steering losses ?

[maccollo]

Are you saying that an idealized two burn ascent, with a horizontal instantaneous burn to set Ap and another at Ap to circularize, would show gravity losses?

Absolutely. If it didn't then steering + gravity + drag losses would not account for the difference between deltaV expended and final velocity.

Let's say you launch on from Mun is KSP. On our first attempt we do an impulse acceleration of 578 m/s straight towards the horizon (and we don't hit any mountains). This gives us an apoapsis of about 25 km. When we arive at apoptosis we perform a 14 m/s burn to circularise. We spent a total of 592 m/s and our final velocity is 538 m/s

On our second attempt we launch at a 45 degree angle. The first burn is 334 m/s, and the circulation burn is 320 m/s. So we spent 654 m/s of delta V, and our velocity 538 m/s.

So the first ascent cost 62 m/s less than the second one.

Now if we use gravity losses the way you define it, and we want to account for the difference between our spent deltaV and final velocity using gravity losses, steering losses and drag losses, then there is no way to explain this difference. If you define gravity losses to simply be velocity that is exchanged for potential energy then not only can we account for the losses in this ascent, we can do it for any ascent and also for any operations that we perform after getting to orbit. Current will equal delta V spent + initial velocity - gravity losses - steering losses - drag losses, and this will always be the case as long as we don't cross into another SOI.

Not sure what is meant by a "constant altitude ascent", can you clarify?

Like in this video between 1:07 and 1:40 https://youtu.be/Hmkxg5etEtE?t=1m7s

The trajectory isn't perfectly flat, but let's say it is. Is this steering losses or gravity losses? They way I see it and they way mechjeb calculates it, it isn't gravity losses. It's simply steering losses.

Nor should you, IMO.

The point is that if you can't do that then the whole concept is worthless, because you can't use it to calculate your actual losses.

Edited August 13, 2015 by maccollo

[Empiro]

Absolutely. If it didn't then steering + gravity + drag losses would not account for the difference between deltaV expended and final velocity.

Let's say you launch on from Mun is KSP. On our first attempt we do an impulse acceleration of 578 m/s straight towards the horizon (and we don't hit any mountains). This gives us an apoapsis of about 25 km. When we arive at apoptosis we perform a 14 m/s burn to circularise. We spent a total of 592 m/s and our final velocity is 538 m/s

On our second attempt we launch at a 45 degree angle. The first burn is 334 m/s, and the circulation burn is 320 m/s. So we spent 654 m/s of delta V, and our velocity 538 m/s.

So the first ascent cost 62 m/s less than the second one.

Now if we use gravity losses the way you define it, and we want to account for the difference between our spent deltaV and final velocity using gravity losses, steering losses and drag losses, then there is no way to explain this difference. If you define gravity losses to simply be velocity that is exchanged for potential energy then not only can we account for the losses in this ascent, we can do it for any ascent and also for any operations that we perform after getting to orbit. Current will equal delta V spent + initial velocity - gravity losses - steering losses - drag losses, and this will always be the case as long as we don't cross into another SOI.

You can argue that this really isn't gravity loss though -- it's an inefficiency in your plotted maneuvers. It's no different than if you perform a Hohmann transfer when a bi-elliptic transfer would be more efficient. In addition, the concept of a "final" velocity is not well defined if you're not in a circular orbit. Plus, it leads to odd situations where you have negative gravity drag -- for example, if you're in orbit at the altitude of Minmus (270 m/s), then lowering yourself to LKO costs about 1100 m/s, but you'll end up with a velocity of about 2300 m/s.

The wikipedia entry also doesn't define gravity drag in the same way you do: https://en.wikipedia.org/wiki/Gravity_drag.

Edited August 13, 2015 by Empiro

[maccollo]

You can argue that this really isn't gravity loss though -- it's an inefficiency in your plotted maneuvers. It's no different than if you perform a Hohmann transfer when a bi-elliptic transfer would be more efficienthttps://en.wikipedia.org/wiki/Gravity_drag.

Well yeah, some ascent profiles are more efficient than others. But you have to be able to account for the difference. In the case of the a bieleptical transfer vs a Hohnmann transfer the difference will still be accounted for in the gravity losses.

In addition, the concept of a "final" velocity is not well defined if you're not in a circular orbit.

It doesn't matter. If the orbit ends up elliptical then you will loose or gain velocity as you gain or loose altitude, gravity losses account for this. What I'm trying to get at is that this is a more useful and simple way of dealing with the problem of delta V losses. Gravity losses are simply reduced to change in velocity as a result of altitude and starting velocity, cosine losses are always just that. Gravity losses are no articulately mixed in with steering losses even if is done to avoid smashing into the planet. When added together with drag losses the sum will ways account for the difference between your current velocity and spent delta V, so it doesn't stop being relevant the moment you cut the engine.

The wikipedia entry also doesn't define gravity drag in the same way you do: https://en.wikipedia.org/wiki/Gravity_drag.

That article defines gravity losses to include aerodynamic drag.

[Red Iron Crown]

Absolutely. If it didn't then steering + gravity + drag losses would not account for the difference between deltaV expended and final velocity.

Nor should it. Some of the delta-V spent is used to attain higher altitude. Equating delta-V expended to actual speed is a fool's errand, IMO.

Let's say you launch on from Mun is KSP. <snip>

All this is predicated on the idea that current speed must equal the delta-v spent (including gravity, drag and steering losses in the general case), a position I disagree with.

Like in this video between 1:07 and 1:40 https://youtu.be/Hmkxg5etEtE?t=1m7s

The trajectory isn't perfectly flat, but let's say it is. Is this steering losses or gravity losses? They way I see it and they way mechjeb calculates it, it isn't gravity losses. It's simply steering losses.

The point is that if you can't do that then the whole concept is worthless, because you can't use it to calculate your actual losses.

The way I think of it is like so:

Steering losses occur when not pointing perfectly prograde or retrograde. Instantaneous burns can still experience steering losses.

Gravity losses occur when not pointing perpendicular to the direction of gravity. Instantaneous burns do not experience gravity losses.

In your linked video the rocket is experiencing both.

To me it is silly to call the reduction in speed when falling to apoapsis "losses", because nothing has been lost. By my definition, gravity losses can only occur when the vessel is under thrust. Really though, I think we are just using different definitions here.

[Kussris]

To be pedantic, I shouldn't have said the orbital velocity wouldn't change as that is a vector. If you burn to normal, the orbital speed (magnitude of velocity) won't change but the velocity vector will, as you will be going in a slightly different direction (i.e. you've changed inclination).

If you burn, you will change your velocity! Steering losses don't mean your burns are having no/reduced effect, it's that you will end up spending more delta-v than necessary to end up in the final orbit.

Thanks! Now I understand it much better.

Edited August 18, 2015 by Kussris

[Skalou]

Nice curves,

I'm looking for this kind of data for a rocket that go into orbit,

do someone has some great links please?

here is mine:

Orbital Mechanics: Theory and Applications

Introduction to Astronautics Sissejuhatus kosmonautikasse Vladislav Pustõnski 2009 Tallinn University of Technology. (english) slides 8-9

After a few test with mechjeb, it appears that the angle alpha taken into acount is from the surface prograde, and i don't understand why it is the surface and not the orbital? (exept for the drag), on an airless body at the moment we leave the ground we don't take care anymore of the surface thing no? (the body could even start spinning backward it won't change your orbitale flight),... so why?

 

[Temstar]

On 12/8/2015 at 9:40 AM, Red Iron Crown said:

To me it is silly to call the reduction in speed when falling to apoapsis "losses", because nothing has been lost. By my definition, gravity losses can only occur when the vessel is under thrust. Really though, I think we are just using different definitions here.

I don't think that's very useful way of thinking though. Imagine a big cannon firing at 45 degree angle into the sky. As the cannon ball leave the muzzle it has equal magnitude of vertical and horizontal velocity. At the AP of the cannonball's flight it's vertical velocity component is now zero while it's horizontal component is still the same as when it left the cannon (assuming no drag).

The upward velocity has been lost to gravity. Of course the cannonball gained gravitational potential energy equal to vertical velocity component lost. But we all know being 100km above the surface does not mean you're in orbit, only the horizontal velocity component is useful for you to achieve orbit.

Similarly when you launch a rocket your engines add both horizontal and vertical speed to the whole vehicle depending on your climb angle. Assuming your aim is a circular orbit, by the time you have entered your target orbit your vertical speed is by definition zero - ie every single m/s of vertical speed your engine has added to your vehicle during its burn has been lost to gravity drag. The only thing that keeps you in your circular orbit is the horizontal speed that your vehicle has accumulated.

We know experimentally that launching horizontal on an airless body cost the least amount of delta-V, even with an instantaneous acceleration from a cannon. Thus we know gravity drag is present even when you have instantaneous acceleration followed by coasting, as long as some component of the acceleration is done against the direction of acceleration from gravity. The only way to avoid losing velocity to gravity is to never accelerate against it - ie firing a cannon on mountain top flat, or accelerate a spacecraft against the horizon.

But what about launching into an elliptical orbit you ask? Well let me ask you: if you start off in a circular orbit and you want to go into an elliptical orbit (say going to the Mun), which method of acceleration is more efficient? Thrust prograde (and thus against the horizon, since we're starting from a circular orbit), or thrust radial?

As for steering loss, imagine a rocket with two engines on two extremes, facing opposite directions. In other words a giant aeolipile:



When you light up those two engines the whole contraction will start spinning furiously, but it won't move an inch. So all the work the engines are doing goes into rotating the vehicle instead of translating the vehicle.

So that's what steering loss is. Any time when the thrust of the engine is not aligned with the CoM of the rocket, some of the work the engine is doing goes into rotating the rocket instead of accelerating it in a direction. This loss of potential acceleration is steering loss. An aeolipile has 100% steering loss hence it doesn't go anywhere.

Edited January 28, 2016 by Temstar

[Skalou]

Yep there is this kind of losses but they don't depend of your trajectory.

The main steering losses i know are due to burning out of prograde (maybe both surface or orbitale are OK and it depends of the frame you take care of the trajectory) :

blue: initial velocity

orange: increment of velocity

purple: final velocity

Edited January 28, 2016 by Skalou

[Red Iron Crown]

On 1/27/2016 at 0:55 AM, Temstar said:

I don't think that's very useful way of thinking though. Imagine a big cannon firing at 45 degree angle into the sky. As the cannon ball leave the muzzle it has equal magnitude of vertical and horizontal velocity. At the AP of the cannonball's flight it's vertical velocity component is now zero while it's horizontal component is still the same as when it left the cannon (assuming no drag).

The upward velocity has been lost to gravity. Of course the cannonball gained gravitational potential energy equal to vertical velocity component lost. But we all know being 100km above the surface does not mean you're in orbit, only the horizontal velocity component is useful for you to achieve orbit.

Similarly when you launch a rocket your engines add both horizontal and vertical speed to the whole vehicle depending on your climb angle. Assuming your aim is a circular orbit, by the time you have entered your target orbit your vertical speed is by definition zero - ie every single m/s of vertical speed your engine has added to your vehicle during its burn has been lost to gravity drag. The only thing that keeps you in your circular orbit is the horizontal speed that your vehicle has accumulated.

We know experimentally that launching horizontal on an airless body cost the least amount of delta-V, even with an instantaneous acceleration from a cannon. Thus we know gravity drag is present even when you have instantaneous acceleration followed by coasting, as long as some component of the acceleration is done against the direction of acceleration from gravity. The only way to avoid losing velocity to gravity is to never accelerate against it - ie firing a cannon on mountain top flat, or accelerate a spacecraft against the horizon.

But what about launching into an elliptical orbit you ask? Well let me ask you: if you start off in a circular orbit and you want to go into an elliptical orbit (say going to the Mun), which method of acceleration is more efficient? Thrust prograde (and thus against the horizon, since we're starting from a circular orbit), or thrust radial?
 

I think of it in terms of energy, always. Burning prograde adds energy, if gravity is reducing the vessel's acceleration then that is the gravity loss. If the vessel is pointing away from prograde then steering losses occur (sometimes this is desirable, as when changing inclination).

A vessel coasting toward apoapsis in vacuum is losing speed but not losing energy, I don't think of that as gravity loss.

With the idealized two-instantaneous-burn ascent, no gravity loss occurs. 

 

Quote

As for steering loss, imagine a rocket with two engines on two extremes, facing opposite directions. In other words a giant aeolipile:



When you light up those two engines the whole contraction will start spinning furiously, but it won't move an inch. So all the work the engines are doing goes into rotating the vehicle instead of translating the vehicle.

So that's what steering loss is. Any time when the thrust of the engine is not aligned with the CoM of the rocket, some of the work the engine is doing goes into rotating the rocket instead of accelerating it in a direction. This loss of potential acceleration is steering loss. An aeolipile has 100% steering loss hence it doesn't go anywhere.

What you are describing as steering loss here I think of as a subset of cosine losses. Though the vessel doesn't need to rotate to experience cosine losses, of course.