var p=3.1415; /* short pi */ //var G=0.00000000006673; var G=6.673E-11; var C=299792458 ; function calc1(n) { var dist=document.c1.dist[n][document.c1.dist[n].selectedIndex].value/1; var time=document.c1.time[n][document.c1.time[n].selectedIndex].value/1; var acc=document.c1.acc[n][document.c1.acc[n].selectedIndex].value/1; var a=document.c1.A[n].value/1; if (n<4) { a=a*acc; } else { a=a/acc; } var d=document.c1.D[n].value/1; d=d/dist; var t=document.c1.T[n].value/1; t=t*time; var sw=n/1; switch (sw) { case 0: document.c1.D[n].value=(t*t*a/2)*dist; // calculate the velocity var v=Math.round(a*t*1000000)/1000000; v=v*dist/time; // calculate the units var numer=document.c1.dist[n][document.c1.dist[n].selectedIndex].text; var denom=document.c1.time[n][document.c1.time[n].selectedIndex].text; denom=denom.substring(0,denom.length-1); var un=v+" "+numer+"/"+denom; // need to convert back to correct unig var vid=document.getElementById('fspeed'); vid.innerHTML=un; break; case 1: document.c1.D[n].value=(a*t*t/(4*p*p))*dist; /* p,r,g d=at^2/4p^2 */ break; case 2: document.c1.D[n].value=(Math.pow(((t*t)/(4*p*p/(G*a))),1/3))*dist; break; case 3: document.c1.D[n].value=(2*G*a/(t*t))*dist; break; case 4: document.c1.A[n].value=(d*Math.sqrt(1-(t*t/(C*C))))*acc; break; case 5: document.c1.A[n].value=(d/Math.sqrt(1-(t*t/(C*C))))*acc; break; } return false; } function calc2(n) { var dist=document.c1.dist[n][document.c1.dist[n].selectedIndex].value; var time=document.c1.time[n][document.c1.time[n].selectedIndex].value; var acc=document.c1.acc[n][document.c1.acc[n].selectedIndex].value; var a=document.c1.A[n].value/1; if (n<4) { a=a*acc; } else { a=a/acc; } var d=document.c1.D[n].value/1; d=d/dist; var t=document.c1.T[n].value/1; t=t*time; var sw=n/1; switch (sw) { case 0: document.c1.T[n].value=Math.sqrt((d*2)/a)/time; var v=Math.round(a*t*1000000)/1000000; // calculate the units var numer=document.c1.dist[n][document.c1.dist[n].selectedIndex].text; var denom=document.c1.time[n][document.c1.time[n].selectedIndex].text; denom=denom.substring(0,denom.length-1); var un=v+" "+numer+"/"+denom; var vid=document.getElementById('fspeed'); vid.innerHTML=un; break; case 1: document.c1.T[n].value=2*p*Math.sqrt(d/a)/time; /* p,r,g t=2psqr(r/a)*/ break; case 2: document.c1.T[n].value=2*p*Math.sqrt(d*d*d/(G*a))/time; break; case 3: document.c1.T[n].value=Math.sqrt(2*G*a/d)/time; break; case 4: document.c1.T[n].value=C*(Math.sqrt(1-(a*a/(d*d))))/time; break; case 5: document.c1.T[n].value=C*(Math.sqrt(1-(d*d/(a*a))))/time; break; } return false; } function calc3(n) { /* p,r,g a = 4 pi**2 r / t^2 */ var dist=document.c1.dist[n][document.c1.dist[n].selectedIndex].value; var time=document.c1.time[n][document.c1.time[n].selectedIndex].value; var acc=document.c1.acc[n][document.c1.acc[n].selectedIndex].value; var a=document.c1.A[n].value/1; if (n<4) { a=a*acc; } else { a=a/acc; } var d=document.c1.D[n].value/1; d=d/dist; var t=document.c1.T[n].value/1; t=t*time; var sw=n/1; switch (sw) { case 0: document.c1.T[n].value=Math.sqrt((d*2)/(a))/time; break; case 1: document.c1.A[n].value=(4*d*p*p/(t*t))/acc; /* p,r,g a = 4 pi**2 r / t^2 */ break; case 2: document.c1.T[n].value=2*p*Math.sqrt(d*d*d/(G*a))/time; break; case 3: document.c1.T[n].value=Math.sqrt(2*G*a/d)/time; break; case 4: document.c1.T[n].value=C*(Math.sqrt(1-(a*a/(d*d))))/time; break; case 5: document.c1.T[n].value=C*(Math.sqrt(1-(d*d/(a*a))))/time; break; } return false; } function setPlanet(n,v,m) { document.c1.dist[n].selectedIndex=0; document.c1.acc[n].selectedIndex=m; document.c1.D[n].value=v; document.c1.A[n].value=1; calc2(n); return false; }
Newtonian Constant Acceleration in g's.
Use this to calculate how far a spaceship will go in a certain time at a certain acceleration. A ship going at a rate of 1g for 100 seconds will travel 49000 meters. If you put in distance traveled it will tell you the time needed to travel that distance at the specified g force. G force here is the force of gravity at the earth's surface. This does not take into account relativistic effects, so use this for interplanetary travel.
Acceleration Time Distance v=0

Acceleration in g's on a cylinder such as a space station or space colony.
Use this to find out how fast a space station must spin (period of rotation time in seconds) to maintain a certain force of gravity. You can vary the radius to get a new acceleration or vary the period to get a new radius. Changing the g force will recalculate the period. G force here is the force of gravity at the earth's surface.

Period Radius Acceleration

Period and radius of a circular orbit around a planet.
This gives you the distance from the center of planet and the time it takes for one circular orbit. The units are in terms of planet mass. I've given Earth, Jupiter and the sun for examples. (remember the earth is 6,378 km in radius so subtract that out to get distance from the surface.) Click these links to pre load set diameters Earth, Venus, Mars, Jupiter, The Moon

Period Radius Planet Mass

Escape velocity at a planet's surface
This is the speed that will allow an object to leave the gravitational field of a planet. Anything slower will be slowed down enough by gravity to fall backwards to the planet. Radius is is the radius of the planet. Click these links to pre load set diameters Earth, Venus, Mars, Jupiter, The Moon

Velocity Radius Planet Mass

Relativistic effects

Length Contraction.

The length of things shrinks in the direction of the motion. This is measurable at relativistic speeds. Enter the Rest length and a Velocity and get the contracted length. Enter the Contracted length and get the Velocity. How long will a yard stick be at 1/2 the speed of light? The answer is about 2.6 feet. This calculation is based on an Earth Observer.
Velocity Earth length Moving Length

Time Dilation.
At relativistic speeds time slows down. Enter the Rest interval and a Velocity and get the dilated time. For example how long will ten seconds appear to be when traveling at 1/2 C? Answer=11.57 seconds. This is a Little confusing. Time slows down for a fast moving space traveller which means that it looks like more seconds pass on Earth than for the traveller. This calculation is based on an observer from earth.

Velocity Moving Time Earth Time

Long Relativistic Journeys
Going to a star requires that you speed up and then slow down. It is a double calculation. Half of the trip you accelerate, but if you need to slow down for the second half so that you reach the star, not splat against it. A traveler experiences the trip at one time rate and an observer on one of the stars will experience much longer time. At 1 G acceleration, it is possible to travel to a near galaxy within a generation, but when you will come back it will be millions of years later.
Use this calculator to figure the time it takes to get to a star at constant acceleration and then constant deceleration. Counter to what we might expect, a voyager can make it to a star 20 light years away in 6 years. This is not a violation of Relativity, but a side effect.
Note: unlike the other functions I didn't include the calculation where you plug in time and get distance. That acosh got me.
Note 2: If you try for really long distances the trip time shows infinity - this is due to limitations of JavaScript.

Acceleration Distance
Trip Time Earth Time

Size of Black Holes

The Swarzchilde radius is calculated as r=2*G*M/c^2. The actual size of black holes is tiny.


Mass Radius

Conversions

Here are some conversions of units which might be used in space travel

Length Conversions

 

Energy Conversions