Request confirmation: Launch site and inclination change delta-V map?

[Jestersage]

Now I am certain my poor piloting factor into it. However, does launch sites and or inclination affect the amount of delta-V need to go from launch to orbit? It feels like it, but how do I confirm it's actually due to the inclination and not poor piloting?

The following are the ones I have done:

• Polar, from KSC
• Toward 90 degrees, Woomerang, resulting in 45 degree
• Toward 90 degree, Dessert, resulting in 6.5 degree
• And of course standard KSC launch profile to 90 degree.

I just feel even from KSC, a polar launch seems to be more difficult.

Edited February 24, 2019 by Jestersage

[Aegolius13]

The lowest delta-v path to orbit is always to launch at the equator and head east, because you're using Kerbin's rotational velocity to your advantage.  If you launch west from the equator, the opposite is true.  You can see this effect by looking at the difference between your "orbit" and "surface velocities." 

When you launch from an inclined launch site, you can't instantly get into an equatorial orbit - the minimum inclination of your eventual orbit is going to be the same as the latitude you started from.  So there is no way to get the same bonus from Kerbin's rotation as if you had started at the equator.  This means that the minimum delta-v required to reach orbit is higher at higher latitudes.

When you launch into a polar orbit,  orbital velocity does not come into effect in this way, so starting latitude should not really make a difference.  Though you will get some Coriolis effect distorting your trajectory.  

Finally, if you launch from an inclined site, and want to end up in an equatorial orbit, it will take extra delta-v.  But that's not strictly necessary; just depends on where you want your craft to end up.  

 

[Starman4308]

The big factors here are Kerbin's rotation and that, when in a stable orbit, you always return to the place of your last maneuver.

 

At the equator, Kerbin rotates by about 175 m/sec. As such, when launching due east from said equator, you gain 175 m/sec of bonus velocity.

Launching due north, you don't actually get a 90 degree orbit; you get something like 85-ish degrees inclination. This is because, while you're adding northwards velocity... you're not cancelling out the eastwards spin of Kerbin. To actually get a perfectly polar orbit, you have to launch slightly west of north.

Launching due west, not only do you lose the 175 m/sec spin advantage, you have to pay it all over again just to hit zero horizontal velocity. As such, it's a 350 m/sec penalty to launch west.

 

At a higher-latitude site, you start with less velocity from Kerbin's spin, all the way down to zero were you to launch from the poles. This can help when launching to polar or retrograde orbits, as you have less eastwards velocity to cancel out... but it is a hindrance if you do want to go for an eastwards or mostly-eastwards orbit.

 

One other significant advantage of equatorial launch sites is the ability to access low-inclination orbits without plane-change maneuvers. You always return to the last place you made a maneuver. So, if you make your ascent-to-orbit at 40 degrees latitude, even traveling due east, your resulting orbit will have at least one point at +/- 40 degrees latitude, meaning an inclination of at least 40 degrees. While there are multiple tricks to make this less of an issue, it'll still always be a fundamental limitation: no equatorial orbits without some sort of dog-leg or plane-change maneuver.

This is part of why the real-world Korou is a fairly attractive launch site; being dead-on at the equator, Korou not only gets a substantial benefit from Earth's spin, but it also minimizes the plane-change to get to geostationary Earth orbit.

[Jestersage]

Thanks for confirming. No wonder I feel I had been building overpowered craft sometimes even though I did not change it -- I forgot to change to Woomerang/Dessert for testing.