I am Using Haversine Formula to find the great circle distance in my work
But I want to know The most Exact formula for small distance like 10 meters only
Continue using the haversine formula. The [spherical] law of cosines formula is known to be inaccurate at short distances. See https://en.wikipedia.org/wiki/Great-circle_distance ... also this SO question: MySQL WordPress Query Returning a Distance of Zero for Some Records
There are a couple of alternative options.
One is to convert points from geodetic to Cartesian system of coordinates and then use Euclidean distance in space. Relative error due to curvature is about 10^-9 for distances below 1 km and 10^-3 for distances below 1000 km.
Another option is to project locations on plane that is tangent to surface of the Earth and then calculate Euclidean distance on the plane. But this solution will not work near poles and additional care should be taken at 180th meridian. A careful implementation for ellipsoid datum can use just one computation if sine and one computation of square root aside of arithmetic operations, and can potentially be faster than Haversine formula for spherical datum, at the same time being much more accurate.
Related
I'm trying to decide whether it makes cpu processing time sense to use the more complex Haversine formula instead of the faster Pythagorean's formula but while there seems to be a pretty unanimous answer on the lines of: "you can use Pythagora's formula for acceptable results on small distances but haversine is better", I can not find even a vague definition on what "small distances" mean.
This page, linked in the top answer to the very popular question Calculate distance between two latitude-longitude points? claims:
If performance is an issue and accuracy less important, for small distances Pythagoras’ theorem can be used on an equirectangular projection:*
Accuracy is somewhat complex: along meridians there are no errors, otherwise they depend on distance, bearing, and latitude, but are small enough for many purposes*
the asterisc even says "Anyone care to quantify them?"
But this answer claims that the error is about 0.1% at 1000km (but it doesn't cite any reference, just personal observations) and that for 4km (even assuming the % doesn't shrink due to way smaller distance) it would mean under 4m of error which for public acces GPS is around the open-space best gps accuracy.
Now, I don't know what the average Joe thinks of when they say "small distances" but for me, 4km is definitely not a small distance (- I'm thinking more of tens of meters), so I would be grateful if someone can link or calculate a table of errors just like the one in this answer of Measuring accuracy of latitude and longitude? but I assume the errors would be higher near the poles so maybe choose 3 representative lattitudes (5*, 45* and 85*?) and calculate the error with respect to the decimal degree place.
Of course, I would also be happy with an answer that gives an exact meaning to "small distances".
Yes ... at 10 meters and up to 1km meters you're going to be very accurate using plain old Pythagoras Theorem. It's really ridiculous nobody talks about this, especially considering how much computational power you save.
Proof:
Take the top of the earth, since it will be a worst case, the top 90 miles longitude, so that it's a circle with the longitudinal lines intersecting in the middle.
Note above that as you zoom in to an area as small as 1km, just 50 miles from the poles, what originally looked like a trapezoid with curved top and bottom borders, essentially looks like a nearly perfect rectangle. In other words we can assume rectilinearity at 1km, and especially at a mere 10M.
Now, its true of course that the longitude degrees are much shorter near the poles than at the equator. For example any slack-jawed yokel can see that the rectangles made by the latitude and longitude lines grow taller, the aspect ratio increasing, as you get closer to the poles. In fact the relationship of the longitude distance is simply what it would be at the equator multiplied by the cosine of the latitude of anywhere along the path. ie. in the image above where "L" (longitude distance) and "l" (latitude distance) are both the same degrees it is:
LATcm = Latitude at *any* point along the path (because it's tiny compared to the earth)
L = l * cos(LATcm)
Thus, we can for 1km or less (even near the poles) calculate the distance very accurately using Pythagoras Theorem like so:
Where: latitude1, longitude1 = polar coordinates of the start point
and: latitude2, longitude2 = polar coordinates of the end point
distance = sqrt((latitude2-latitude1)^2 + ((longitude2-longitude1)*cos(latitude1))^2) * 111,139*60
Where 111,139*60 (above) is the number of meters within one degree at the equator,
because we have to convert the result from equator degrees to meters.
A neat thing about this is that GPS systems usually take measurements at about 10m or less, which means you can get very accurate over very large distances by summing up the results from this equation. As accurate as Haversine formula. The super-tiny errors don't magnify as you sum up the total because they are a percentage that remains the same as they are added up.
Reality is however that the Haversine formula (which is very accurate) isn't difficult, but relatively speaking Haversine will consume your processor at least 3 times more, and up to 31x more computational intensive according to this guy: https://blog.mapbox.com/fast-geodesic-approximations-with-cheap-ruler-106f229ad016.
For me this formula did come useful to me when I was using a system (Google sheets) that couldn't give me the significant digits that are necessary to do the haversine formula.
So for example, I have two pairs of coordinates: (103.82 W, 32.024 N) and (104.2 W, 32.587 N). I'm trying to figure out how to calculate the distance between these two points and then represent that distance in DMS form. When I use the distance formula, I obtain a decimal number that does not accurately represent my desired number when I convert it to DMS format. I'm supposed to use the distance formula, so I'm not sure what I'm missing here or what I could be doing wrong. I don't believe I need to use the haversine formula for this.
Any help would be greatly appreciated!
The Rosetta Code web site has working code examples in various computer languages (Java, JavaScript, Python, and many others) to compute the distance between two lat/lon coordinates using the Haversine formula.
You can verify your distance results using this online distance calculator for comparison.
Given a coordinate in decimal degrees you can convert that number into the DMS components or vice versa. The formula has been answered in a related answer and there are online converters to convert between DMS and decimal degrees.
Path image
I need to calculate distance from subscriber position to Position B in the image. I have the GPS co-ordinate of the subscriber and the "B" position. How can I calculate the distance?
Simple case: Express lat and long values in decimal form and use the standard geometry distance formula if subscriber is less than 100 miles from position B. distance = sqrt((lat1-lat2)^2 - (long1-long2)^2).
More general case: Use the haversine formulas using a great circle to calculate distances from points on a sphere for more accurate measurements if position B might be a continent or two away from the subscriber. Let's call the subscriber position A and say and say he is at lat[a], long[a] and the fixed point B is at lat[b], long[b]. Let r represent the radius of the earth (about 3961 miles).
distance = 2*r*arcsin(sqrt(sin^2((lat[b]-lat[a])/2) + cos(lat[a])*cos(lat[b])*sin^2((long[b]-long[a])/2)))
If you specify r in miles, your answer will come out in miles. If you use kilometers use 6373 for a good number for the earth's radius, and of course the answer will come out in kilometers.
Exact case: The haversine formula will not provide a perfect answer because the earth is not a perfect sphere. Even apart from the mountains and the canyons, the earth has a larger radius at the equator than it does at the poles. The radius at the equator is the equator is about 3963 miles, and at the poles it is about 3950 miles. So you really need to devise your own lookup table (or borrow one from google maps) if you are measuring distances halfway around the globe and you have to be exact.
The haversine formula will be accurate to less than half of a percentage point. In 1000 miles your answer will be accurate to within 5 miles.
Haversine formula: https://en.wikipedia.org/wiki/Haversine_formula
Radius of the earth: https://en.wikipedia.org/wiki/Earth_radius
I am currently creating a feature and patterning it across a flat plane to get the maximum number of features to fit on the plane. I do this frequently enough to warrant building some sort of marcro for this if possible. The issue that I run into is I still have to manually set the spacing between the parts. I want to be able to create a feature and have it determine "best" fit spacing given an area while avoiding overlaps. I have had very little luck finding any resources describing this. Any information or links to potentially helpful resources on this would be much appreciated!
Thank you.
Before, you start the linear pattern bit:
Select the face2 of that feature2, get the outer most loop2 of edges. You can test for that using loop2.IsOuter.
Now:
if the loop has one edge: that means it's a circle and the spacing must superior to the circle's radius
if the loop has more that one edge, that you need to calculate all the distances between the vertices and assume that the largest distance is the safest spacing.
NOTA: If one of the edges is a spline, then you need a different strategy:
You would need to convert the face into a sketch and finds the coordinates of that spline to calculate the highest distances.
Example: The distance between the edges is lower than the distance between summit of the splines. If the linear pattern has the a vertical direction, then spacing has to be superior to the distance between the summit.
When I say distance, I mean the distance projected on the linear pattern direction.
I have a set of GPS Coordinates and I want to find the speed required for a UAV to travel between them. Doing this by calculating distance in x y z and then dividing by time to travel - m/s.
I know the great circle distance but I assume this will be incorrect since they are all relatively close together(within 10m)?
Is there an accurate way to do this?
For small distances you can use the haversine formula without a relevant loss of accuracy in comparison to Vincenty's f.e. Plus, it's designed to be accurate for very small distances. This can be read up here if you are interested.
You can do this by converting lat/long/alt into XYZ format for both points. Then, figure out the rotation angles to move one of those points (usually, the oldest point) so that it would be at lat=0 long=0 alt=0. Rotate the second position report (the newest point) by the same rotation angles. If you do it all correctly, you will find X equals the east offset, Y equals the north offset, and Z equals the up offset. You can use Pythagorean theorm with X and Y (north and east) offsets to determine the horizontal distance traveled. Normally, you just ignore the altitude differences and work with horizontal data only.
All of this assumes you are using accurate formulas to convert lat/lon/alt into XYZ. It also assumes you have enough precision in the lat/lon/alt values to be accurate. Approximations are not good if you want good results. Normally, you need about 6 decimal digits of precision in lat/lon values to compute positions down to the meter level of accuracy.
Keep in mind that this method doesn't work very well if you haven't moved far (greater than about 10 or 20 meters, more is better). There is enough noise in the GPS position reports that you are going to get jumpy velocity values that you will need to further filter to get good accuracy. The math approach isn't the problem here, it's the inherent noise in the GPS position reports. When you have good reports, you will get good velocity.
A GPS receiver doesn't normally use this approach to know velocity. It looks at the way doppler values change for each satellite and factor in current position to know what the velocity is. This works reasonably well when the vehicle is moving. It is a much faster way to detect changes in velocity (for instance, to release a position clamp). The normal user doesn't have access to the internal doppler values and the math gets very complicated, so it's not something you can do.