is totally accounted for in the equation I already gave.  25. Originally Posted by Harold14370
I don't know Leszek. Over 5 miles you might need to account for the Coriolis effect.
Of course if you really want to pick nits, you can also account for the curavature of the Earth and the fact that the projectile actually follows a path that is a section of an ellipse.

As follows:

Use the radius of the Earth,Mass of the Earth and the initial velocity to find the semi-major axis(a) of the ellipse through energy conservation. But also so solving for a This allows us to find the period of the orbit. Now the Horizontal component of the velocity is and the horizonatal velocity can be used to find the areal velocity(A) of the orbit by but areal velocity is also equal to where e is the eccentricity of the orbit.

so solving for e gives now, given that where r is the radial distance of from the focus of the ellipse at an angle of phi from the perigee, then substituting the radius of the Earth for r and taking the arc cos gives us the angle between perigee and the point where the orbit interescts the surface of the Earth. Since the peak of the trajectory occurs at apogee, then the fraction of the Earth's circumference the projectile covers in its trajectory is and this value, multiplied by the circumference of the Earth, gives you the distance traveled by the projectile over the surface of the Earth.   Bookmarks
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