Enter the latitude and longitude of two points to get the distance between them along the Earth's surface in km and miles. You can switch the input between decimal degrees and DMS.
Fill in every field with numbers. In decimal form, south latitudes and west longitudes are negative (for example, 77.0366° W → −77.0366). In DMS form, choose N/S and E/W from the drop-downs.
Result
Enter the latitude and longitude of the two points in the fields on the left and press "Calculate". The result will appear here.
Table of Contents
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What you can do on this page
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What is this calculation used for?
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How to Use
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Formula
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Symbols and terms
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Good to know before you start
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How to calculate it in Excel
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How to calculate it in Google Sheets
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How to calculate it in Python
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How to write it in LaTeX and other math languages (copy and paste)
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How to have ChatGPT do the calculation
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DataChef Features
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Related Features
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NumberChef Calculators List
What you can do on this page
- Enter the latitude and longitude of two points, and you get the distance between them along the Earth's surface (in miles and km) on the spot
- You can enter decimal degrees (for example, 40.7128) or degrees, minutes and seconds (DMS, for example, 40°42′46.08″). Switch with the radio buttons
- The distance is calculated with Lambert's formula, which treats the Earth as an ellipsoid (a slightly flattened sphere). This is closer to the real Earth than the haversine formula, which treats the Earth as a perfect sphere
- A plain-language explanation of the formulas and copy-and-paste formulas for Excel, Google Sheets and Python are also on this page
The result is the approximate length of the shortest path along the Earth's surface (the geodesic), not the travel distance along roads or railways. In decimal form, enter south latitudes and west longitudes as negative numbers (all of the US has negative longitudes).
What is this calculation used for?
Flight routes and flight distances (aviation)
Airplanes fly close to the shortest path along the curve of the Earth (the great-circle route), so the flight distance between cities is estimated as a distance along the Earth's surface, like the one on this page (New York to London is about 3,470 miles, or 5,585 km). Airline mileage charts are based on this kind of distance too.
On a flat map, flights from the US to Asia seem to take a detour over Alaska and near the Arctic. Calculating this distance shows why that path is actually the shortest.
Location services on your phone ("near me" searches)
When a map or delivery app sorts places by "nearest first", it calculates the distance between your location and each store from their latitudes and longitudes. In programming, the haversine formula (sphere model) is the standard tool for this, and ellipsoid formulas like Lambert's formula on this page are used when more accuracy is needed.
Behind the apps you use every day, this kind of formula is calculated over and over.
GPS distance in running and hiking apps (sports)
A GPS watch or running app records your position every few seconds, calculates the distance between each pair of points with this kind of formula, and adds them up. Each step is only tens of feet, but the latitude-and-longitude-to-distance conversion is repeated thousands of times to show "you ran 6.2 miles today".
Voyage planning and nautical miles (shipping and logistics)
At sea, distance is measured in nautical miles (1 nautical mile = 1.852 km, about 1.15 miles, roughly the length of 1 minute of latitude). The route from the departure port to the destination is planned as a distance along the curve of the Earth, and arrival times and fuel estimates start from this distance calculation.
Shipping carries much of the world's trade, and latitude and longitude distances are part of what keeps it running.
"X miles from the epicenter" in earthquake reports (safety)
Earthquake reports such as "the epicenter was 12 miles southwest of the city" come from the same kind of calculation: the distance between the latitude and longitude of the epicenter (the point on the surface directly above the quake) and those of each town or monitoring station.
Knowing in numbers how far your home is from active faults and past epicenters is a good first step in preparing for earthquakes.
Formula
Central angle \(c\) between the two points (Lambert's formula, part 1)
Standard notation
(the usual math form)
\(U\)
\(=\)
\(\arctan\bigl((1 - f)\tan \varphi\bigr)\)
\(c\)
\(=\)
\(2\arcsin\sqrt{\sin^2\!\frac{U_2 - U_1}{2} + \cos U_1 \cos U_2 \sin^2\!\frac{\lambda_2 - \lambda_1}{2}}\)
In words
(symbols replaced with words)
②
\(U\): reduced latitude
\(=\)
①
latitude \(\varphi\) shrunk by the flattening \(f\)
④
\(c\): central angle between the two points
\(=\)
③
angle made from the differences in reduced latitude and longitude
The formula in words
①
For each of the two points, calculate the
latitude \(\varphi\) shrunk by the flattening \(f\), \(\arctan((1-f)\tan\varphi)\)
②
to get the
reduced latitude \(U\)
(\(U_1\) and \(U_2\)). Next, calculate the
③
angle made from the reduced latitude difference \(U_2 - U_1\) and the longitude difference \(\lambda_2 - \lambda_1\)
④
and you get the
central angle \(c\) between the two points
(the angle between the two points as seen from the center of the Earth)
Quick example
The central angle between New York (40.7128° N, 74.0060° W) and Washington, D.C. (38.8976° N, 77.0366° W) is
central angle \(c\)
\(=\)
angle from the differences in reduced latitude and longitude of New York and D.C.
\(U_1 = \arctan\bigl((1 - \tfrac{1}{298.257})\tan 40.7128^\circ\bigr) \approx 40.6177^\circ, \quad U_2 \approx 38.8036^\circ\)
\(c = 2\arcsin\sqrt{0.00066425} \approx 0.051552 \ \mathrm{rad} \ (\approx 2.9537^\circ)\)
Key idea
The Earth is not a perfect sphere. Its equatorial radius is about 3,963 miles (6,378 km) and its polar radius is about 3,950 miles (6,357 km), so it is squashed from north to south by about 1/300. This shape is called an ellipsoid of revolution. The flattening \(f = \dfrac{1}{298.257}\) measures how squashed it is, and if you put the latitude straight into a formula for a sphere, the result is off by this amount. So first each latitude is converted into the reduced latitude \(U\), a latitude adjusted so that the ellipsoid can be treated like a sphere. Then the standard formula for the central angle between two points on a sphere (a haversine-type formula) is used.
When you program it, writing \(c = 2\,\mathrm{atan2}(\sqrt{a},\ \sqrt{1-a})\) (where \(a\) is the expression under the root) gives the same value as \(2\arcsin\sqrt{a}\), but is more numerically stable. The calculator on this page uses this form.
Correction for the ellipsoid and the distance \(d\) (Lambert's formula, part 2)
Standard notation
(the usual math form)
\(d\)
\(=\)
\(3963.15\)
\(\times\)
\(\bigl(\)
\(c\)
\(-\)
\(\frac{f(X + Y)}{2}\)
\(\bigr)\)
In words
(symbols replaced with words)
④
\(d\): distance between the two points
\(=\)
③
equatorial radius, 3963.15 mi
\(\times\)
\(\bigl(\)
①
\(c\): central angle
\(-\)
②
flattening correction \(f(X+Y) \div 2\)
\(\bigr)\)
The formula in words
①
From the
central angle \(c\) (in radians)
②
subtract the
flattening correction \(f(X+Y) \div 2\) (\(X\) and \(Y\) are in-between values for the correction)
③
multiply by the
equatorial radius, 3963.15 mi
④
and you get the
distance \(d\) between the two points
(in miles; use 6378.1 km for the radius to get km)
Quick example
For New York and Washington, D.C. (central angle c ≈ 0.051552 rad)
distance \(d\)
\(=\)
equatorial radius (3963.15 mi)
\(\times\)
\(\bigl(\)
central angle (0.051552 rad)
\(-\)
flattening correction (about 0.0000386)
\(\bigr)\)
\(X \approx 0.0000093, \quad Y \approx 0.023014, \quad \frac{f(X + Y)}{2} \approx 0.0000386\)
\(d = 3963.15 \times (0.051552 - 0.0000386) \approx 204.2 \ \mathrm{mi} \quad (204.2 \times 1.60935 \approx 328.6 \ \mathrm{km})\)
Key idea
The base is the arc length rule for a sector: arc length = radius × central angle (in radians). \(3963.15 \times c\) alone is the arc length on a sphere with the equatorial radius, which is about 204.3 miles from New York to Washington, D.C. The term \(\dfrac{f(X+Y)}{2}\) then corrects for the flattened shape (here it shortens the distance by about 0.15 miles, or about 800 feet, to 204.2 miles). To get the answer in km, use the equatorial radius 6378.1 km in place of 3963.15 mi.
The exact definitions of the in-between values \(X\) and \(Y\) are \(X = \dfrac{(c - \sin c)\sin^2 P \cos^2 Q}{\cos^2(c/2)}\) and \(Y = \dfrac{(c + \sin c)\sin^2 Q \cos^2 P}{\sin^2(c/2)}\), where \(P = \dfrac{U_1 + U_2}{2}\) and \(Q = \dfrac{U_2 - U_1}{2}\). They look complicated, but all of them are calculated step by step from the reduced latitudes and the central angle found in the first formula. The error of Lambert's formula is said to be about 10 meters (about 30 feet) even over distances of thousands of miles.
Converting DMS to decimal degrees
Standard notation
(the usual math form)
\(D\)
\(=\)
\(d\)
\(+\)
\(\frac{m}{60}\)
\(+\)
\(\frac{s}{3600}\)
In words
(symbols replaced with words)
④
\(D\): angle in decimal degrees
\(=\)
①
\(d\): degrees
\(+\)
②
minutes \(m \div 60\)
\(+\)
③
seconds \(s \div 3600\)
The formula in words
①
To the
degrees \(d\)
②
add the
minutes \(m\) divided by 60
③
and the
seconds \(s\) divided by 3600
④
and you get the
angle in decimal degrees \(D\)
(for south latitudes and west longitudes, put a minus sign in front at the end)
Quick example
The latitude of Washington, D.C., 38°53′51.36″ N, in decimal degrees is
angle in decimal degrees \(D\)
\(=\)
degrees (38)
\(+\)
minutes (53) ÷ 60
\(+\)
seconds (51.36) ÷ 3600
\(38 + \frac{53}{60} + \frac{51.36}{3600} = 38 + 0.883333\ldots + 0.014266\ldots \approx 38.8976\)
Key idea
Degrees, minutes and seconds count in base 60, just like hours, minutes and seconds of time. 1 degree = 60 minutes and 1 minute = 60 seconds (so 1 degree = 3600 seconds). That is why you divide minutes by 60 and seconds by 3600 to turn them into degrees.
For south latitudes and west longitudes, put a minus sign in front of the whole converted value (for example, 77°2′11.76″ W → −77.0366°). In decimal form, enter this negative value in the calculator. In DMS form, choosing "S" or "W" from the drop-down does the same thing.
Lambert's formula finds the distance between two latitude and longitude points in three moves - treat the Earth as a slightly flattened ellipsoid, find the central angle between the two points, and apply "arc length = radius × central angle" with a correction for the flattening. DMS counts in base 60 like time, so minutes ÷ 60 and seconds ÷ 3600 turn it into decimal degrees.
Symbols and terms
Symbols
| \(\varphi\) | phi | Latitude: the angle that tells how far north or south of the equator (0°) a place is. North is positive and south is negative (−90° to 90°). |
| \(\lambda\) | lambda | Longitude: the angle that gives the east-west position, measured from the prime meridian (0°) through the old Royal Observatory in Greenwich, England. East is positive and west is negative (−180° to 180°). |
| \(f\) | eff | The flattening of the Earth: how squashed the ellipsoid is, as a ratio. This calculation uses \(f = \dfrac{1}{298.257}\). |
| \(U\) | you | The reduced latitude: the latitude on the ellipsoid converted with \(U = \arctan((1-f)\tan\varphi)\) so that it can be used in formulas for a sphere. |
| \(c\) | see | The central angle: the angle between the two points as seen from the center of the Earth (in radians). Through "arc length = radius × central angle", it is the base of the distance. |
| \(X\), \(Y\) | ex, why | In-between values used to correct for the flattened shape. They are calculated from the central angle \(c\), the average of the reduced latitudes \(P = \dfrac{U_1+U_2}{2}\) and half their difference \(Q = \dfrac{U_2-U_1}{2}\). |
| \(d\) | dee | The distance between the two points, from the first letter of "distance". This page shows it in miles and km (kilometers). |
| ° ′ ″ | degree, minute, second | The symbols for degrees, minutes and seconds of angle. 1° = 60′ and 1′ = 60″. Example - 40°42′46.08″ (40 degrees, 42 minutes, 46.08 seconds). |
Terms
| latitude | The angle that gives the north-south position, with the equator at 0°. The North Pole is 90° N and the South Pole is 90° S. It is a basic idea of maps and location data, taught in elementary and middle school geography. |
| longitude | The angle that gives the east-west position, with the prime meridian (the line through Greenwich, England) at 0°. It goes up to 180° east and 180° west. The contiguous US lies between about 67° W and 125° W. |
| DMS | A way of writing angles in base 60 with degrees, minutes and seconds. 1 degree = 60 minutes and 1 minute = 60 seconds. It is common on USGS topographic maps, nautical charts and GPS devices. |
| ellipsoid | Here, an ellipsoid of revolution - the solid you get by spinning an ellipse around its axis. The Earth's rotation makes it bulge at the equator, so it is modeled as an ellipsoid rather than a sphere. |
| flattening | The ratio that shows how squashed an ellipsoid is, defined as (equatorial radius − polar radius) ÷ equatorial radius. For the Earth it is about 1/300. |
| reduced latitude | The latitude of a point on the ellipsoid after it is moved onto the helper circle (sphere) the ellipse is based on. It is found with \(U = \arctan((1-f)\tan\varphi)\) and is used in geodesy (the science of measuring the Earth's shape). It is also called the parametric latitude. |
| central angle | The angle between two points as seen from the center of a circle or sphere. The sector rule "arc length = radius × central angle (in radians)" is the base of the distance calculation on this page. |
| geodesic | The shortest path between two points on a curved surface. On the Earth it is close to the great-circle route, and airline routes follow roughly this shape. This page finds the approximate length along the geodesic. |
| haversine formula | A well-known formula for the distance between two points that treats the Earth as a perfect sphere. Lambert's formula on this page is an improved version for the ellipsoid, and it uses a haversine-type formula for the central angle in the first step. |
| Lambert's formula | A method for the distance between two points that models the Earth as an ellipsoid. Its error is said to be about 10 meters (about 30 feet) even over thousands of miles, which is more accurate than treating the Earth as a sphere. |
| mile | A unit of distance used mainly in the US and the UK (the statute mile). 1 mile = 5,280 feet ≈ 1.609 km (exactly 1.609344 km by the international definition). This calculator converts with 1 mile = 1.60935 km. It is a different unit from the nautical mile below. |
| nautical mile | A unit of distance used at sea and in aviation. 1 nautical mile = 1.852 km (about 1.15 miles or 6,076 feet). It was chosen to be about the length of 1 minute (1/60 of a degree) of latitude. |
Good to know before you start
Here is what helps you use the calculation on this page with real understanding, not just by pressing the button.
To just find a distance, knowing how to read latitude and longitude is enough. To understand what is inside the formulas, go back as far as trigonometric ratios and radians.
| Latitude and longitude (Grades 4–6 social studies) |
|
| Angles and base 60 (Grades 3–4 time, Grade 4 angles) |
|
| Arc length of a sector (Geometry) |
|
| Trigonometric ratios sin, cos and tan (Geometry and Algebra 2) |
|
| Radians (Algebra 2 and Precalculus) |
|
| The shape of the Earth (middle school Earth science) |
|
How to calculate it in Excel
Copy the whole table below and paste it into cell A1 in Excel. It works as is.
Table to convert DMS to decimal degrees
| Degrees | 38 |
| Minutes | 53 |
| Seconds | 51.36 |
| Decimal degrees (add − for S and W) | =B1+B2/60+B3/3600 |
Table to find the distance with Lambert's formula
| Point 1 latitude (decimal; S is −) | 40.7128 |
| Point 1 longitude (decimal; W is −) | -74.0060 |
| Point 2 latitude | 38.8976 |
| Point 2 longitude | -77.0366 |
| Flattening f | =1/298.257 |
| Reduced latitude U1 | =ATAN((1-B5)*TAN(RADIANS(B1))) |
| Reduced latitude U2 | =ATAN((1-B5)*TAN(RADIANS(B3))) |
| Longitude difference Δλ (radians) | =RADIANS(B4-B2) |
| Haversine term a | =SIN((B7-B6)/2)^2+COS(B6)*COS(B7)*SIN(B8/2)^2 |
| Central angle c (radians) | =2*ATAN2(SQRT(1-B9),SQRT(B9)) |
| Average P | =(B6+B7)/2 |
| Half difference Q | =(B7-B6)/2 |
| Correction value X | =(B10-SIN(B10))*SIN(B11)^2*COS(B12)^2/COS(B10/2)^2 |
| Correction value Y | =(B10+SIN(B10))*SIN(B12)^2*COS(B11)^2/SIN(B10/2)^2 |
| Distance (miles) | =6378.1/1.60935*(B10-B5*(B13+B14)/2) |
| Distance (km) | =B15*1.60935 |
The first table converts DMS to decimal degrees. B4 shows 38.8976.
The second table is every step of Lambert's formula. B1 to B4 are the inputs (decimal degrees), B6 onward are the in-between values of formulas 1 and 2, B15 is the distance in miles (about 204.15), and B16 is the distance in km (about 328.56). The calculator on this page rounds these to 204.2 mi and 328.6 km.
RADIANS(…) converts degrees to radians. In Excel, ATAN2(x, y) takes its arguments in the order (x-coordinate, y-coordinate), so note that c is written as 2×ATAN2(√(1−a), √a). 6378.1/1.60935 is the equatorial radius in miles (about 3963.15).
The second table is every step of Lambert's formula. B1 to B4 are the inputs (decimal degrees), B6 onward are the in-between values of formulas 1 and 2, B15 is the distance in miles (about 204.15), and B16 is the distance in km (about 328.56). The calculator on this page rounds these to 204.2 mi and 328.6 km.
RADIANS(…) converts degrees to radians. In Excel, ATAN2(x, y) takes its arguments in the order (x-coordinate, y-coordinate), so note that c is written as 2×ATAN2(√(1−a), √a). 6378.1/1.60935 is the equatorial radius in miles (about 3963.15).
How to calculate it in Google Sheets
Copy the whole table below and paste it into cell A1 in Google Sheets. It works as is.
Table to convert DMS to decimal degrees
| Degrees | 38 |
| Minutes | 53 |
| Seconds | 51.36 |
| Decimal degrees (add − for S and W) | =B1+B2/60+B3/3600 |
Table to find the distance with Lambert's formula
| Point 1 latitude (decimal; S is −) | 40.7128 |
| Point 1 longitude (decimal; W is −) | -74.0060 |
| Point 2 latitude | 38.8976 |
| Point 2 longitude | -77.0366 |
| Flattening f | =1/298.257 |
| Reduced latitude U1 | =ATAN((1-B5)*TAN(RADIANS(B1))) |
| Reduced latitude U2 | =ATAN((1-B5)*TAN(RADIANS(B3))) |
| Longitude difference Δλ (radians) | =RADIANS(B4-B2) |
| Haversine term a | =SIN((B7-B6)/2)^2+COS(B6)*COS(B7)*SIN(B8/2)^2 |
| Central angle c (radians) | =2*ATAN2(SQRT(1-B9),SQRT(B9)) |
| Average P | =(B6+B7)/2 |
| Half difference Q | =(B7-B6)/2 |
| Correction value X | =(B10-SIN(B10))*SIN(B11)^2*COS(B12)^2/COS(B10/2)^2 |
| Correction value Y | =(B10+SIN(B10))*SIN(B12)^2*COS(B11)^2/SIN(B10/2)^2 |
| Distance (miles) | =6378.1/1.60935*(B10-B5*(B13+B14)/2) |
| Distance (km) | =B15*1.60935 |
The same formulas as in Excel (RADIANS, ATAN2, SQRT and so on) work as is in Google Sheets. Copy the whole table, paste it into cell A1, and replace B1 to B4 with your own latitudes and longitudes.
How to calculate it in Python
import math
# Latitude and longitude of the two points (decimal degrees; south and west are negative)
lat1, lon1 = 40.7128, -74.0060 # point 1 (near New York City Hall)
lat2, lon2 = 38.8976, -77.0366 # point 2 (near the White House, Washington, D.C.)
flattening = 1 / 298.257 # flattening of the Earth, f
equator_radius_km = 6378.1 # equatorial radius in km
equator_radius_mi = equator_radius_km / 1.60935 # equatorial radius in miles (about 3963.15)
# Formula 1, first half: reduced latitudes (latitudes shrunk by the flattening)
u1 = math.atan((1 - flattening) * math.tan(math.radians(lat1)))
u2 = math.atan((1 - flattening) * math.tan(math.radians(lat2)))
delta_lon = math.radians(lon2 - lon1)
# Formula 1, second half: central angle c between the two points (haversine-type formula)
a = math.sin((u2 - u1) / 2) ** 2 + math.cos(u1) * math.cos(u2) * math.sin(delta_lon / 2) ** 2
central_angle = 2 * math.atan2(math.sqrt(a), math.sqrt(1 - a))
# Formula 2: correction for the flattened shape (X and Y), then the distance
p = (u1 + u2) / 2
q = (u2 - u1) / 2
x = (central_angle - math.sin(central_angle)) * math.sin(p) ** 2 * math.cos(q) ** 2 / math.cos(central_angle / 2) ** 2
y = (central_angle + math.sin(central_angle)) * math.sin(q) ** 2 * math.cos(p) ** 2 / math.sin(central_angle / 2) ** 2
distance_mi = equator_radius_mi * (central_angle - flattening * (x + y) / 2)
distance_km = distance_mi * 1.60935
print(f"Distance between the two points: {distance_mi:.1f} miles ({distance_km:.1f} km)")
Runs with the standard library only. Change the latitudes and longitudes at the top and run it (convert DMS to decimal first with "degrees + minutes ÷ 60 + seconds ÷ 3600"). If you enter exactly the same coordinates for both points, it divides by 0 and stops with an error.
How to write it in LaTeX and other math languages (copy and paste)
Central angle \(c\) between the two points (Lambert's formula, part 1)
U₁ = atan((1 − f)·tanφ₁), U₂ = atan((1 − f)·tanφ₂), c = 2·asin(√(sin²((U₂ − U₁)/2) + cosU₁·cosU₂·sin²((λ₂ − λ₁)/2)))
U_1 = \arctan\bigl((1-f)\tan\varphi_1\bigr),\ U_2 = \arctan\bigl((1-f)\tan\varphi_2\bigr),\quad c = 2\arcsin\sqrt{\sin^2\frac{U_2-U_1}{2} + \cos U_1 \cos U_2 \sin^2\frac{\lambda_2-\lambda_1}{2}}
<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
<mrow>
<msub><mi>U</mi><mi>i</mi></msub>
<mo>=</mo>
<mi>arctan</mi>
<mo>⁡</mo>
<mrow>
<mo>(</mo>
<mrow><mo>(</mo><mn>1</mn><mo>−</mo><mi>f</mi><mo>)</mo></mrow>
<mi>tan</mi><mo>⁡</mo><msub><mi>φ</mi><mi>i</mi></msub>
<mo>)</mo>
</mrow>
<mo>,</mo>
<mspace width="1em"/>
<mi>c</mi>
<mo>=</mo>
<mn>2</mn>
<mi>arcsin</mi>
<mo>⁡</mo>
<msqrt>
<mrow>
<msup><mi>sin</mi><mn>2</mn></msup>
<mo>⁡</mo>
<mfrac>
<mrow><msub><mi>U</mi><mn>2</mn></msub><mo>−</mo><msub><mi>U</mi><mn>1</mn></msub></mrow>
<mn>2</mn>
</mfrac>
<mo>+</mo>
<mi>cos</mi><mo>⁡</mo><msub><mi>U</mi><mn>1</mn></msub>
<mi>cos</mi><mo>⁡</mo><msub><mi>U</mi><mn>2</mn></msub>
<msup><mi>sin</mi><mn>2</mn></msup>
<mo>⁡</mo>
<mfrac>
<mrow><msub><mi>λ</mi><mn>2</mn></msub><mo>−</mo><msub><mi>λ</mi><mn>1</mn></msub></mrow>
<mn>2</mn>
</mfrac>
</mrow>
</msqrt>
</mrow>
</math>
U_1 = arctan((1-f) tan(varphi_1)), U_2 = arctan((1-f) tan(varphi_2)), c = 2 arcsin(sqrt(sin^2((U_2-U_1)/2) + cos(U_1) cos(U_2) sin^2((lambda_2-lambda_1)/2)))
u1 = ArcTan[(1 - f) Tan[phi1]]; u2 = ArcTan[(1 - f) Tan[phi2]]; c = 2 ArcSin[Sqrt[Sin[(u2 - u1)/2]^2 + Cos[u1] Cos[u2] Sin[(lambda2 - lambda1)/2]^2]]
U1 := arctan((1 - f)*tan(phi1)); U2 := arctan((1 - f)*tan(phi2)); c := 2*arcsin(sqrt(sin((U2 - U1)/2)^2 + cos(U1)*cos(U2)*sin((lambda2 - lambda1)/2)^2));
u1 = atan((1 - f)*tan(phi1)); u2 = atan((1 - f)*tan(phi2)); c = 2*asin(sqrt(sin((u2 - u1)/2)^2 + cos(u1)*cos(u2)*sin((lambda2 - lambda1)/2)^2));
U_1 = atan((1 - f) tan(φ_1)), U_2 = atan((1 - f) tan(φ_2)), c = 2 asin(√(sin^2((U_2 - U_1)/2) + cos(U_1) cos(U_2) sin^2((λ_2 - λ_1)/2)))
Correction for the ellipsoid and the distance \(d\) (Lambert's formula, part 2)
d = 3963.15·(c − f·(X + Y)/2), X = (c − sin c)·sin²P·cos²Q ÷ cos²(c/2), Y = (c + sin c)·sin²Q·cos²P ÷ sin²(c/2), P = (U₁ + U₂)/2, Q = (U₂ − U₁)/2
d = 3963.15\left(c - \frac{f(X + Y)}{2}\right),\quad X = \frac{(c - \sin c)\sin^2 P \cos^2 Q}{\cos^2(c/2)},\quad Y = \frac{(c + \sin c)\sin^2 Q \cos^2 P}{\sin^2(c/2)},\quad P = \frac{U_1 + U_2}{2},\ Q = \frac{U_2 - U_1}{2}
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d = 3963.15 (c - f(X + Y)/2), X = ((c - sin c) sin^2(P) cos^2(Q))/(cos^2(c/2)), Y = ((c + sin c) sin^2(Q) cos^2(P))/(sin^2(c/2)), P = (U_1 + U_2)/2, Q = (U_2 - U_1)/2
p = (u1 + u2)/2; q = (u2 - u1)/2; x = (c - Sin[c]) Sin[p]^2 Cos[q]^2/Cos[c/2]^2; y = (c + Sin[c]) Sin[q]^2 Cos[p]^2/Sin[c/2]^2; d = 3963.15 (c - f (x + y)/2)
P := (U1 + U2)/2; Q := (U2 - U1)/2; X := (c - sin(c))*sin(P)^2*cos(Q)^2/cos(c/2)^2; Y := (c + sin(c))*sin(Q)^2*cos(P)^2/sin(c/2)^2; d := 3963.15*(c - f*(X + Y)/2);
p = (u1 + u2)/2; q = (u2 - u1)/2; x = (c - sin(c))*sin(p)^2*cos(q)^2/cos(c/2)^2; y = (c + sin(c))*sin(q)^2*cos(p)^2/sin(c/2)^2; d = 3963.15*(c - f*(x + y)/2);
d = 3963.15(c - f(X + Y)/2), X = (c - sin c) sin^2(P) cos^2(Q)/cos^2(c/2), Y = (c + sin c) sin^2(Q) cos^2(P)/sin^2(c/2), P = (U_1 + U_2)/2, Q = (U_2 - U_1)/2
Converting DMS to decimal degrees
D = d + m/60 + s/3600
D = d + \frac{m}{60} + \frac{s}{3600}
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D = d + m/60 + s/3600
d + m/60 + s/3600
deg := d + m/60 + s/3600; # D (the differential operator) is reserved in Maple, so the decimal degrees are named deg
D = d + m/60 + s/3600;
D = d + m/60 + s/3600
How to have ChatGPT do the calculation
You are a geography calculation assistant. Do the following calculation by actually running Python code, and base your answer only on the numbers from the execution result (do not answer by mental math or guessing). Find the distance between two points from their latitudes and longitudes using Lambert's formula, which models the Earth as an ellipsoid with flattening f = 1/298.257. Steps: convert each latitude to the reduced latitude U = atan((1-f)·tanφ); a = sin²((U2-U1)/2) + cosU1·cosU2·sin²((λ2-λ1)/2); central angle c = 2·atan2(√a, √(1-a)); P = (U1+U2)/2; Q = (U2-U1)/2; X = (c-sin c)·sin²P·cos²Q/cos²(c/2); Y = (c+sin c)·sin²Q·cos²P/sin²(c/2); distance (km) = 6378.1·(c - f·(X+Y)/2). Point 1: 40.7128° N, 74.0060° W (near New York City Hall) Point 2: 38.8976° N, 77.0366° W (near the White House, Washington, D.C.) Give the distance in miles and km (1 mile = 1.60935 km), and show the code you used and the numbers from the execution result.
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