Trajectories, Part 1
"Vacuum:"
By Donna Cline
Copyright © 2003
Gravitational Forces | Vacuum Trajectory | Envelope
Flat-Fire Approximation | Uphill/downhill | Example
will
not go into the proofs and deriving each formula from one another. If that
is what you are looking for there are a number of books that will give you that
type of information. "Mathematics for Exterior Ballistics" by Gilbert Ames Bliss,
1944, "Exterior Ballistics" by McShane, Kelley, and Reno, University of Denver
Press, 1953, "Sierra Reloading Manual 4th Edition" by Sierra Bullets, L.P., 1995,
"Modern Exterior Ballistics" by Robert L. McCoy, Schiffer Publishing Ltd., 1999,
and "A Ballistics Handbook" by Geoffery Kolbe, Pisces Press, 2000, just to name a
few.
It was Sir Isaac Newton (1642-1727) that, while standing on the shoulders of giants, built his great theory of motion.
Newton's First Law of Motion:
Every body continues in its state of rest or of uniform speed in a straight line unless it is compelled to change that state by forces acting on it. The tendency of a body to maintain its state of rest or uniform motion in a straight line is called inertia. As a result, Newton's first law is often called the law of inertia.
Newton's Second Law of Motion:
The Acceleration of an object is directly proportional to the net force acting on it and is inversely proportional to its mass. The direction of the acceleration is in the direction of the applied net force. A net force exerted on an object may make its speed increase or if it is in a direction opposite to the motion, it will reduce the speed. If the net force acts sideways on a moving object, the direction as well as the magnitude of the velocity changes. Thus, a net force gives rise to acceleration: a = F ÷ m or F = m * a. Acceleration is the velocity that is changing and is equal to velocity divided by time, a = (V_{2} - V_{1}) ÷ (t_{2} - t_{1}). This same equation can give you time, t = V ÷ a and velocity, V = a * t. Therefore: F = m * a = m * V ÷ t or F * t = M * V.
Newton's Third Law of Motion:
Whenever one object exerts a force on a second object, the second object exerts an equal and opposite force on the first object. Or to put it another way; for every action there is an equal and opposite reaction. This is what makes jets and rockets fly and it is also what makes a gun recoil and you feel it as a kick.
These three laws are what govern all motion in the universe and we have known about them ever since Sir Isaac Newton first published his great work, the Principia, in 1687.
The curved path of a projectile
is called a trajectory. The equations for a vacuum trajectory where gravity is the
only force acting on the projectile is quite simple. The trajectory of a projectile
in a vacuum would inscribe nearly a parabolic path.
The general shape of the trajectory is called a parabolic because on a flat world and in a vacuum, an unresisting medium, absence of an atmosphere, the path of a projectile is actually a parabola, as was first proved by Galileo in 1638. Galileo noticed that, owing to the curvature of the earth, the force of gravity did not cat in parallel lines, but acted instead in lines that converged to the center of the earth. As a consequence, the usual path of a projectile fired on the earth in the absence of air would be a portion of an ellipse. If the shot were fired horizontally from an eminence at any ordinary small arms velocity the elliptical path would intersect the earth’s surface and the projectile would fall to earth striking the surface. But if the velocity were to be 26,000 fps, the projectile would never return to earth, but would be itself a satellite with a circular orbit passing through its original point of firing once every seventeen revolutions. A circle is an ellipse that has both of its foci (plural of focus) at the same point in space and whose equation of the graph is r2 = x2 + y2. If the velocity were greater than (>) 26,000 fps and less than (<) 36,000 fps, the orbit would be elliptical. An ellipse is the set of all points in the plane the sum of whose distances from two fixed points (foci), each called a focus, is a constant and whose equation of the graph is 1 = (x2/a2) + (y2/b2). If the velocity was exactly at 36,000 fps it would tolly-trop into space in a parabola. The geometric definition of a parabola is the set of points in the plane equidistant from a fixed point called the focus and a fixed line called the directrix and whose equation of the graph is y = ax2 + bx + c. And if the velocity were greater than (>) 36,000 fps the trajectory would be a hyperbola. The geometric definition of a Hyperbola is the set of all points in the plane, the difference of whose distances from two fixed points, called the foci, is a constant and whose equation of the graph is 1 = (x2/a2) - (y2/b2). The hyperbola also has lines called asymptotes associated with it. Asymptotes are lines that the hyperbola approaches close to but never touches for ever larger values of x and y. The only time the trajectory would be a straight line is if it was fired straight upward or downward.
For all practical purposes we will
never shoot a bullet in a vacuum trajectory. So, why am I even going into showing a
vacuum trajectory? Well, atmospheric trajectories display many of the same properties
of a vacuum trajectory, but lack the convenience of a simple analytical solution.
A vacuum trajectory is the simplest trajectory, only dealing with the force of
gravity, which will show the basic fundamentals of trajectories without a bunch
of clutter and the mathematical methods presented here will form the framework
on which the higher approximations are based on.
Simply put, in a vacuum, the
projectile leaves the bore, at the origin, with the barrel at a positive angle from
a straight line to the target is called the inclination (slope) angle of departure.
As the projectile climbs in height or gains altitude to the summit of its flight
this is the ascending branch of its trajectory. The height between the origin and
the summit of the projectile's flight path is it's maximum ordinate and the summit
lays half way between the origin and the target. From the summit till the point of
impact, the projectile's path is on the descending branch of the trajectory. The
angle from the projectile's path on the descending branch of the trajectory to the
straight line to the origin is the inclination angle of fall. The projectile's
inclination angle of departure is the same angle as the inclination angle of fall.
And the vertical velocity up ward and horizontal velocity at the point of origin
will be the same as the vertical velocity down ward and horizontal velocity at the
point of impact.
Gravitational Forces on a Projectile:
Let us first deal with the celestial bodies other than the earth itself. For our purposes we can consider all bodies of our solar system to be spherical and composed of concentric homogeneous spherical shells. Text on celestial mechanics shows that the gravitational field of such bodies, outside of their surfaces, is the same as though their masses were concentrated at their centers. We shall also suppose that all the bodies of our solar system moves in a circular orbit about their respective centers. For our purposes we can consider the orbit of the earth’s center about the sun’s center to have a radius of 93,000,000 miles and the radius of the earth to be about 4,000 miles. The angular velocity of the earth’s center about the sun is about:
Equation 1-1:
Angular velocity (in radians/sec.) = 2* / seconds in a year
Angular velocity (in radians/sec.) = 2* 3.14159 26536 / 31,557,600
Angular velocity (in radians/sec.) = 6.28318 53072 / 31,557,600
Angular velocity =~ 0.00000019910212776637 or 1.991 * 10^{-7} rad. / sec.
Centrifugal acceleration of the earth’s center is:
Equation 1-2:
Centrifugal acceleration (a_{c} in feet/sec²) = (1.991 * 10^{-7})² * radius
of the earth’s orbit in feet
Centrifugal acceleration (a_{c} in feet/sec²) = (1.991 * 10^{-7})² * (93,000,000 mi. * 5280 ft.)
Centrifugal acceleration (a_{c} in feet/sec²) = 3.9641657 * 10^{-14} * 491040000000 ft.
a_{c} =~ 0.0194656394 or 0.019 ft. / sec.²
It is not difficult to see that for all points on the surface of the earth, this difference has its greatest value at the point “P” nearest to the sun. Centrifugal acceleration of the earth at point “P”:
Equation 1-3:
Centrifugal acceleration (a_{p} in feet/sec²) = (1.991 * 10^{-7})² * radius
of the point “P” orbit in feet
Centrifugal acceleration (a_{p} in feet/sec²) = (1.991 * 10^{-7})² * (92,996,000 mi. * 5280 ft.)
Centrifugal acceleration (a_{p} in feet/sec²) = 3.9641657 * 10^{-14} * 491018880000 ft.
a_{p} =~ 0.0194648022 or 0.019 ft. / sec.²
At this point the acceleration a_{p} has the same direction as a_{c}, and its magnitude is greater in the ratio of 93,000,000² to 92,996,000², since the radius of the earth is about 4,000 miles. The ratio differs from unity by about 1 / 11,624, and | a_{p} | is about 0.019 ft. per sec. per sec., so | a_{p} – a_{c} | cannot exceed 0.0000018 ft / sec², which is entirely negligible for ballistics purposes. Like wise for the moon’s attraction, though somewhat larger, is also negligible, while the effects of the other planets are far smaller.
Vacuum Trajectory:
The Negitive Force of Gravity on a Projectile:
Where the acceleration due to gravity is pulling the projectile in the downward direction towards the ground to end it's flight.
The maximum gun elevation that will give the maximum range is found by differentiating the equation:
Equation 1-4:
X_{w} = (V_{o}² * sin (2 * Ø_{o})) ÷ g
with respect to Ø_{o}, setting the deivative equal to zero, and solving for Ø_{o}:
Equation 1-5:
dX ÷ dØ_{o} = (2 * V_{o}²) ÷ cos (2 * Ø_{o}) = 0
The solution of this equation gives us:
Equation 1-6:
0 = cos (2 * Ø_{o})
So that:
90º = 2 * Ø_{o}
45º = Ø_{o}
Therefore,
by setting Ø_{o} equal to 45º you can find out how far the
projectile will go. And if X is made in small enough increments for a
given Ø_{o} you can plot the trajectory of the projectile.
Remember that the equations along with the initial conditions dictate the
projectile's trajectory.
The first thing we need to
find is the inclination angle of departure:
Equation 1-4:
X_{w} = (V_{o}² * sin (2 * Ø_{o})) ÷ g
Where X_{w} is the range to the target in feet,
V_{o} is muzzle velocity in fps,
and g is the acceleration due to gravity (32.1734 ft per sec²).
By rearranging the formula to give Ø_{o} we have:
X_{w} = (V_{o}² * sin (2 * Ø_{o})) ÷ g
(V_{o}² * sin (2 * Ø_{o})) ÷ g = X_{w}
(V_{o}² * sin (2 * Ø_{o})) = X_{w} * g
sin (2 * Ø_{o}) = (X_{w} * g) ÷ V_{o}²
2 * Ø_{o} = csc ((X_{w} * g) ÷ V_{o}²)
Ø_{o} = csc ((X_{w} * g) ÷ V_{o}²) ÷ 2; (csc is
the same as the inverse sin X or sin^{-1} X.)
OR
Equation 1-7:
Ø_{o} = sin^{-1} ((X_{w} * g) ÷ V_{o}²) ÷ 2
Let's set our initial conditions, for we know our: muzzle velocity (V_{o}) = 2800 fps and range (X_{w}) = 300 yards or 900 ft.
Ø_{o} = sin^{-1} ((X_{w} * g) ÷ V_{o}²) ÷ 2
Ø_{o} = sin^{-1} ((900 ft * 32.1734 ft/sec²) ÷ 2800² ft²/sec²) ÷ 2
Ø_{o} = sin^{-1} ((900 ft * 32.1734 ft/sec²) ÷ 2800² ft²/sec²) ÷ 2
Ø_{o} = sin^{-1} ((900 * 32.1734) ÷ 2800²) ÷ 2
Ø_{o} = sin^{-1} (28956.06 ÷ 2800²) ÷ 2
Ø_{o} = sin^{-1} (28956.06 ÷ 7840000) ÷ 2
Ø_{o} = sin^{-1} (0.003693375) ÷ 2
Ø_{o} = 0.2116152808 ÷ 2
Ø_{o} = 0.1058076404º
There are actually two elevation angles which satisfy the equation:
Equation 1-4:
X_{w} = (V_{o}² * sin (2 * Ø_{o})) ÷ g
Only the lower angle solution is given by the equation:
Equation 1-7:
Ø_{o} = sin^{-1} (X_{w} * g) ÷ V_{o}²) ÷ 2
The higher-angle solution (denoted by Ø'_{o}) is given as:
Ø'_{o} = 90º - Ø_{o} = 90º - sin^{-1} (X_{w} * g ÷ V_{o}²) ÷ 2
Ø'_{o} = 90º - 0.1058076404º = 90º - sin^{-1} (900 ft * 32.1734 ft/sec² ÷ 2800² ft²/sec²) ÷ 2
Ø'_{o} = 89.89419236º = 90º - sin^{-1} (900 * 32.1734 ÷ 2800²) ÷ 2
Ø'_{o} = 89.89419236º = 90º - sin^{-1} (28956.06 ÷ 7840000) ÷ 2
Ø'_{o} = 89.89419236º = 90º - sin^{-1} (0.003693375) ÷ 2
Ø'_{o} = 89.89419236º = 90º - 0.2116152808 ÷ 2
Ø'_{o} = 89.89419236º = 90º - 0.1058076404
Ø'_{o} = 89.89419236º = 89.89419236º
The higher-angle is commonly
encountered in the use of mortars. Mortars are often used to attack targets
inaccessible to direct weapons fire, such as on the other side of a hill. In
addition, mortars are Relatively large, heavy, low pressure, low velocity weapons,
and the vacuum trajectory is often a good approximation to the actual flight of
heavy, low velocity projectiles.
Now we're on our way.
Range (X_{w}) is 900 ft and at that range our rifle is zeroed. We would expect the height
(Y_{w}) to be zero at that range. Let's test it by using the formula that does not
use Time Of Flight (T_{w}) as a variable:
Equation 1-8:
[Y_{w} = X_{w} * tan Ø_{o} - {(g * X_{w}²) ÷ (2
* V_{o}² * cos² Ø_{o})}]
Y_{w} = 900 ft * tan 0.1058076404º - ((32.1734 ft/sec² * 900² ft²) ÷ (2 * 2800² ft²/sec² * cos² 0.1058076404))
Y_{w} = 900 ft * tan 0.1058076404º - ((32.1734 ft * 900²) ÷ (2 * 2800² * cos² 0.1058076404))
Y_{w} = 900 ft * 0.0018466938 - ((32.1734 ft * 810000) ÷ (2 * 7840000 * 0.9999982949²))
Y_{w} = 1.662024417 ft - (26060454 ÷ (15680000 * 0.9999965897))
Y_{w} = 1.662024417 ft - (26060454 ÷ 15679946.53)
Y_{w} = 1.662024417 ft - 1.662024418
Y_{w} = -0.0000000005ft or rounded to 0.00
Well, that looks like it's a zero to me. Sence we know the range, X_{w}, and we found the inclination angle of departure, Ø_{o}. We're ready to find the Time Of Flight, T_{w}, with the formula:
Equation 1-9:
X_{w} = T_{w} * V_{o} * cos Ø_{o}
T_{w} = X_{w} ÷ (cos Ø_{o} * V_{o})
T_{w} = 900 ft ÷ (cos 0.1058076404º * 2800 ft/sec)
T_{w} = 900 ft ÷ (0.9999982949 * 2800 ft/sec)
T_{w} = 900 ft ÷ 2799.995226 ft/sec
T_{w} = 900 ÷ 2799.995226 sec
T_{w} = 0.3214291195 sec (There, the Time Of Flight.)
We have another formula for Y_{w} but this is dealing with T_{w}:
Equation 1-10:
Y_{w} = T_{w} * V_{o} * sin Ø_{o} - (g * T_{w}² ÷ 2)
Y_{w} = 0.3214291195 sec * 2800 ft/sec * sin 0.1058076404º - (32.1734 ft/sec² * 0.3214291195² sec² ÷ 2)
Y_{w} = 0.3214291195 * 2800 ft * 0.0018466906 - (32.1734 ft * 0.1033166789 ÷ 2)
Y_{w} = 900.0015346 ft * 0.0018466906 - (3.324048836 ft ÷ 2)
Y_{w} = 1.662024417 ft - 1.662024418 ft
Y_{w} = -0.000000001 ft or rounded to 0.00
And here is the formula for X_{w} this also is dealing with T_{w}:
Equation 1-10:
X_{w} = T_{w} * V_{o} * cos Ø_{o} - (g * T_{w}² ÷ 2)
X_{w} = 0.3214291195 sec * 2800 ft/sec * cos 0.1058076404º - (32.1734 ft/sec² * 0.3214291195² sec² ÷ 2)
X_{w} = 0.3214291195 * 2800 ft * 0.9999982949 - (32.1734 ft * 0.10331667886254528025 ÷ 2)
X_{w} = 900.0015346 ft * 0.9999982949 - (3.32404883571621431959535 ft ÷ 2)
X_{w} = 900.00000000738335354 ft - 1.662024417858107159797675 ft
X_{w} = 898.337975589525246 ft or rounded to 898 ft. (This one is
not as close as I would like it to be, but it is quite good, to less than 1%.)
The angle between the inclination angle of origin on the ascending branch and the inclination angle of fall on the descending branch of the trajectory at impact can be verified by differentiating the equation:
Equation 1-8:
Y_{w} = X_{w} * tan Ø_{o} -
[(g * X_{w}²) ÷ (2 * V_{o}² * cos² Ø_{o})]
with respect to X:
Equation 1-11:
dY ÷ dX = tan Ø_{o} - ((g * X_{w})
÷ (V_{o}² * cos² Ø_{o}))
At impact:
Equation 1-4:
X_{w} = sin (2 * Ø_{o}) * (V_{o}² ÷ g)
and substituting into the above equation, we have:
Equation 1-12:
(dY ÷ dX) _{I} = tan Ø_{I} = tan Ø_{o} - (sin (2
* Ø_{o}) ÷ cos² Ø_{o})
With the help of the trigonometric identity we get [tan Ø_{I} = -tan
Ø_{o}]. Thus the angle of fall on level ground is always the negative
of the angle of departure, for any vacuum trajectory. This result can be
generalized to show that the ascending and descending branches of any vacuum
trajectory are symmetric about a vertical line passing through the summit.
You already
know enough to find the trajectory summit. But, here are some other formulas
to finding the trajectory summit. The Time Of flight to the summit (T_{s}) is:
Equation 1-13:
T_{s} = (V_{o} * sin Ø_{o}) ÷ g
T_{s} = (2800 ft/sec * sin 0.1058076404º) ÷ 32.1734 ft/sec²
T_{s} = (2800 sec * .0018466906) ÷ 32.1734
T_{s} = 5.170733815 ÷ 32.1734
T_{s} = 0.1607145597 (To check it multiply by 2 and compare to T_{w}, we're 0.0000000001 off. I think it checks out good.)
The Range to the summit (X_{s}) is:
Equation 1-14:
X_{s} = [V_{o}² * sin (2 * Ø_{o})] ÷ (2
* g)
X_{s} = (2800² ft²/sec² * sin (2 * 0.1058076404º)) ÷ (2
* 32.1734 ft/sec²)
X_{s} = (7840000 ft²/sec² * sin 0.2116152808) ÷ 64.3468 ft/sec²
X_{s} = (7840000 ft²/sec² * 0.003693375) ÷ 64.3468 ft/sec²
X_{s} = 28956.06 ft ÷ 64.3468
X_{s} = 450 ft (That's half of T_{w}, 900 ft, checks out good.)
The height of the projectile at the summit (Y_{s}), maximum ordinates, is:
Equation 1-15:
Y_{s} = (V_{o}² * sin² Ø_{o}) ÷ (2
* g) = g * T_{w}² ÷ 8
Y_{s} = (2800² ft²/sec² * sin² 0.1058076404º) ÷ (2
* 32.1734 ft/sec²) = 32.1734 ft/sec² * 0.3214291195² sec² ÷ 8
Y_{s} = (7840000 ft²/sec² * 0.0018466906²) ÷ 64.3468 ft/sec² = 32.1734 ft/sec² * 0.1033166789 sec² ÷ 8
Y_{s} = (7840000 ft²/sec² * 0.0000034103) ÷ 64.3468 ft/sec² = 32.1734 ft/sec² * 0.1033166789 sec² ÷ 8
Y_{s} = 26.73648819 ft ÷ 64.3468 = 3.324048836 ft ÷ 8
Y_{s} = 0.4155061043 ft = 0.4155061045 ft (These are with in 0.0000000002 ft of each other, checks out good.)
Envelope of Vacuum Trajectories:
For a fixed muzzle velocity, every vacuum trajectory will, at some point, be tangent to a curve that is defined as the envelope of trajectories. This envelope of trajectories defines the danger space, to aircraft and ground personnel, associated with a firing range. Although actual projectiles will not travel as far nor as high as in the vacuum envelope, the curve of an actual envelope is strikingly similar in appearance to that of Figure below.
Let's define an envelope of vacuum trajectory for a .30-06, 180 grain bullet,
with a muzzle velocity of 2800 fps. We'll work the formula out for a muzzle
angle of 90º to the horizon, 65º to the horizon, and 45º to the horizon.
The equation for this envelope is:
Equation 1-16:
Y_{e} = (V_{o}² ÷ (2 *
g)) - (g * X_{w}² ÷ (2 * V_{o}²))
First we need to find the range of a projectile with a velocity of 2800 fps at an angle of 65º and 45º. The range of an angle of 90º to the horizon is zero.
Equation 1-4:
X_{w} = (V_{o}² * sin (2 * Ø_{o})) ÷ g
X_{w} = (2800² ft²/sec² * sin (2 * 65º)) ÷ 32.1734 ft/sec²
X_{w} = (2800² ft²/sec² * sin (130º)) ÷ 32.1734 ft/sec²
X_{w} = (2800² ft * sin (130º)) ÷ 32.1734
X_{w} = (7840000 ft * 0.7660444431) ÷ 32.1734
X_{w} = 6005788.434 ft ÷ 32.1734
X_{w} = 186669.3739 ft = 35.35404809 miles
AND
Equation 1-4:
X_{w} = (V_{o}² * sin (2 * Ø_{o})) ÷ g
X_{w} = (2800² ft²/sec² * sin (2 * 45º)) ÷ 32.1734 ft/sec²
X_{w} = (2800² ft²/sec² * sin (90º)) ÷ 32.1734 ft/sec²
X_{w} = (2800² ft * sin (90º)) ÷ 32.1734
X_{w} = (7840000 ft * 1) ÷ 32.1734
X_{w} = 7840000 ft ft ÷ 32.1734
X_{w} = 243679.5614 ft = 46.15143208 miles
We now plug all three of our X_{w} into the envelope of vacuum trajectory formula.
First:
X_{w} of zero for 90º:
Equation 1-16:
Y_{e} = (V_{o}² ÷ (2 * g)) - (g * X_{w}² ÷ (2
* V_{o}²))
Y_{e} = (2800² ft²/sec² ÷ (2 * 32.1734 ft/sec²))
- (32.1734 ft/sec² * 0² ft² ÷ (2 * 2800² ft²/sec²))
Y_{e} = (7840000 ft²/sec² ÷ (2 * 32.1734 ft/sec²))
- (32.1734 ft/sec² * 0 ft² ÷ (2 * 7840000 ft²/sec²))
Y_{e} = (7840000 ft²/sec² ÷ 64.3468 ft/sec²) - (32.1734 ft * 0 ÷ 15680000)
Y_{e} = (7840000 ft ÷ 64.3468) - (0 ÷ 15680000 ft)
Y_{e} = 121839.7807 ft - 0 ft
Y_{e} = 121839.7807 ft = 23.07571604 miles; (X_{w} = 0 ft, Y_{e} = 121839.7807 ft)
Second:
X_{w} of 186669.3739 ft for 65º:
Equation 1-16:
Y_{e} = (V_{o}² ÷( 2 * g)) - (g * X_{w}² ÷ (2
* V_{o}²))
Y_{e} = (2800² ft²/sec² ÷ (2 * 32.1734 ft/sec²))
- (32.1734 ft/sec² * 186669.3739² ft² ÷ (2 * 2800² ft²/sec²))
Y_{e} = (7840000 ft²/sec² ÷ (2 * 32.1734 ft/sec²))
- (32.1734 ft/sec² * 34845455152.21800121 ft² ÷ (2 * 7840000
ft²/sec²))
Y_{e} = (7840000 ft ÷ (2 * 32.1734)) - (32.1734 ft * 34845455152.21800121 ÷ (2
* 7840000))
Y_{e} = (7840000 ft ÷ 64.3468) - (1121096766794.370640129814 ft ÷ 15680000)
Y_{e} = 121839.7807 ft - 71498.5182904573 ft
Y_{e} = 50341.2624095427 ft = 9.53433 miles; (X_{w} = 186669.3739 ft, Y_{e} = 50341.2624095427 ft)
Third:
X_{w} of 243679.5614 ft for 45º:
Equation 1-16:
Y_{e} = (V_{o}² ÷ (2 * g)) - (g * X_{w}² ÷ (2
* V_{o}²))
Y_{e} = (2800² ft²/sec² ÷ (2 * 32.1734 ft/sec²))
- (32.1734 ft/sec² * 243679.5614² ft² ÷ (2 * 2800² ft²/sec²))
Y_{e} = (7840000 ft²/sec² ÷ (2 * 32.1734 ft/sec²))
- (32.1734 ft/sec² * 59379728644.09636996 ft² ÷ (2 * 7840000
ft²/sec²))
Y_{e} = (7840000 ft ÷ (2 * 32.1734)) - (32.1734 ft * 59379728644.09636996 ÷ (2
* 7840000))
Y_{e} = (7840000 ft ÷ 64.3468) - (1910447761557.970149271064 ft ÷ 15680000)
Y_{e} = 121839.7807 ft - 121839.7807 ft
Y_{e} = 0 ft = 0 miles; (X_{w} = 243679.5614 ft, Y_{e} = 0 ft)
If this were plotted, the plot would look something like this:
The reason why you only see two trajectories is that the third is overlaid on the 'Y' axes.
Flat-Fire Approximation to the Vacuum Trajectory:
The equation for motion in a vacuum is:
Equation 1-8:
Y_{w} = X_{w} * tan Ø_{o} - (g * X_{w}²
÷ (2 * V_{o}² * cos² Ø_{o}))
This equation can be rewritten as:
Equation 1-17:
Y_{w} = X_{w} * tan Ø_{o} - ((g * X_{w}²
÷ (2 * V_{o}²)) * sec² Ø_{o})
Back before the turn of the century up till the end of World War 2 (WWII) the numeric calculation were very time consuming and costly, therefore the approximations were the fastest and less costly way to get satisfactory results to the problems of motion. The flat-fire Approximation was fine assuming that the height of Y was everywhere vary close to the line from the firearm bore to the target, X, as long as one could live with a little less accuracy. The derivative of the trajectory height with respect to elevation angle, at a fixed range, is:
Equation 1-18:
dY ÷ dØ_{o} = X_{w} * (1 -
((g * X_{w} ÷ V_{o}²) * tan Ø_{o})
* sec²
Ø_{o}
If tan² Ø_{o} « 1 then sec² Ø_{o} may be replaced by unity with one more than a one % error. Tan² Ø_{o} « being restricted to a vacuum trajectory for which Ø_{o} < 5 degrees, i.e., for which tan Ø_{o} < 0.1, may be treated as a flat-fire trajectory, we have:
Equation 1-19:
[Y_{w} = X_{w} * tan Ø_{o} -
(g * X_{w}² ÷ (2* V_{o}²))]
Equation 1-20:
[dY ÷ dØ_{o} = X_{w} (1 - ((g * X_{w} ÷ V_{o}²) * tan Ø_{o})]
And for shorter ranges, where (g * X_{w} ÷ V_{o}²) * tan Ø_{o} « 1, then the equation may be further approximated as:
Equation 1-21:
dY ÷ dØ_{o} = X_{w}
The above equation is usually called the "rigid trajectory" approximation, because a change in elevation angle produces a change in trajectory height that increases in direct proportion to an increasing range, and the trajectory appears to rotate "rigidly" about the origin. The rigid trajectory approximation is valid for a vacuum trajectory if:
Equation 1-22:
[(g * X_{w} ÷ V_{o}²)
* tan Ø_{o}
« 1]
The error in trajectory height from the flat-fire approximation is the difference between:
Equation 1-19:
Y_{w} = X_{w} * tan Ø_{o} -
(g * X_{w}² ÷ (2* V_{o}²))
AND
Equation 1-8:
Y_{w} = X_{w} * tan Ø_{o} - (g * X_{w}²
÷ (2 * V_{o}² * cos² Ø_{o})
That is, the approximation:
Equation 1-19:
Y_{w} = X_{w} * tan Ø_{o} -
(g * X_{w}²
÷ (2* V_{o}²))
is to high by the amount:
Equation 1-23:
[E_{y} = ((g * X_{w}²) ÷ (2 * V_{o}²)) * tan² Ø_{o}]
and
this vertical error in the flat-fire approximation to the vacuum trajectory
increases rapidly with increased range and elevation angle.
For example, let us use the same example
as we did before and that was our initial conditions is a muzzle
velocity (V_{o}) = 2800 fps and range (X_{w}) = 300 yards or
900 ft. We will find the angle of departure by the formula:
Equation 1-4:
X_{w} = (V_{o}² * sin (2 * Ø_{o})) ÷ g
and after solving for Ø_{o}, our equation looks like:
Equation 1-7:
Ø_{o} = sin^{-1} ((X_{w} * g) ÷ V_{o}²) ÷ 2
Ø_{o} = sin^{-1} ((900 ft * 32.1734 ft/sec²) ÷ 2800² ft²/sec²) ÷ 2
Ø_{o} = sin^{-1} ((900 ft * 32.1734 ft/sec²) ÷ 2800² ft²/sec²) ÷ 2
Ø_{o} = sin^{-1} ((900 * 32.1734) ÷ 2800²) ÷ 2
Ø_{o} = sin^{-1} (28956.06 ÷ 2800²) ÷ 2
Ø_{o} = sin^{-1} (28956.06 ÷ 7840000) ÷ 2
Ø_{o} = sin^{-1} (0.003693375) ÷ 2
Ø_{o} = 0.2116152808 ÷ 2
Ø_{o} = 0.1058076404º
Now the same shot but zeroed at a range (X_{w}) of 600 yards or 1800 ft, for a comparison.
Equation 1-7:
Ø_{o} = sin^{-1} ((X_{w} * g) ÷ V_{o}²) ÷ 2
Ø_{o} = sin^{-1} ((1800 ft * 32.1734 ft/sec²) ÷ 2800² ft²/sec²) ÷ 2
Ø_{o} = sin^{-1} ((1800 ft * 32.1734 ft/sec²) ÷ 2800² ft²/sec²) ÷ 2
Ø_{o} = sin^{-1} ((1800 * 32.1734) ÷ 2800²) ÷ 2
Ø_{o} = sin^{-1} (57912.12 ÷ 2800²) ÷ 2
Ø_{o} = sin^{-1} (57912.12 ÷ 7840000) ÷ 2
Ø_{o} = sin^{-1} (0.00738675) ÷ 2
Ø_{o} = 0.4232334483 ÷ 2
Ø_{o} = 0.2116167241º
The error for each of these ranges resulting from using the flat-fire approximation is:
Equation 1-23:
E_{y} = ((g * X_{w}²) ÷ (2 * V_{o}²)) * tan² Ø_{o}
E_{y} = ((32.1734 ft/sec² * 900² ft²) ÷ (2 * 2800² ft²/sec²)) * tan² 0.1058076404º
E_{y} = ((32.1734 ft * 900²) ÷ (2 * 2800²)) * tan² 0.1058076404º
E_{y} = ((32.1734 ft * 810000) ÷ (2 * 7840000)) * 0.0018466938²
E_{y} = (26060454 ÷ 15680000) * 0.0000034103
E_{y} = 166201875 * 0.0000034103
E_{y} = .0000056679 ft = .0000680154 inches
Equation 1-23:
E_{y} = ((g * X_{w}²) ÷ (2 * V_{o}²)) * tan² Ø_{o}
E_{y} = ((32.1734 ft/sec² * 1800² ft²) ÷ (2 * 2800² ft²/sec²)) * tan² 0.2116167241º
E_{y} = ((32.1734 ft * 1800²) ÷ (2 * 2800²)) * tan² 0.2116167241º
E_{y} = ((32.1734 ft * 3240000) ÷ (2 * 7840000)) * 0.0036934254²
E_{y} = (104241816 ÷ 15680000) * 0.0000136414
E_{y} = 6.648075 * 0.0000034103
E_{y} = 0.000090689 ft = 0.0010882679 inches
By doubling the range we nearly doubled the
angle of departure. But the error jumped by a factor of 2^{4} power or 16 times.
If we take the primary equation and add the error
equation.
Equation 1-8:
Y_{w} = X_{w} * tan Ø_{o} -
((g * X_{w}²)
÷ (2 * V_{o}² * cos² Ø_{o}))
PLUS
Equation 1-23:
E_{y} = ((g * X_{w}²) ÷ (2 * V_{o}²)) * tan² Ø_{o}
We will get the same result as if we used the Flat-Fire Approximation to the Vacuum Trajectory alone. Let's see, we will use one of the examples from above:
Equation 1-8:
Y_{w} = X_{w} * tan Ø_{o} -
((g * X_{w}²) ÷ (2 * V_{o}² * cos² Ø_{o}))
Y_{w} = 1800 ft. * tan 0.2116167241º - (32.1734 ft/sec² * 1800² ft.² ÷ (2
* 2800² ft²/sec² * cos² 0.2116167241º))
Y_{w} = 1800 ft. * tan 0.2116167241º - (32.1734 ft * 1800² ÷ (2
* 2800² * cos² 0.2116167241º))
Y_{w} = 1800 ft. * 0.0036934254 - (32.1734 ft * 1800² ÷ (2
* 2800² * 0.9999931794²))
Y_{w} = 6.64816572 ft. - (32.1734 ft * 3240000 ÷ (2 * 7840000
* 0.9999863588))
Y_{w} = 6.64816572 ft. - (104241816 ft.
÷ (15680000 * 0.9999863588))
Y_{w} = 6.64816572 ft. - (104241816 ft.
÷ 15679786.11)
Y_{w} = 6.64816572 ft. - 6.648165689 ft.
Y_{w} = 0.000000031 ft.
Equation 1-23:
E_{y} = ((g * X_{w}²) ÷ (2 * V_{o}²)) * tan² Ø_{o}
E_{y} = ((32.1734 ft/sec² * 1800² ft.²) ÷ (2 * 2800² ft²/sec²))
* tan² 0.2116167241º
E_{y} = ((32.1734 ft * 1800²) ÷ (2 * 2800²))
* 0.0036934254²
E_{y} = ((32.1734 ft * 3240000) ÷ (2 * 7840000))
* 0.0000136414
E_{y} = (104241816 ft. ÷ 15680000) * 0.0000136414
E_{y} = 6.648075 ft. * 0.0000136414
E_{y} = 0.0000906891 ft.
And we just add the two together and we get:
Equation 1-24:
Y_{w} + E_{y}
0.000000031 ft. + 0.0000906891 ft.
0.0000907201 ft.
Uphill/downhill Firing in a Vacuum:
Special
Note: The uphill/downhill section is a very difficult subject to understand
and especially hard to simplify were it makes any sense. For this reason
I have used almost verbatim form the text in "Modern Exterior Ballistics"
by Robert L. McCoy, pages 47 - 51.
The vacuum
trajectory of uphill and downhill Firing for all practical purposes is the
same. But I will show the more interested reader some very important differences.
First of all, we will only be adding one term, "A", to the above formulas of
motion and that is an angle called the "superelevation." This is where the
(X,Y) plain is inclined at an angle + or - A, relative to the horizontal. To
form a new coordinate system that is offset by "A", (X_{a},Y_{a}). This
angle is positive for uphill firing, and negative for downhill firing.
The gravitational acceleration vector now has an angle 'A' components to it, -g * sin A, -g * cos A. This changes our equations of motion slightly to the form of:
Equation 1-25:
X_{a} = T_{w} * V_{o} * cos Ø_{o} - (g * T_{w}^{2} * sin A ÷ 2)
and
Equation 1-26:
Y_{a} = T_{w} * V_{o} * sin Ø_{o} - (g * T_{w}^{2} cos A ÷ 2)
Note: If angle 'A' is set to zero these two equations reduce to the original formula of:
Equation 1-27:
X_{w} = T_{w} * V_{o} * cos Ø_{o}
and
Equation 1-28:
Y_{w} = T_{w} * V_{o} * sin Ø_{o} - (g * T_{w}^{2} ÷ 2)
The
elimination of time from the equations is more complex than it is for the
level ground trajectory because both X_{a} and Y_{a} vary
quadratically with time. There are two special cases of the uphill-downhill
vacuum trajectory problem that reduces to relatively simple analytical
forms, and these special cases readily illustrate the interesting nature
of the problem.
The first case involves
an expression for the dependence of slant range to impact, R_{s},
on the two angles, A and Ø_{o}. Through some Mathematical wizardry
the equation:
Equation 1-26:
Y_{a} = T_{w} * V_{o} * sin Ø_{o} - (g * T_{w}^{2} cos A ÷ 2)
is transformed into:
Equation 1-29:
R_{s} ÷ R = [1 - tan Ø_{o} * tan A] * sec A.
R_{s} is the range at the uphill/downhill angle A, and R is the same range along the level ground. If R_{s} / R is > 1 than the slant range to impact along the incline will exceed the level ground range, and if R_{s} / R is < 1 than the slant range will be shorter than the level ground range. The equation:
Equation 1-29:
R_{s} ÷ R = [1 - tan Ø_{o} * tan A] * sec A
Illustrates
some very interesting properties of vacuum trajectory for uphill and
downhill firing.
Since,:
Equation 1-30:
tan (-A) = -tan A
and
Equation 1-31:
sec (-A) = sec A,
than
Equation 1-32:
R_{s} ÷ R > 1
Equation 1-33:
for -90^{o} < A < 0.
This means that the slant range will always exceed the ground range for downhill firing. This makes sense, because gravity ends up aiding the projectile vice retarding it during uphill firing. For uphill firing where
Equation 1-34:
90^{o} > A >0
with very small Ø_{o}
Equation 1-32:
R_{s} ÷ R > 1.
However, for larger superelevation angles:
Equation 1-35:
R_{s} ÷ R < 1,
and the slant range will then be less than the level ground rang. There is one particular value of Ø_{o} for which:
Equation 1-36:
R_{s} ÷ R = 1
at any given positive angle 'A', and this critical value is when:
R_{s} ÷ R
is set to 1 in the equation:
Equation 1-29:
R_{s} ÷ R = [1 - tan Ø_{o} * tan A] * sec A
and solving for Ø_{o-cr} of:
Equation 1-37:
Ø_{o-cr} = tan^{-1} * [(1 - cos A) * cot A]
Where
Ø_{o-cr} = critical superelevation angle for R_{s} ÷ R = 1.
For uphill vacuum
trajectories if:
Equation 1-38:
Ø_{o} < Ø_{o-cr}
than
Equation 1-32:
R_{s} ÷ R > 1,
if Ø_{o} = Ø_{o-cr}
than
Equation 1-36:
R_{s} ÷ R = 1,
and
if Ø_{o} > Ø_{o-cr}
than
Equation 1-35:
R_{s} ÷ R < 1.
The table along with the two figures below illustrate the interesting effect of various angles of 'A' on the vacuum trajectory.
A (Degrees) |
Ø_{o cr} |
0 |
0 |
15 |
7.25 |
30 |
13.06 |
45 |
16.33 |
60 |
16.10 |
75 |
11.23 |
90 |
0 |
The equation Ø_{o-cr} = tan^{-1} * [(1 - cos A) * cot A] is indeterminate at A = 0, and L' Hospital's rule must be used to find Ø_{o-cr} at A = 0 |
The flat-fire approximation to the equation:
Equation 1-29:
R_{s} ÷ R = [1 - tan Ø_{o} * tan A] *
sec A
provides another interesting and useful result. For
Equation 1-39:
|tan Ø_{o} * tan A| << 1
and thus reduces to the simple form of:
R_{s} * R =~ sec A, or (R_{s} * cos A) ÷ R =~ 1 ("=~" means
approximately equal to).
The second form of these two equations is referred to as the rifleman's rule for uphill/downhill firing; if the slant range to the target is R_{s}, the rifle sights should be set for the equivalent horizontal range R_{s} * cos A, in order to hit the target.
However,
the flat-fire approximation for uphill/downhill firing is more restrictive
than for the level ground case. For moderate 'A' angles, tan A is of order
unity, which requires that tan Ø_{o} << 1, which is more
restrictive than the corresponding condition, tan^{2} Ø_{o}
<< 1, for flat-fire across level ground.
The second special case
of the uphill-downhill problem is an expression for the height of impact,
Y_{a}, of the vacuum trajectory, at a slant range equal to the
corresponding level ground impact range. Solving the equation:
Equation 1-25:
X_{a} = T_{w} * V_{o} * cos
Ø_{o} - (g * T_{w}^{2} * sin A ÷ 2)
for time by means of the quadratic formula we get:
Equation 1-40:
t = ((V_{o} * cos Ø_{o}) ÷ (g * sin
A)) * [1 ± the squar root of (1 - ((2 * g * X_{a} * sin
A) ÷ (V_{o}^{2} * cos^{2} Ø_{o}))
The root corresponding to the negative sign before the radical in the above equation is the correct solution. After some algebraic manipulation, the equation may be written in the alternative form:
Equation 1-41:
t = (2 * X_{a}) ÷ ((V_{o} * cos
Ø_{o}) * [1 + the squar root of (1 - ((2 * g * X_{a} *
sin A) ÷ (V_{o}^{2} * cos^{2} Ø_{o})))])
Now, for:
Equation 1-42:
X = R = (V_{o}^{2} ÷ g) * sin (2 * Ø_{o}):
Equation 1-43:
[t = (4 * V_{o} * sin Ø_{o}) ÷ (g * (1 + v))]
Where
Equation 1-44:
[v = the square root of(1 - (4 * tan Ø_{o} * sin A))]
Substituting the equation:
Equation 1-43:
[t = (4 * V_{o} * sin Ø_{o}) ÷ (g * (1 + v))]
into the equation:
Equation 1-28:
[Y_{w} = T_{w} * V_{o} * sin Ø_{o} - (g
* T_{w}^{2} ÷ 2)]
We get the equation:
Equation 1-45:
Y_{a1} = ((4 * V_{o}^{2} *
(sin Ø_{o})^{2}) ÷ (g * (1 + v))) * [1 - ((2 * cos A)
÷ (1 + v))]
Before we explore the properties of equations:
Equation 1-44:
[v = the square root of(1 - (4 * tan Ø_{o} * sin A))]
and
Equation 1-45:
Y_{a1} = ((4 * V_{o}^{2} *
(sin Ø_{o})^{2}) ÷ (g * (1 + v))) * [1 - ((2 * cos A)
÷ (1 + v))
We must first address the restriction implied in the v equation. The parameter v can be real only if the quantity 4 * tan Ø_{o} * sin A < 1. For downhill firing, sin (-A) = -sin A, thus v is always real for any A < 0. However, for firing uphill, the parameter v can be real only if tan Ø_{o} < (csc A) ÷ 4. This inequality defines another restriction on Ø_{o}; the projectile cannot reach an uphill target if the superelevation angle exceeds a maximum value, given by the equation:
Equation 1-46:
Ø_{o max} = tan^{-1}[(csc A) ÷ 4]
The angle Ø_{o max} is the superelevation angle above which the uphill vacuum trajectory can never reach the target. The figure below illustrates the variation of the two critical superelevation angles, Ø_{o cr} and Ø_{o max}, with uphill angle of site, A, and shows how the two curves bound the various solution regions for vacuum trajectories.
Several interesting properties of equations:
Equation 1-44:
[v = the square root of(1 - (4 * tan Ø_{o} * sin A))]
and
Equation 1-45:
Y_{a1} = (((4 * V_{o}^{2} *
(sin Ø_{o})^{2}) ÷ (g * (1 + v))) * [1 - ((2 * cos A)
÷ (1 + v))]
are listed below:
(a)
If A = 0, then v = 1, and Y_{a1} = 0. The impact height vanishes, as it should, for level ground firing.
(b)
If, Ø_{o} = Ø_{o cr} = tan^{-1} [(1 - cos A) * cot A], v = (2 * cos A) - 1, and Y_{a1} = 0. The impact height correctly vanishes fot the critical superelevation angle, wher R_{s} ÷ R = 1.
(c)
For flat-fire (tan Ø_{o} << 1), v [=] 1, and the approximate impact height is given by the equation:
Equation 1-47:
[Y_{a1} = ((2 * V_{o}^{2} * (sin Ø_{o})^{2}) ÷ g) * [1 - cos A]]
Now, cos (-a) = cos A, and we observe that for flat-fire, the trajectory will always intersect the target above center, and the projectile will strike equally high for either uphill or downhill firing.
"Example:"
An air gun fires a heavy projectile at a muzzle velocity of 250 feet per second. Sight settings have been obtained for ranges between 50 yards and 400 yards, on level ground. Determine the impact locations on target placed a the same range, but along a 30 degree uphill incline.
R (Yards) |
R (Feet) |
50 |
150 |
100 |
300 |
200 |
600 |
300 |
900 |
400 |
1200 |
The first step in the solution is to find the gun elevation angles required to hit the level ground target, using equation:
Equation 1-7:
Ø_{o} = sin^{-1} ((X_{w} * g)
÷ V_{o}²) ÷ 2
R (Yards) |
R (Feet) |
Ø_{o} (Degrees) |
50 |
150 |
2.214284152 |
100 |
300 |
4.441937084 |
200 |
600 |
8.995410874 |
300 |
900 |
13.80002990 |
400 |
1200 |
19.07525171 |
For A = 30 degrees use equation:
Equation 1-29:
R_{s} ÷ R = [1 - tan Ø_{o} * tan A] * sec A
to fine the ratio of slant range to level ground range as illustrated in Table 1-4.
R (Yards) |
R (Feet) |
Ø_{o} (Degrees) |
R_{s} ÷ R |
R_{s} |
50 |
150 |
2.214284152 |
1.128923338 |
56.446 |
100 |
300 |
4.441937084 |
1.102912457 |
110.291 |
200 |
600 |
8.995410874 |
1.049165647 |
209.833 |
300 |
900 |
13.80002990 |
0.990951130 |
297.285 |
400 |
1200 |
19.07525171 |
0.924168947 |
369.668 |
The projectile will hit high on the 50, 100, and 200 yard uphill targets, and low on the 300 and 400 yard targets. To find how high or low the impacts will be, we will use the exact parametric equations:
Equation 1-44:
v = the square root of(1 - (4 * tan Ø_{o} * sin A))
and
Equation 1-45:
Y_{a1} = ((4 * V_{o}^{2} * (sin Ø_{o})^{2}) ÷ (g * (1 + v))) * [1 - ((2 * cos A) ÷ (1 + v))]
The flat-fire approximation to the impact height use equation:
Equation 1-47:
Y_{a1} = ((2 * V_{o}^{2} *
(sin Ø_{o})^{2}) ÷ g) * [1 - cos A]
The flat-fire approximation equation is included for a comparison. Table 1-5 illustrates these three equations below.
R (Yards) |
50 |
100 |
200 |
300 |
400 |
v = the square root |
0.9605562963 |
0.9190406717 |
0.8266772807 |
0.7132683761 |
0.5553424413 |
Y_{a1} = ((4 * V_{o}^{2} * |
8.2749906450 |
28.399015430 |
64.645577100 |
-33.94890446 |
-727.4769881 |
Y_{a1} = ((2 * V_{o}^{2} * |
9.324423086 |
37.46675126 |
152.7011399 |
355.3998250 |
667.1241391 |
At 50 yard range, flat-fire is a valid assumption for the low velocity air gun, and the error in equation:
Equation 1-47:
Y_{a1} = ((2 * V_{o}^{2} *
(sin Ø_{o})^{2}) ÷ g) * [1 - cos A]
is just over one inch in impact height. For the longer ranges, all of which violate the flat-fire restriction, the accuracy of equation:
Equation 1-47:
Y_{a1} = ((2 * V_{o}^{2} *
(sin Ø_{o})^{2}) ÷ g) * [1 - cos A]
degrades rapidly, and at the two longest ranges, it predicts
a high impact on the target, when in fact, the impact will be low.
Try repeating the
calculations of the above example for firing downhill along a minus 30
degree incline.
Enough detail has been
included in this section to demonstrate that firing uphill and downhill
is not a trivial problem, even with the simplifying assumption of a
vacuum trajectory. The fact that actual atmosphere trajectories behave
in a remarkably similar fashion is sufficient reason to understand the
behavior of uphill and downhill vacuum trajectories.
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