• 0
Sign in to follow this  
cephalo

Accurate burns

Question

When I am executing a burn on a maneuver node, do I want to start my burn at t-0 or do I split the time before and after the node? The wikipedia article on 'delta-v' seems to prefer splitting the burn time around the direction change, but I wonder if the games estimate is based on that or burning upon reaching the node. It seems like an almost arbitrary design decision, trading off an easier user interface for a simpler estimate calculation.

Does anyone know for sure when the burn is expected to happen to match the estimate provided by the nodes in game?

Edited by cephalo

Share this post


Link to post
Share on other sites

13 answers to this question

Recommended Posts

  • 0

I may be wrong, but you basically want half your burn to occur before the node, and half after it. Right when you hit the node, you should be halfway through the maneuver.

Share this post


Link to post
Share on other sites
  • 0

Most burns are executed most accurately by splitting the burn time and starting using half of the total time before the node. That is assuming that the burn is short, taking place to orbit, while in orbit to intercept another celestial body, and under a couple of minutes at most. Long interplanetary burns from orbit are best split up into several shorter burns during each Periapsis.

As always, some maneuvers can be more time and delta V critical then others.

Share this post


Link to post
Share on other sites
  • 0

When in low kerbin orbit (about 85km) I like to try to do about 2/3 of burn before the node.

Generally doing your burn lower in the gravity well to be more efficient and starting early enables me to get more of the burn done before I rise into higher orbits.

I also like to burn along the horizon just ahead of my blue indicator so that I'm burning horizontally. When the blue indicator reaches and passes the horizon I then follow it (out of the brown and into the blue). This helps minimize losses due to gravity (but it may add steering losses). But overall 50/50 is fine also. Just make sure you don't accidentally end up in the atmosphere because you started your burn early...

Share this post


Link to post
Share on other sites
  • 0

The burn time given by the game is simply the time required to accelerate the planned amount of dV, at the ships current max acceleration, given in the ship info in map view. From what I've observed it doesn't take into account the change in mass as fuel is used, so the longer the burn is the less accurate the displayed time is.

I've searched an answer for this question before, and while I'm on my phone right now and don't have the link, I found that burning in the same direction as prograde at the maneuver, half dV before the node and half after, is the best approximation of an instantaneous burn. In other words, point at the blue maneuver marker and start at half the expected burn time to be the most accurate.

Share this post


Link to post
Share on other sites
  • 0

It's not quite as simple as splitting the burn time in half across the node, technically you want to split the velocity change over the node, since as you burn your craft gets lighter and thus acceleration increases.

Share this post


Link to post
Share on other sites
  • 0

For reasonably simple gameplay, just split it 50:50, it might not be perfect, but it works for practical purposes. If you need sensitive 0.1m/s adjustment at the end of the burn, or for very small burns, set the thrust limiter very low, probably to 5%, for 20x the throttle control (assuming medium to high TWR). In most cases, there will be at least some small error in the net vector of the burn, requiring a small mid-course correction burn (which people like NASA routinely allow for, it's very realistic to do a later correction burn or even series of burns). Over very long distance, a tiny error magnifies greatly. It's often easier not to worry about the tiny error and just correct for it later.

For advanced gameplay, what the other folks said above is entirely relevant. But, there will almost certainly still be a net error at the end of the burn, as it's near impossible to get the vector absolutely perfect, and the burn duration millisecond perfect.

I mostly prefer to keep it simple myself, as long as it gets the job done in a reasonable overall manner. With experience, it's possible to guesstimate the split a little better than 50:50, but without going too technical and doing loads of maths. I do the initial main burn, check the result at least puts me in the roughly correct area at the destination, then move on to plotting a basic mid-course correction to see if it's worthwhile doing one. After any mid-course correction, I'll do a final correction burn on entry to the target SoI, if need be. For longer trips, more correction burns may be worthwhile, but if the dV for any plotted correction burn is less than 1.0m/s, I'll tend to leave it until later in the trip.

Share this post


Link to post
Share on other sites
  • 0

nekogod and alistone are on the right track. The fact is that dV is a function of work, which can be represented as "the area under the curve". Since fuel flow rate is constant, the "curve" is a straight line with positive slope... which gives a triangle. So to find the perfect time, you'd need to find the point, t, at which the dV on the left and right are equal. This is analogous to splitting a right triangle into two odd-shaped portions. On the left is a shorter similar triangle, and on the right is a taller more narrow quadrilateral. The fact that it's narrow is representative of the fact that it takes less time to burn the same amount of dV (due to increased acceleration).

Doing the integration shows that the midway point is depends only on total burn time, the proper midway point in terms of time is given by:

T-midpoint=(.5*(T^2))^1/2, where T is the total burn time and T-midpoint is found in seconds after starting the burn.

So to get burn start time, you'd subtract T-midpoint from t=0 to get the proper time to start (EG: -35 seconds).

This formula gives a linear relationship between total burn time and midpoint time. T/(sqrt(2)), or approximately .707*T.

So aiming for about 70% of total burn time before the node will be a very good approximation.

Of course, if you know the acceleration at the beginning of the burn, you can use the same integrals to solve for total burn length as well (but simple geometry is easier and faster).

Last but not least, bear in mind that maneuver dV is an estimation based on assumed "impulse" maneuvers in which all dV is burned at once. The longer it takes you, the less efficient the burn will be and the further off the estimate will be. Also, burning longer than intended can throw off timing significantly. Keeping burn times short reduces the penalties for incorrect burn midpoints and helps keep estimates accurate. I'd say that if your burns are long enough where burning 50/50 produces noticeably different results than burning 70/30, you're taking far too long anyhow.

Share this post


Link to post
Share on other sites
  • 0

I like to do 50/50 with about 2 seconds early. The Maneuver nodes do not account for how much your TWR will change during your burn, so most successful burns actually stop about a second before the total time that the node SAID it was going to need.

Also, there is somehting called "suicide burns" in which you develop a trajectory that will intercept a planetary body, such as Mun, create a maneuvering node AT where your trjectory intercepts the ground, and then pull your maneuver until your apoapsis disappears. Executing this burn as close to perfect as possible will save you the most fuel, but is obviously very dangerous if you mistime it :). For these burns you want to start your burn when your burn time MATCHES your time to node (or about 10 seconds after, again, for the above reason). If you burn too late, you will "get" to your node with only half your speed lost :)

Share this post


Link to post
Share on other sites
  • 0
I like to do 50/50 with about 2 seconds early. The Maneuver nodes do not account for how much your TWR will change during your burn, so most successful burns actually stop about a second before the total time that the node SAID it was going to need.

Also, there is somehting called "suicide burns" in which you develop a trajectory that will intercept a planetary body, such as Mun, create a maneuvering node AT where your trjectory intercepts the ground, and then pull your maneuver until your apoapsis disappears. Executing this burn as close to perfect as possible will save you the most fuel, but is obviously very dangerous if you mistime it :). For these burns you want to start your burn when your burn time MATCHES your time to node (or about 10 seconds after, again, for the above reason). If you burn too late, you will "get" to your node with only half your speed lost :)

Burning just before 50/50 is still correct for suicide burns; the time to node (time to impact) for a suicide burn assumes that you're moving at your speed from orbit. Since you're cancelling all your speed with the suicide burn, you're not going to be going at that speed for the whole burn.

If you were to burn when the time to node is the same as the burn time, then you would have only covered half the distance to the ground when the burn competes and you've killed all your velocity.

The same effect also causes some wierdness for large inclination changes (greater than 60 degrees) in low orbits. For those, you'll want to burn closer to 33/66 (1/3 of the burn time till node).

Edited by Reddeyfish

Share this post


Link to post
Share on other sites
  • 0

The same effect also causes some weirdness for large inclination changes (greater than 60 degrees) in low orbits. For those, you'll want to burn closer to 33/66 (1/3 of the burn time till node).

The best way to do large inclination changes is not to use maneuver node, rather you should stay pointed at your normal marker (the purple triangles) all the way through the burn for maximum efficiency, for timing you can find the delta-v by using the formula

DVi = 2* v * sin(IncChange/2)

to find delta v and then making a manuver node with the same delta-v to figure out the time. just split that time 50/50 and your good to go!

Source (better explanation than me): http://en.wikipedia.org/wiki/Orbital_inclination_change#Circular_orbit_inclination_change

Share this post


Link to post
Share on other sites
  • 0

I may be wrong in my math, but the equation listed is for a constant heading burn, and so a straight line velocity vector. Following the normal marker gives you a curved path, and so is less efficient.

Share this post


Link to post
Share on other sites
  • 0

I believe adinfinitum is correct, consider that if you add two vectors to implement a change, you are essentially negating each by some ratio (depending on the angle)... so the total energy required is greater. Following the normal or anti-normal markers is like adding an infinite series of vectors with changing magnitude and direction. Those that come after the target point will be detracting from some of the work done by some of the earlier vectors. It gets pretty messy mathwise, but suffice it to say that it is most efficient to decide how much work you need to do and in what direction, and do that directly... anything else will create losses.

Share this post


Link to post
Share on other sites
  • 0
The best way to do large inclination changes is not to use maneuver node, rather you should stay pointed at your normal marker (the purple triangles) all the way through the burn for maximum efficiency, for timing you can find the delta-v by using the formula
DVi = 2* v * sin(IncChange/2)

to find delta v and then making a manuver node with the same delta-v to figure out the time. just split that time 50/50 and your good to go!

Source (better explanation than me): http://en.wikipedia.org/wiki/Orbital_inclination_change#Circular_orbit_inclination_change

No maneuver is still better. Think of your speed vector in a 3D space, imagine you have two vectors, one is your current speed, one your target speed, which is of the same length but pointing to a different direction. Following maneuver will spend you the amount of dV equal to the exact line distance between the two vectors. Following your "always normal" way is going to cost you the arc length that connects the two vectors - and that's larger than a line distance, for sure.

Of course, if the burn time is extremely long then this analysis may be completely off, but for relatively short burn, this argument seems valid to me.

Share this post


Link to post
Share on other sites
This thread is quite old. Please consider starting a new thread rather than reviving this one.

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.
Note: Your post will require moderator approval before it will be visible.

Guest
Answer this question...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Sign in to follow this