Blaine Fields
Well-Known Member
According to my references, in the Northern hemisphere the lateral or horizontal component is always to the right irrespective of shot direction. Does anyone have a source which would challenge this claim?
Coriolis effect
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Blaine, That is ridiculous. Common sence tells you that that would be impossible, as the entire premise of coriolis effect, deals with and varies with direction.
Now, not that this would ever matter from a small arms ballistics point of view, but, the effect from 100% at a north/south shot to 0% at an east/west shot could be roughly estimated by multipling the resultant vector times the sine of the angle from north/south. Gee, don't forget to add in the gyroscopic stability cause that's gonna throw a wrench in the equasion!
Regardless of the hemisphere, a shot fired tward the equator would impact west of it's intended point of aim, and a shot fired away from the equator would impact to the east.
Shots fired due east or west at the equator will have no resultant effect, and shots fired due east or west at any point away from the equator will exhibit a very small effect, spelled VERY SMALL, because of thier altitude varying. If the altitude is increasing, then, so is the southward component of motion due to the orbit not being geo-syncronous.
Since you have put me through this, my challenge to you is to either refute this, or, take a 210gr Berger 30 Cal bullet shot at it's advertised BC, at 2800fps, at an altitude of sea level, From the equator, due north/west (45deg), an impact at 1000meters exactly on target, with the target at exactly the same sea level altitude, and a nice easy theoretical GS of 1.0 . All indoors with no wind, 70 deg F, 70% Hum, 29.5Bar, and you tell me, how far the impact is affected by coriolis effect as opposed to the same shot fired due south east. Should be an easy one! You only need to calculate 1 of the 2 and multiply the result by 2. Simple!
Now, not that this would ever matter from a small arms ballistics point of view, but, the effect from 100% at a north/south shot to 0% at an east/west shot could be roughly estimated by multipling the resultant vector times the sine of the angle from north/south. Gee, don't forget to add in the gyroscopic stability cause that's gonna throw a wrench in the equasion!
Regardless of the hemisphere, a shot fired tward the equator would impact west of it's intended point of aim, and a shot fired away from the equator would impact to the east.
Shots fired due east or west at the equator will have no resultant effect, and shots fired due east or west at any point away from the equator will exhibit a very small effect, spelled VERY SMALL, because of thier altitude varying. If the altitude is increasing, then, so is the southward component of motion due to the orbit not being geo-syncronous.
Since you have put me through this, my challenge to you is to either refute this, or, take a 210gr Berger 30 Cal bullet shot at it's advertised BC, at 2800fps, at an altitude of sea level, From the equator, due north/west (45deg), an impact at 1000meters exactly on target, with the target at exactly the same sea level altitude, and a nice easy theoretical GS of 1.0 . All indoors with no wind, 70 deg F, 70% Hum, 29.5Bar, and you tell me, how far the impact is affected by coriolis effect as opposed to the same shot fired due south east. Should be an easy one! You only need to calculate 1 of the 2 and multiply the result by 2. Simple!
Hell, I think I want in on this one!
As I understand it... in the Norhtern hemisphere Coriolis deflects to the right and it deflects to the left in the southern hemisphere. I find this easiest to understand as when viewed from a downward view above the North Pole... the Earth spins counter-clockwise BUT when this downward view is from above the South Pole the Earth is spinning clockwise.... the Earth spins in opposite directions when comparing North and South hemispheres.
Deflection from Coriolis as I understand it is due to the projectile traveling at the Earth surface speed at it's flight origin and retaining this induced speed as it travels to another part of the earth that may very well have a different surface speed.
For a quick set of numbers (in a special environment, no drag) and an example.... earth nominally 60 miles per degree at the equator, 360 degrees in a circle.....21,600 miles for one revolution...24 hours per rev = 900 mph surface speed at the equator. Shoot a round north or south with a 900 mph origin speed and have it fly for a minute at 5280 fps (one mile per second). In that time it will have flown 1 degree (60 miles) north. At this distance from the equator the earth (our example earth) is only 21,223 mile around so it's surface speed is 884 mph.... Our projectile has an origin speed of 900 mph and when it arrives on degree north it's at a spot that is moving slower that the origin by 6 mph or 8.5 feet per second, with a flight time of 60 seconds * 8.5 FPS we should have a displacement of about 510 feet to the east of the intended target (an arc to the right in the north hemisphere and an arc to the left in the south hemishpere).
As I understand it... in the Norhtern hemisphere Coriolis deflects to the right and it deflects to the left in the southern hemisphere. I find this easiest to understand as when viewed from a downward view above the North Pole... the Earth spins counter-clockwise BUT when this downward view is from above the South Pole the Earth is spinning clockwise.... the Earth spins in opposite directions when comparing North and South hemispheres.
Deflection from Coriolis as I understand it is due to the projectile traveling at the Earth surface speed at it's flight origin and retaining this induced speed as it travels to another part of the earth that may very well have a different surface speed.
For a quick set of numbers (in a special environment, no drag) and an example.... earth nominally 60 miles per degree at the equator, 360 degrees in a circle.....21,600 miles for one revolution...24 hours per rev = 900 mph surface speed at the equator. Shoot a round north or south with a 900 mph origin speed and have it fly for a minute at 5280 fps (one mile per second). In that time it will have flown 1 degree (60 miles) north. At this distance from the equator the earth (our example earth) is only 21,223 mile around so it's surface speed is 884 mph.... Our projectile has an origin speed of 900 mph and when it arrives on degree north it's at a spot that is moving slower that the origin by 6 mph or 8.5 feet per second, with a flight time of 60 seconds * 8.5 FPS we should have a displacement of about 510 feet to the east of the intended target (an arc to the right in the north hemisphere and an arc to the left in the south hemishpere).
Bzzzz, wrong. Sorry Dave.
Your example assumes that the shot is being takn from the equator. That is in contradiction to the original posters question. What is asked is if the shot is taken from the northern hemisphere, will it always have an effect in one direction regardless of the direction of the shot and the answer is NO it will not.
Stand at a latitude any other than the equator and as per your view, the relationship of east/west right/left vary depending weather you are standing facing north or south. Ignore the common viewpoint of looking down from above the poles, you are standing on the earth. Somewhere. If you are in the N Hemis, facing north, east is to your right. If you are facing south, east is to your left. Now your 2 shots have opposite, but not equal effects.
Your example assumes that the shot is being takn from the equator. That is in contradiction to the original posters question. What is asked is if the shot is taken from the northern hemisphere, will it always have an effect in one direction regardless of the direction of the shot and the answer is NO it will not.
Stand at a latitude any other than the equator and as per your view, the relationship of east/west right/left vary depending weather you are standing facing north or south. Ignore the common viewpoint of looking down from above the poles, you are standing on the earth. Somewhere. If you are in the N Hemis, facing north, east is to your right. If you are facing south, east is to your left. Now your 2 shots have opposite, but not equal effects.
Blaine Fields
Well-Known Member
According to Prof. Pejsa ("Modern Practical Ballistics") and Robert McCoy ("Modern Exterior Ballistics") the following describes lateral deflection resulting from a turning earth (i.e., the Coriolis effect):
Deflection = (K*R^2*sin L)/V where K is a constant, R is the range, L is the latitude and V is the average velocity over the range R.
According to McCoy, all deflections are to the right in the Northern hemisphere and math-wise the formula presented will not change in sign.
Intuitively this is hard to accept. Hence, my post: does anyone have a source to the contrary?
Deflection = (K*R^2*sin L)/V where K is a constant, R is the range, L is the latitude and V is the average velocity over the range R.
According to McCoy, all deflections are to the right in the Northern hemisphere and math-wise the formula presented will not change in sign.
Intuitively this is hard to accept. Hence, my post: does anyone have a source to the contrary?
As I understand it... After all, I'm not a rocket scientist
.
Okay, northern hemisphere only. Shots will always be deflected to the right in varying amounts with the exception of a shot EXACTLY due east or west. The amount of the deflection is dependant upon origin and target surface speed and the projectile time of flight. We're in essence shooting at a moving target from a moving position... the target is always moving east and the origin position is always moving east...when the speed of the two positions match there is zero deflection.
Shots fired at intevals of the hours on a clock face... 12:00 being due north toward the geographic pole, 3:00 due east, 6:00 due south and 9:00 due west.
Shot fired due north...greater surface speed at origin than at target position... projectile if deflected to the right (east) as seen by the shooter (due north and south have the greatest deflection for this effect). Repeat at 1:00...same shot angle and resultant distance travel but less eastward (right as viewed by the shooter) deflection because the target location is a nearer the same surface speed.... Repeat at 2:00 position and again lesser eastward deflection (right as viewed by shooter) as the target and origin surface speeds are closer. Shoot at 3:00, same origin and target surface speed so no deflection... note that in theory the target can only be EXACTLY due east. Now we swing to the 4:00 position and shoot toward the south east....the origin surface speed is LESS than the target surface speed so the projectile is deflect to the west of the target (to the right as viewed by the shooter). Swing to 5:00 and again the projectile is deflected to the west (the right as viewed by the shooter) but to more extent than a 4:00 position shot. Shoot at 6:00 position and the effect is the greatest again and the projectile is delected to the west (the right again).
The due east and due west "no deflection" zones are only possible if the origin and target are on exactly the same latitude. A pretty skinny possibility for a shot at a distance of 60 miles as in my previous example.
Okay, northern hemisphere only. Shots will always be deflected to the right in varying amounts with the exception of a shot EXACTLY due east or west. The amount of the deflection is dependant upon origin and target surface speed and the projectile time of flight. We're in essence shooting at a moving target from a moving position... the target is always moving east and the origin position is always moving east...when the speed of the two positions match there is zero deflection.
Shots fired at intevals of the hours on a clock face... 12:00 being due north toward the geographic pole, 3:00 due east, 6:00 due south and 9:00 due west.
Shot fired due north...greater surface speed at origin than at target position... projectile if deflected to the right (east) as seen by the shooter (due north and south have the greatest deflection for this effect). Repeat at 1:00...same shot angle and resultant distance travel but less eastward (right as viewed by the shooter) deflection because the target location is a nearer the same surface speed.... Repeat at 2:00 position and again lesser eastward deflection (right as viewed by shooter) as the target and origin surface speeds are closer. Shoot at 3:00, same origin and target surface speed so no deflection... note that in theory the target can only be EXACTLY due east. Now we swing to the 4:00 position and shoot toward the south east....the origin surface speed is LESS than the target surface speed so the projectile is deflect to the west of the target (to the right as viewed by the shooter). Swing to 5:00 and again the projectile is deflected to the west (the right as viewed by the shooter) but to more extent than a 4:00 position shot. Shoot at 6:00 position and the effect is the greatest again and the projectile is delected to the west (the right again).
The due east and due west "no deflection" zones are only possible if the origin and target are on exactly the same latitude. A pretty skinny possibility for a shot at a distance of 60 miles as in my previous example.
Blaine
I can't see any other way for it to work... deflection is to the right in varying amounts in the northern hemisphere. Perhaps we should consult the folks that built the naval gunfire program for the British Navy... Heard it was a problem for them during the Falklands fiasco.
I can't see any other way for it to work... deflection is to the right in varying amounts in the northern hemisphere. Perhaps we should consult the folks that built the naval gunfire program for the British Navy... Heard it was a problem for them during the Falklands fiasco.
Refering to the Ohio St U Physics Dept there is an example that actually even refers to artillery projectiles in section 3.1. This explaination is contrary to Dave's. My guess on this conclusion, though it is not mentioned specifically, is that in addition to the eastward velocity component, there must also be an acceleration due to yaw during rotation of the origin. The yaw, independant of the eastward velocity (they always use east in the example) is going to produce a lateral velocity that, relatively speaking will increase as it nears the equator, reducing the effect of the southward shots "right hand" motion. Conversely, it also reduces the apparent eastward velocity when fired northward.
I guess another way to think of it is, those velocities would only be unaffected if you fired the shot dead verticle at which time the projectile would act like a geo-synchronous sattelite and fall right back on the shooter. It would continue to rotate at exactly the same speed as the earth, then gravity would set in and bring it straight back down to the same spot. (regardless of the latitude)
I guess another way to think of it is, those velocities would only be unaffected if you fired the shot dead verticle at which time the projectile would act like a geo-synchronous sattelite and fall right back on the shooter. It would continue to rotate at exactly the same speed as the earth, then gravity would set in and bring it straight back down to the same spot. (regardless of the latitude)
4mesh063
In reference to section 3.1 in the article
"3.1. I Feel The Earth Move Under My Feet: North/South Motion
Note first that all points on the Earth have the same rotational velocity, w (they go around once per day). Also, places at different latitudes have different linear speeds. A point near the equator may go around a thousand miles in an hour, while one near the North Pole could be moving only a few dozen miles in an hour.
Normally, objects in contact with the ground travel the same speed as the ground they stand on. As a result, the Coriolis force generally doesn't have a noticeable effect to people on the ground; the speed of the point you're standing on and the speed of the point you're stepping onto are too close for you to tell the difference. Or, looking back at the Coriolis Force equation above, if the velocity relative to the rotating frame (the Earth) is zero, so is the Coriolis force.
However, when an object moves north or south and is not firmly connected to the ground (air, artillery fire, etc), then it maintains its initial eastward speed as it moves. This is just an application of Newton's First Law. An object moving east continues going east at that speed (both direction and magnitude remain the same) until something exerts a force on it to change its velocity. Objects launched to the north from the equator retain the eastward component of velocity of other objects sitting at the equator. But if they travel far enough away from the equator, they will no longer be going east at the same speed as the ground beneath them.
The result is that an object traveling away from the equator will eventually be heading east faster than the ground below it and will seem to be forced east by some mysterious force. Objects traveling towards the equator will eventually be going more slowly than the ground beneath them and will seem to be forced west."
I do not see this as contrary to my statement(s) perhaps I explained poorly and/or there was an interpretion problem. As in my statement(s), both target and origin site have eastward but dissimilar speeds. If the speed of the target is less than the speed of the origin site ( a northerly shot in the northern hemisphere) the projectile will be deflected to the east and if the target has greater eastward speed than the origin site the projectile will be deflected to the west (a southerly shot in the northern hemishpere). Whether the varying shot deflection is eastward or westward it's always to the right for Coriolis in the northern hemisphere.
In reference to section 3.1 in the article
"3.1. I Feel The Earth Move Under My Feet: North/South Motion
Note first that all points on the Earth have the same rotational velocity, w (they go around once per day). Also, places at different latitudes have different linear speeds. A point near the equator may go around a thousand miles in an hour, while one near the North Pole could be moving only a few dozen miles in an hour.
Normally, objects in contact with the ground travel the same speed as the ground they stand on. As a result, the Coriolis force generally doesn't have a noticeable effect to people on the ground; the speed of the point you're standing on and the speed of the point you're stepping onto are too close for you to tell the difference. Or, looking back at the Coriolis Force equation above, if the velocity relative to the rotating frame (the Earth) is zero, so is the Coriolis force.
However, when an object moves north or south and is not firmly connected to the ground (air, artillery fire, etc), then it maintains its initial eastward speed as it moves. This is just an application of Newton's First Law. An object moving east continues going east at that speed (both direction and magnitude remain the same) until something exerts a force on it to change its velocity. Objects launched to the north from the equator retain the eastward component of velocity of other objects sitting at the equator. But if they travel far enough away from the equator, they will no longer be going east at the same speed as the ground beneath them.
The result is that an object traveling away from the equator will eventually be heading east faster than the ground below it and will seem to be forced east by some mysterious force. Objects traveling towards the equator will eventually be going more slowly than the ground beneath them and will seem to be forced west."
I do not see this as contrary to my statement(s) perhaps I explained poorly and/or there was an interpretion problem. As in my statement(s), both target and origin site have eastward but dissimilar speeds. If the speed of the target is less than the speed of the origin site ( a northerly shot in the northern hemisphere) the projectile will be deflected to the east and if the target has greater eastward speed than the origin site the projectile will be deflected to the west (a southerly shot in the northern hemishpere). Whether the varying shot deflection is eastward or westward it's always to the right for Coriolis in the northern hemisphere.
Brent
Well-Known Member
Phil,
The way I understood Dave's explanation and the article you posted the link to, which Dave pasted above, says the same thing?
POI will always be to the right in the northern, and to the left in the southern hemisphere.
The result is that an object traveling away from the equator will eventually be heading east faster than the ground below it and will seem to be forced east by some mysterious force. Objects traveling towards the equator will eventually be going more slowly than the ground beneath them and will seem to be forced west."
Now, I'll bet it has a varying degree of deviation depending on what latitude you're shooting from as well. In the same hemisphere and shooting the same direction and the same distance that is...
The way I understood Dave's explanation and the article you posted the link to, which Dave pasted above, says the same thing?
POI will always be to the right in the northern, and to the left in the southern hemisphere.
The result is that an object traveling away from the equator will eventually be heading east faster than the ground below it and will seem to be forced east by some mysterious force. Objects traveling towards the equator will eventually be going more slowly than the ground beneath them and will seem to be forced west."
Now, I'll bet it has a varying degree of deviation depending on what latitude you're shooting from as well. In the same hemisphere and shooting the same direction and the same distance that is...
A...let me see if I understand the practical application of this...If I'm North of the Equator and am shooting North, then I should use a left hand twist bbl to counter this effect. And if shooting South being South of the Equator a right hand twist, correct? But what happens if I fire one off say Northeast or Southwest? I guess I'll have to take a compas reading and choose the rifle with the proper twist.
Good shooting,
db
Good shooting,
db
Boyd Heaton
Well-Known Member
You guys have "WAY TO MUCH TIME ON YOUR HANDS"
Darryl Cassel
Well-Known Member
Hello fellows.
As I have mentioned before, thats why when LR hunting, "we" take a "sighter shot" first.
It will eliminate the variables such as rotation speed of the earth, direction we are shooting and the wind, all in one operation. No need to worry about the long math equasions , just the elevation and windage knobs and how they effect the impact of the bullet from point A to point B.
Since we are not putting a rocket into orbit or going to the moon or shooting 1000 miles or more with a projectile, the sighter shot method leaves more time to enjoy our hunting and allows clean kills in the process.
Speaking of enjoy, I have to go send a few bullets across the mountains today out to 1500 to 1800 yards.
I try to keep it simple and it has always Worked for me.
Later
DC
As I have mentioned before, thats why when LR hunting, "we" take a "sighter shot" first.
It will eliminate the variables such as rotation speed of the earth, direction we are shooting and the wind, all in one operation. No need to worry about the long math equasions , just the elevation and windage knobs and how they effect the impact of the bullet from point A to point B.
Since we are not putting a rocket into orbit or going to the moon or shooting 1000 miles or more with a projectile, the sighter shot method leaves more time to enjoy our hunting and allows clean kills in the process.
Speaking of enjoy, I have to go send a few bullets across the mountains today out to 1500 to 1800 yards.
I try to keep it simple and it has always Worked for me.
Later
DC
