# Projectile Speed, Crosswind Relationship

1. Jan 3, 2010

What equations can demonstrate if greater projectile speed can or cannot contribute to greater accuracy (or directional stability) of a projectile, while traveling horizontally within an envirnoment containing lateral crosswinds?

In controlled and powered horizontal flight (aircraft or missile), can greater speeds reduce the degree to which a crosswind can deflect an aircraft or missile from its intended course or groundtrack?

This example seems to entail relationships between force (crosswind and projectile, speed and mass) and duration of applied lateral forces (crosswind).

2. Jan 3, 2010

### Astronuc

Staff Emeritus
The effect of a given crosswind would be determined by the time required to travel down range.

The lateral deflection z would be proportional to vz t, assuming the lateral wind speed vz is constant, and t = time of flight. Traveling faster downrange decreases the time, so lateral deflection is decreased.

An aircraft simply points at an angle to the line from origin to destination in order to offset the crosswind.

A projectile can be aimed at an angle to the target to compensate for the drift due to crosswind. It get's a bit complicated if the lateral wind speed is variable or a function of altitude (height).

Range - http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra4

Time of flight - http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra6

3. Jan 12, 2010

It seems varying projectile speed (force) would generate greater or less groundtrack displacement created by a given crosswind, once air mass (density) and projectile mass are known.

It seems this example could be expressed via force vector trigonometry.

True or false?

4. Jan 15, 2010

### zgozvrm

At the very basic level, use vectors: The length of the vectors representing the force of the projectile and the wind. The angle of the vectors representing the direction of travel of the projectile and wind.

The longer the projectile's vector, the smaller the angle of offset between it and the resultant vector (with the wind vector staying constant). Therefore, the projectile is thrown off it's original course by a lesser amount.

Of course, the problem is complicated by other factors such as wind shear, aerodynamics, projectile density, wind density, etc. But, I think that the use of vectors is generally sufficient to prove what you are trying to show.

5. Jan 15, 2010

### DocZaius

Hello,

My intuition could be wrong on this, but I would have thought that the lateral deflection z wouldn't be proportional to vzt. I would think that if we were to follow the projectile's trajectory from above it, we would see the projectile accelerate in the z direction as it travels through the lateral wind. As a result, the z deflection would not have a linear relationship with the time spent in the lateral wind. What am I missing?

6. Jan 15, 2010

### zgozvrm

I think the OP is assuming no deceleration or deviation of flight path of the projectile due to gravity and/or air friction.

7. Jan 15, 2010

### DocZaius

My statement concerned lateral wind, which was adressed in post #2 Wouldn't lateral wind deviate the flight path?

It seems that you are under the impression that I was talking about either the effect of gravity (up and down), or air friction (forwards and backwards) on the projectile. I was instead talking about the effect of crosswind (left and right) on the projectile, which seems to be the central topic of this thread.

I'm actually glad you mentioned air friction though, because I think that would actually be a better way to take lateral wind into account. Treat the lateral wind speed as the initial relative lateral velocity in the drag force equation F= -bv (in the lateral direction!)

Last edited: Jan 15, 2010
8. Jan 15, 2010

### zgozvrm

Yes, I'm agreeing with you as to whether lateral wind will deviate the flight path.

Your previous post stated that you didn't think that the deflection was proportional to the speed of the projectile. I was simply stating that, in reality, you are correct and this "in-proportionality" would be due to things like wind shear, aerodynamics, air density, etc. But, I don't think the OP was concerned with those variables.

The question is whether the deflection of the projectile's path is lessened merely by increasing it's velocity. To simplify things, look at the basic forces acting upon the object and you can see that the statement is true.

Yes, there are other forces acting on it as well. But the OP was talking about varying only one of them ... the velocity (speed) of the projectile.

For example, if we were to shoot an arrow at 250 feet per second at a target 60 yards away with absolutely no wind, it is of course acted upon by several other forces: gravity, air density, the aerodynamics of the arrow (smoothness of the shaft, angle and straightness of the feathers), etc.

If we add a cross-wind to the situation (of, say 2 mph), the arrow will definitely be deflected from it's intended path (all other variables remaining the same). If that arrow was now shot at a velocity of 325 feet per second from the same position, at the same angle and elevation in respect to the target with the same 2 mph cross-wind, the arrow will surely miss its mark by a lesser amount.

Now if we repeat this experiment by changing only the distance to the target, the result is the same: The faster the arrow, the closer to its intended path through a cross-wind. (But NOT by a proportional amount!)

9. Jan 15, 2010

### DocZaius

I thought that the OP asked for equations, which is why I tried to go just one degree of complication further than simply saying "greater velocity means less time spent in crosswind which means less cross deflection." But maybe you're right that the level of detail in the answer I sought wasn't in proportion with the question's requested level of detail. (previous sentence said in a friendly joking manner)

10. Jan 15, 2010

### zgozvrm

Yes, the OP did ask for equations, I simply stated that he could use vectors, leaving the trigonometry to him.

So, if we assume all other variables to remain the same, we can basically take them out of consideration.

There are three types of cross-winds that can occur:
1) directly perpendicular to the intended (aimed) flight of the projectile
2) at an angle less than perpendicular to the intended flight of the projectile
3) at an angle more than perpendicular to the intended flight of the projectile

Using my arrow example, let's assume that the arrow is shot due North (90 degrees) at 250 feet per second (fps) and we'll take each of the three cross-wind situations into account.

First, lets assume a 2 mph cross-wind that is blowing West to East (0 degrees). 2 mph converts to approximately 2.933 fps. So we have a resultant vector of magnitude (speed) = $$\sqrt{250^2 + 2.933^2} \approx 250.017$$ fps. The angle of deflection can be found by:

$$\tan^{-1}\left( \frac{2.933}{250} \right) \approx 0.672^\circ$$

Now, if we change the arrow speed to 325 fps, we get a resulting speed of $$\sqrt{325^2 + 2.933^2} \approx 325.013$$ fps and an angle of:

$$\tan^{-1}\left( \frac{2.933}{325} \right) \approx 0.517^\circ$$
which is less deflection.

Now, let's do the same thing with a SW to NE wind (45 degrees):
At 250 fps (without showing all of the math), we have a resulting speed of approx. 252.083 fps and a deflection angle of about 0.471 degrees.
Increasing the arrow speed to 325 fps, we have a resulting speed of approx. 327.081 fps and a deflection angle of about 0.363 degrees.
Again, the increased speed lessened the deflection angle caused by the cross-wind.

Lastly, let's use a NW to SE wind (315 degrees):
At 250 fps: 247.934 fps @ 0.479 degrees
At 325 fps: 322.932 fps @ 0.368 degrees

So you can see that in each case, the angle of deflection is lessened by simply increasing the velocity of the projectile. Of course these values are not completely accurate since other forces do act upon the flight of the arrow. But those other forces act upon the arrow equally at a fixed distance, so it isn't necessary to use the complicated math required to calculate the deflection angles simply to prove the point of the question.

11. Jan 18, 2010

In this calculation, it seems the resultant plane direction (61.39°) is incorrect by 180°. True? Suggestions?

Square roots are poorly represented by: (1/2)

P (plane): Bearing 241° (traveling southwest) @ 683 f/s; W (wind): traveling south @ 14 f/s (10 mph)

[683 f/s cos(241°), 683 f/s sin(241°)] = -331, -597
[-14 cos(90°), -14 sin(90°)] = 0, 10

-331 + 0 = -331
-597 + -14 = -611

||P + W|| = 331² + 611² = 465, 970(1/2) = 694 f/s

tan −1(611/331) = 61.55°

This calculation "seems" correct but may also suffer from a similar error.

P (plane): Bearing 52° (traveling northeast) @ 799 f/s; W (wind): traveling south) @ 14 f/s (10 mph)

[799 f/s cos(52°), 799 f/s sin(52°)] = 491, 629
[-14 cos(90°), -14 sin(90°)] = 0, -10

491 + 0 = 491
629 + -14 = 615

||P + W|| = 491² + 615² = 624,242(1/2) = 786 f/s

tan −1(615/491) = 51.39°

Last edited: Jan 19, 2010