Astronomy

Calculate time when star is above altitude 30°

Calculate time when star is above altitude 30°



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To find the best observation time for an object, I'd like to calculate the time when it is 30° or more above the horizon. Local Sideral Time would be sufficient.

To include that in my program, I need the formula.

Example: On June 4th, Jupiter has the coordinates RA= 9h 19m 28.0s Dekl= 16° 32' 0"

It rises at 10:32 and sets at 00:05.

After rise, when is it at altitude 30°, and after transit, when is it at altitude 30° again ?


I found this formula at http://www.stjarnhimlen.se/comp/riset.html. Although it's for the sun, it seems to be what I'm looking for.

$$cos ( ext{LHA}) = frac{sin ( ext{h}) - sin ( ext{lat}) imes sin ( ext{Decl})}{cos ( ext{lat}) imes cos ( ext{Decl})}$$

Applied to the sample assuming a latitude of 45° I get.


Is this the correct approach?


The answer is there on the stjarnhimlen.se site and also stargazing.net.

Now we can compute the Sun's altitude above the horizon:

sin(h) = sin(lat) * sin(Decl) + cos(lat) * cos(Decl) * cos(LHA) LHA = LST - RA

h=Altitude=30$^circ$, LHA = Local Hour Angle, lat= Your latitude on Earth, Decl = Object's declination, Ra = Object's right ascension, and LST = Local Standard Time. Just need to solve for LST.


You're not saying what programming language you're using. If it's Python, or if you could link Python libraries from it, then PyEphem would provide everything you need.

http://rhodesmill.org/pyephem/


Yes, this is the correct approach. The $h$ in the equation is the altitude above the horizon of the object at which you consider it to be rising or setting. This is typically non-zero, because of atmospheric refraction, and, in the case of the Sun or the Moon, because of their finite diameters. In your case, the object 'rises' when it climbes above $h=30^circ$ and 'sets' when it drops below that altitude.

If $left|cos(LHA) ight|>1$, there is no solution, because your object never crosses the $h=30^circ$ line. $LHA$ is the local hour angle, and you can find the local sidereal time $ heta$ using

$LHA = heta - alpha,$

where $alpha$ is the right ascension of your object.


Air Pressure at Altitude Calculator

The following table and graph illustrate the relationship between altitude and pressure using the default values for pressure and temperature at sea level. Using ISA standards, the defaults for pressure and temperature at sea level are 101,325 Pa and 288 K.


Altitude above Sea Level Absolute Atmospheric Pressure
feet miles meters kPa atm psia
-5000 -0.95 -1524 121.0 1.19 17.55
-4000 -0.76 -1219 116.9 1.15 16.95
-3000 -0.57 -914 112.8 1.11 16.36
-2000 -0.38 -610 108.9 1.07 15.79
-1000 -0.19 -305 105.0 1.04 15.24
-500 -0.09 -152 103.2 1.02 14.96
0 0.00 0 101.3 1.00 14.70
500 0.09 152 99.5 0.98 14.43
1000 0.19 305 97.7 0.96 14.17
1500 0.28 457 96.0 0.95 13.92
2000 0.38 610 94.2 0.93 13.66
2500 0.47 762 92.5 0.91 13.42
3000 0.57 914 90.8 0.90 13.17
3500 0.66 1067 89.1 0.88 12.93
4000 0.76 1219 87.5 0.86 12.69
4500 0.85 1372 85.9 0.85 12.46
5000 0.95 1524 84.3 0.83 12.23
6000 1.14 1829 81.2 0.80 11.78
7000 1.33 2134 78.2 0.77 11.34
8000 1.52 2438 75.3 0.74 10.92
9000 1.70 2743 72.4 0.71 10.51
10000 1.89 3048 69.7 0.69 10.11
15000 2.84 4572 57.2 0.56 8.29
20000 3.79 6096 46.6 0.46 6.75
25000 4.73 7620 37.6 0.37 5.45
30000 5.68 9144 30.1 0.30 4.36
35000 6.63 10668 23.8 0.24 3.46
40000 7.58 12192 18.8 0.19 2.72
45000 8.52 13716 14.7 0.15 2.14
50000 9.47 15240 11.6 0.11 1.68
55000 10.42 16764 9.1 0.09 1.32
60000 11.36 18288 7.2 0.07 1.04
65000 12.31 19812 5.6 0.06 0.82

Weather Conditions
Due to the fact that weather conditions affect pressure and altitude calculations, the pressure and temperature at sea level must be known. The altitude at a given air pressure can be calculated using Equation 1 for an altitude up to 11 km (36,090 feet). This equation can be arranged to also calculate the air pressure at a given altitude as shown in Equation 2.

= static pressure (pressure at sea level) [Pa]
= standard temperature (temperature at sea level) [K]
= standard temperature lapse rate [K/m] = -0.0065 [K/m]
= height about sea level [m]
= height at the bottom of atmospheric layer [m]
= universal gas constant = 8.31432
= gravitational acceleration constant = 9.80665
= molar mass of Earth’s air = 0.0289644 [kg/mol]

Earth’s Atmosphere
Due to the fact that Earth’s atmosphere experiences different rates of heating and cooling through each of its layers, these equations help to model this through the use of the temperature lapse rate, which is the rate at which temperature changes through altitude change. Some layers, such as the stratosphere (from 11km to 20km), have a constant temperature throughout the layer. This requires different equations to determine the altitude or pressure. Equations 3 and 4 specify the calculation for altitude and pressure respectfully in this zero temperature lapse rate layer.

For these equations , , and correspond to the altitude, pressure, and temperature at the bottom of the stratosphere. The pressure at the bottom of the layer is determined from the user provided inputs of the pressure and temperature at sea level knowing that the altitude at the bottom of the layer is 11 km assuming the default pressure was used at sea level, the pressure at the bottom of the stratosphere is 22,632 Pa. The temperature at the bottom of the stratosphere is determined by subtracting 71.5 K from the temperature at sea level.


The seeing conditions from a mathematical and model stand point are described by two fundamental parameters r0 and t0 in the Kolmogorov’s turbulence theory.
It is worth to mention that both t0 and r0 parameters are a function of the wavelength of the light used for imaging, we should come back to this point later. Usually r0 and t0 are given at 550nm for the visible range.
t0 is known as the coherence time and it is equals to the time interval over which the rms wavefront error due to the turbulence is 1 radian, or lambda/(2*pi), roughly lambda/6. In short this is the time over which the image of a star can be considered frozen (constant tilt/tip values, other higher order aberrations may still exist though).
t0 is typically in the order of few ms to 10s ms. It should be pointed out that the seeing is not constant across the scope field of view (FOV), since the light from two astronomical objects may travel on slightly different paths though the turbulence in the Earth’s atmosphere. If the objects are separated by a “large” angle, one object’s wavefront distortion will not be applicable to the other wavefront, because their turbulence induced errors will be different (no correlation anymore). The angle for which the wavefront errors are almost the same is known as the isoplanatic angle, a critical parameter for AO design and operation.

Isoplanatic angles are usually much smaller, few ” across, than the imaging camera field of view for most amateur astronomers setups.

The right image shows the notion of common atmospheric path.
The final seeing effecting an image is the integral of the turbulence across the all atmosphere.
In high altitude turbulence are very different for both objects since the common atmospheric path is minimum. This can be traced to the fact that for a given angle the distance between two turbulent cells is larger, higher you are further apart they become.
Close to the ground turbulence (boundary layer) exhibit a large overlap which could cover eventually the all imaging camera FOV. However in the high altitude the overlap is minimum.

The isoplanatic angle takes in account the all atmospheric path, the integral effect.
It is usually defined as the angle, from a reference star, for which the Strehl’s ratio decreases by 2.72 (or e 1 ) versus the reference star Strehl’s ratio.

This means that if the scope is indeed diffraction limited the Strehl’s ratio should be at, or above, 80%, 100% in a perfect situation.
At the isoplanatic angular distance from the reference star it drops down to

37%, usually in just few arc seconds.
So the isoplanatic patch is quite small.


Calculate time when star is above altitude 30° - Astronomy

I'm trying to figure out how to calculate the location of the sun in the sky (angle above horizon) for a certain date & time of day. For the city & state I live in I know the time of sun rise on a given date. Is there someplace I can get the calculation to figure this out myself or a web site that can do this for me?

The U.S. Naval Observatory has an online tool to tabulate the position of the Sun (or Moon) during one day. The angle above the horizon is the altitude, and the angle around the horizon, measured clockwise from due north, is called the azimuth.

If you are a glutton for punishment, you can calculate it yourself.

Update by Lynn (March 11, 2003): One reader e-mailed us and suggested the Approximate Astronomical Positions site, which will calculate positions of the Sun, Moon, and planets. It also gives equations for many calculations and has links to sites explaining many of the calculations. This is another good place to look if you want to calculate things yourself.

This page was last updated on June 27, 2015.

About the Author

Britt Scharringhausen

Britt studies the rings of Saturn. She got her PhD from Cornell in 2006 and is now a Professor at Beloit College in Wisconson.


How do you calculate Altitude and Azimuth?

Hello I'm doing a project for my astronomy class and I have to calculate the altitude and azimuth of the moon can anyone give me some tips or any help?

It might be helpful to tell us more about what you need to do. Are you supposed to predict the altitude and azimuth, or go outside and measure it?

The altitude and azimuth of an object in the sky refers to the angle of the object above the nearest point on the horizon (altitude), and the angle of that nearest point on the horizon with respect to north (azimuth, such that the north star always has az=0, something directly east has az=90 degrees, south az=180 degrees, etc.). Because the moon, sun, and stars move over the course of the day, their altitude and azimuth change over the course of the day as well.

You can measure the altitude and azimuth of the moon using a sextant and a compass as described above.

You can predict the altitude and azimuth of the moon if you know the date and time. Good luck!


Azimuth / Altitude calculation script

Hi all, somewhat of a newbie stargazer here as well as a very amateur programmer, hoping for some feedback on a tiny library I wrote up to calculate the altitude and azimuth of a fixed star, given terrestrial coordinates.

The script is on github here, relevant parts pasted below.

Example of use if you really want to run it (although I was mostly looking for feedback on the equations themselves used in the 'azimuth' and 'altitude' methods):

So far, it produces output within a few arcminutes of 'live' data from Stellarium, which is good enough for what I want. But I'm wondering if anyone who knows trig or astronomical calculations better than I do (read as: anyone who knows trig) can tell me if I'm missing out on anything, like periodicity that should be taken into account with the arcsin and arctan.

If anyone's curious, I'm hoping to use this as the basis for a text-based based planetarium, because I want to learn and it sounded fun.

#2 Oleg Astro

It doesn't work on my computer:

Attached Thumbnails

#3 dasmith

#4 dasmith

I updated the git repo with a quick and dirty interactive script, mystars_interactive.rb. Just need both mystars.rb and mystars_interactive.rb in the same folder, and run the latter.

I am seeing that, as I suspected, the arctan function is causing some problems. For instance, it's reporting an azimuth of 29 for Formalhaut at my location, whereas the actual azimuth is 209 ( 180 + 29 ). So the equation appears to be solid except I need to add some sort of "if this then add 360 or 180" type logic.

So far I haven't been able to get any false altitude values, which is good.

Edit: well I should have RTFM'd.

From the very source from which I got the equations. Daughter's getting up soon but I'll hopefully have this worked out by tomorrow using a double argument arctan. Who knew spherical trigonometry could be so hard?

The azimuth is found from the formula:
tan A = -sin(LHA) / [tan(δ) cos(φ) - sin(φ) cos(LHA)]

where A is the azimuth angle. When finding the azimuth, the double-argument arctangent function (such as ATAN2 in Fortran or atan2() in C) should be used with the numerator and denominator above entered separately (care must be taken in the order of the arguments). In most implementations, this will give a resulting angle between -pi and +pi radians. In order to express the resulting azimuth in the range between 0 and 360 degrees, the result should be converted to degrees, if necessary, and 360 degrees should be added to any negative values.

Edited by dasmith, 12 August 2017 - 03:52 PM.

#5 dasmith

So like I said this is more a novelty than anything, but I've written up the basics of a little viewer. Right now it just takes user lon and lat and a desired Hipparcos number to center on, and displays a sky map, sized to the current terminal and time of day, showing a 10 degree FOV N-S and scales E-W accordingly.

Edit: you can now scroll around with arrow keys or the keypad and zoom in and out with + and -. Enter exits.

The first thing you'll notice in the screenshots is that everything is distended vertically (or squashed horizonally, depending how you choose to look at it). This is a result of using a terminal font that's taller than it is wide. Once I settle on a good square font, the distortion should disappear.

This is more of a proof of concept than anything right now. I originally wanted to display a perspective view, but that's rather daunting, so my compromise was to have the application draw a single hemisphere, based on lon / lat / date&time similar to what you'd get downloading the sky maps from skymaps.com, and then zoom in on the desired area of sky. Maybe I'll get to a true perspective some time, but this seems interesting enough for now (and dang, I salute the people who actually develop real software like stellarium that shows true perspective view) and I'd rather concentrate on allowing a user to scroll around, select stars, get info, etc.

Anyway here's a few screenshots. The purple constellation lines I drew in after to make it easier to see.

Full sky, demonstrating the vertical skewing, tweaked the settings to not display designations.


Calculate time when star is above altitude 30° - Astronomy

You are lost on a desert island
with a sextant, a chronometer, a carrier pigeon,
and your copy of Smart's Spherical Astronomy.
Explain how you will save yourself.
(Assume that the chronometer is keeping GMT,
and that you know the date.)

Step 1: determine your latitude.
There are (at least) two possible techniques.

1 . Measure the altitude of Polaris above the northern horizon, using the sextant.
This is approximately equal to your latitude.
(Polaris, the "North Star", lies very close to the North Celestial Pole.)

There are various problems with this.
Firstly, if you are in the southern hemisphere, Polaris will be below the horizon!
Secondly, you need to carry out the measurement in nautical twilight,
while it is still light enough to see the horizon,
and Polaris is only a second-magnitude star,
so it may not appear bright enough to measure accurately.
Thirdly, Polaris does not lie exactly at the North Celestial Pole,
so your result could be nearly 1 degree in error.

2 . So, as an alternative,
measure the altitude of the Sun at midday, using the sextant.

Knowing the date, calculate the declination of the Sun
(it varies sinusoidally,
with a period of 1 year starting at the spring equinox,
and an amplitude of 23.4 degrees.)

The midday altitude, when the Sun is on the local meridian,
is composed of:
the height of the celestial equator above the southern horizon (equal to the co-latitude)
plus the height of the Sun above the celestial equator (its declination).
(If you are in the southern hemisphere,
the celestial equator will be closer to the northern horizon
in this case its distance from the southern horizon, the co-latitude,
will be greater than 90°.)
Knowing the altitude and the solar declination,
calculate the co-latitude and hence the latitude.

If the sextant can be read to an accuracy of a few arc-minutes,
you should correct your reading for refraction.
The apparent zenith angle of an object z' is greater than its true zenith angle z
by the value k tan(z') , where k is approximately 1 arc-minute.

Step 2: Determine your longitude .
Again there are (at least) two possible techniques.

1 . During nautical twilight,
if you can locate a star whose celestial coordinates you know,
measure its altitude above the horizon using the sextant,
and note the time (GMT) using the chronometer.

Knowing the star's altitude, its declination, and your latitude (previously determined),
calculate its Hour Angle
by applying the cosine rule to "the" Astronomical Triangle.

Knowing the star's Right Ascension,
calculate the local sidereal time of the observation
(Local Hour Angle = Local Sidereal Time - Right Ascension).

Knowing the date,
calculate the Greenwich Sidereal Time
corresponding to the Greenwich Mean Time of the observation.
GST is equal to GMT at the autumn equinox,
and GST runs faster than GMT by one day in 365.25 days.

The difference between the Local Sidereal Time (from your observation)
and Greenwich Sidereal Time (from the chronometer)
is your longitude east or west of Greenwich.

2 . Failing a star with known coordinates, use the Sun.
Note the time (GMT) when it reaches its greatest altitude:
this is midday, Local Apparent Time.

Use the formulae given in Smart's Spherical Astronomy
to calculate the Equation of Time on that date.
(Or derive it from first principles:
allow firstly for the non-uniform motion of the Sun around the ecliptic (Kepler's Second Law)
then allow for the fact that the ecliptic is tilted to the equator.)

Add or subtract the Equation of Time to your Local Apparent Time,
to obtain Local Mean Time.
The difference between Local Mean Time and GMT
is your longitude east or west of Greenwich.

Step 3 :
Tear a strip of paper from the title-page of Smart's Spherical Astronomy
to write a message giving your latitude and longitude.
Launch it by carrier-pigeon and wait to be rescued!

This question formed part of the final exam at UCLA in 1961.
(Trimble, V., "The Observatory" 118 , 32, 1998).


الامتحان النهائي

انت الآن تائه في منطقة صحراوية و معك جهاز "السدس"، و ساعة ميقاتية دقيقة، و حمام زاجل و نسخة ورقية من موقع "الفلك الموضعي".

اشرح لنا الطريقة لانقاذ نفسك .(بافتراض أن ساعتك الميقاتية مضبوطة وفق غرينتش GMT, و أنك تعرف تاريخ اليوم.)

Step 1: determine your latitude.
There are (at least) two possible techniques.

1 . Measure the altitude of Polaris above the northern horizon, using the sextant.
This is approximately equal to your latitude.
(Polaris, the "North Star", lies very close to the North Celestial Pole.)

There are various problems with this.
Firstly, if you are in the southern hemisphere, Polaris will be below the horizon!
Secondly, you need to carry out the measurement in nautical twilight,
while it is still light enough to see the horizon,
and Polaris is only a second-magnitude star,
so it may not appear bright enough to measure accurately.
Thirdly, Polaris does not lie exactly at the North Celestial Pole,
so your result could be nearly 1 degree in error.

2 . So, as an alternative,
measure the altitude of the Sun at midday, using the sextant.

Knowing the date, calculate the declination of the Sun
(it varies sinusoidally,
with a period of 1 year starting at the spring equinox,
and an amplitude of 23.4 degrees.)

The midday altitude, when the Sun is on the local meridian,
is composed of:
the height of the celestial equator above the southern horizon (equal to the co-latitude)
plus the height of the Sun above the celestial equator (its declination).
(If you are in the southern hemisphere,
the celestial equator will be closer to the northern horizon
in this case its distance from the southern horizon, the co-latitude,
will be greater than 90°.)
Knowing the altitude and the solar declination,
calculate the co-latitude and hence the latitude.

If the sextant can be read to an accuracy of a few arc-minutes,
you should correct your reading for refraction.
The apparent zenith angle of an object z' is greater than its true zenith angle z
by the value k tan(z') , where k is approximately 1 arc-minute.

Step 2: Determine your longitude .
Again there are (at least) two possible techniques.

1 . During nautical twilight,
if you can locate a star whose celestial coordinates you know,
measure its altitude above the horizon using the sextant,
and note the time (GMT) using the chronometer.

Knowing the star's altitude, its declination, and your latitude (previously determined),
calculate its Hour Angle
by applying the cosine rule to "the" Astronomical Triangle.

Knowing the star's Right Ascension,
calculate the local sidereal time of the observation
(Local Hour Angle = Local Sidereal Time - Right Ascension).

Knowing the date,
calculate the Greenwich Sidereal Time
corresponding to the Greenwich Mean Time of the observation.
GST is equal to GMT at the autumn equinox,
and GST runs faster than GMT by one day in 365.25 days.

The difference between the Local Sidereal Time (from your observation)
and Greenwich Sidereal Time (from the chronometer)
is your longitude east or west of Greenwich.

2 . Failing a star with known coordinates, use the Sun.
Note the time (GMT) when it reaches its greatest altitude:
this is midday, Local Apparent Time.

Use the formulae given in Smart's Spherical Astronomy
to calculate the Equation of Time on that date.
(Or derive it from first principles:
allow firstly for the non-uniform motion of the Sun around the ecliptic (Kepler's Second Law)
then allow for the fact that the ecliptic is tilted to the equator.)

Add or subtract the Equation of Time to your Local Apparent Time,
to obtain Local Mean Time.
The difference between Local Mean Time and GMT
is your longitude east or west of Greenwich.

Step 3 :
Tear a strip of paper from the title-page of Smart's Spherical Astronomy
to write a message giving your latitude and longitude.
Launch it by carrier-pigeon and wait to be rescued!

This question formed part of the final exam at UCLA in 1961. (Trimble, V., "The Observatory" 118 , 32, 1998).


What you experience with time will remain constant even as you experience a change in speed. However, the relation of your time with those that you leave behind will change. Trips conducted by astronauts have demonstrated this when their clocks moved slower in direct relation to their speed while leaving Earth.

Time does not really exist but is just a representation of motion. Terms such as speed should then require an outside reference. In the case of the astronauts, they measure speed in relation to the sun and the Earth. From this, they adjust their time as needed.

The closer we are to the speed of light, the shorter is the traveling time. For an observer who isn’t moving, if a star is one light year away, then it would take approximately one year to reach that star traveling at the speed of light. If we travel slower than the speed of light, the journey will take longer.


Known problems

The animation spacecraft is at a different scale to the distance between the observer and destination. Even for the shortest space travel distances, for example the earth to the moon, the spacecraft would occupy less than a pixel. This problem will not be fixed.

As an object moves further into the distance it appears smaller to an observer. This change in perspective distance is not represented in the animation. The reduction in the spacecraft length from the observer's framework at velocities approaching the speed of light is an entirely different concept to perspective distance.

If you set the iterations on the animation to a low number, e.g. less than 20, the animation's spaceship time will not be calculated accurately if the observer and traveler times diverge substantially.

The code is old and the user interface needs to be refreshed. (Also the PHP component is overkill and was only used for learning purposes.) You're encouraged to improve the code and place the travel calculator on your own website it's FLOSS.

A bug fix was made in June 2016. The calculation for the fuel needed for the trip did not take into account conservation of momentum. These two webpages helped me correct the error and I am grateful to the various people contributed the notes that helped me fix this (Physics Stack Exchange users user2096078, Qmechanic and udrv, Don Koks for the Relativistic Rocket, and John F who emailed me) :

Copyright (C) Nathan Geffen 2012 under the GNU Affero General Public License.
This software is available here.
There are probably bugs, bad ones. And there are no doubt errors in the text. I would like this site to be 100% accurate eventually.
Please tell me about bugs and errors by emailing nathangeffen at quackdown dot info or logging issues at the above code repository.

This is the distance from earth to your destination. Either enter a value or search the database for a distance to a space object by typing the first few letters of its name. All objects in the database matching that start with the letters you have typed will appear. Select the one you want. Distances are approximate because the planets' positions change continuosly relative to the earth. If you leave distance blank, it will be calculated --if you enter the observer time elapsed and the traveler's maximum velocity-- using this equation:

where
c = the speed of light,
v = maximum velocity,
t = time elapsed in observer timeframe.

This is the constant acceleration of the traveler's spacecraft. Half way through the journey, the spacecraft starts decelerating at the same rate.

If you leave the acceleration blank, it will be calculated using Newton's laws of motion (depending on which fields have values):

where
s = distance,
v = maximum velocity and
t = time elapsed in observer timeframe

This is increasingly inaccurate as you approach the speed of light, so for large distances, such as to the nearest stars, it is better to enter the acceleration manually.

If a spacecraft accelerates constantly at 1g --or 9.8m/s-- the travelers on board can experience earth-like gravity. Unfortunately interstellar travel at this acceleration will likely never be achieved because of the huge amount of energy required. It is not possible to travel to the nearest stars at this acceleration if the fuel must be carried onboard the spacecraft, no matter what kind of fuel is used.

This is the maximum velocity the spacecraft will reach, from the perspective of an observer on earth. This occurs when the spacecraft is half way to its destination. This is calculated using this equation:

where
c = speed of light,
a = acceleration and
t = time elapsed to end of journey in observer timeframe.

This is the time elapsed for the whole journey from the observer on earth's time frame. This is calculated using this equation:

where
c = speed of light,
d = distance of the journey and
a = acceleration.

This is the time elapsed for the whole journey from the perspective of the spacecraft. This is calculated using this equation:

where
c = speed of light,
d = distance of the journey and
a = acceleration.

This is the mass of the spacecraft excluding its fuel. The default value of 25,000kg is approximately the maximum payload of the Endeavour space shuttle.

Note that if this field is blanked out it is not calculated. This field must have a value if you want energy and fuel mass to be calculated.

Also note that if the fuel mass is calculated to be more than the mass of your spacecraft, then your trip cannot be done unless you extract fuel from space. If your fuel mass is more than half the mass of your spacecraft, you're probably on a one way trip, so take enough food, books and episodes of Star Trek to last the rest of your life.

This is the amount of energy your spacecraft's payload will need to constantly accelerate to half way to your destination and then decelerate at the same rate until you reach your destination. This is calculated using this equation:

where
c = speed of light,
v = maximum velocity and
m = spacecraft mass.

The fuel conversion rate is the the efficiency with which your spacecraft's fuel is converted into energy. At today's fuel conversion rates there is no prospect of sending a spacecraft to another star in a reasonable period of time. Advances in technologies such as nuclear fusion are needed first.

The default fuel conversion rate of 0.008 is for hydrogen into helium fusion. David Oesper explains that this rate assumes 100% of the fuel goes into propelling the spacecraft, but there will be energy losses which will require a greater amount of fuel than this.

If you leave this field blank but enter the fuel mass, it is calculated by dividing the given fuel mass by what the fuel mass would be if it were perfectly efficient (i.e. a conversion rate of 1.0).

e = energy,
m = fuel mass and
c = speed of light.

This is the mass of the fuel needed to for your journey. This is calculated using this equation:

v = maximum velocity and
c = speed of light.

Source: The Relativistic Rocket and Physics Stack Exchange. (Thanks to users user2096078, Qmechanic and udrv. Also thanks to John F for informing me of a bug that has now hopefully been corrected.)

This is the length of the spacecraft at the beginning of the journey. Note that the spacecraft length always stays the same for the people in it. This is calculated using this equation:

where
l = length of traveler from observer's perspective,
v = maximum velocity of traveler and
c = speed of light.

This is the length of the spacecraft from the observer on earth's perspective. Of course spacecrafts are small, so it would be impossible to see a spacecraft from earth on an interstellar voyage. This is calculated using this equation:

where
l0 = original length of spacecraft on earth,
v = maximum velocity of traveler and
c = speed of light.


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Treasury yields rise Friday and end higher for the week, with long-dated debt registering their steepest weekly gains in months, after economic data showed an upward march in the cost of living.

Stocks end mostly higher as S&P 500 sets record, sees strongest weekly gain since February

Stocks ended mostly higher Friday, capping a week of gains that saw the S&P 500 and Nasdaq Composite set records as investors erased a pullback that followed a more hawkish tone from the Federal Reserve on June 16, and took a further rise in inflation in stride. The Dow Jones Industrial Average rose around 239 points, or 0.7%, to finish near 34,436, according to preliminary figures. The S&P 500 advanced around 14 points, or 0.3%, to close near 4,281, finishing at a record for a second straight session and logging a weekly rise of 2.74%, its strongest since the week ended Feb. 5, according to Dow Jones Market Data. The Nasdaq Composite pulled back slightly from the latest in a string of records set Thursday, losing around 9 points, or 0.1%, to end near 14,345. The Dow rose 3.4% for the week, while the Nasdaq gained 2.4%.