parachutes and hitting the ground

[davidpsummers]

I was reading some threads about having problems with hitting the ground in 1.1 before you are slow enough to open your parachute.  I have had this problem on Duna (even with drogue chutes).  The solution given was not to come straight down (which I like to do to land accurately) but to some in with lateral velocity to "give you more time to decelerate".  But I had convinced myself that wasn't the way to go.  Your downward acceleration due to gravity doesn't change with lateral speed and the time you take to cover the distance to the ground, by the same token, shouldn't change?

What I am I missing?

[SteveD80]

On decent through atmosphere you are slowing down to terminal velocity.  if you come in sideways there is more time for the atmosphere to eat away at your velocity. 

[bewing]

48 minutes ago, davidpsummers said:

Your downward acceleration due to gravity doesn't change with lateral speed and the time you take to cover the distance to the ground, by the same token, shouldn't change?

What I am I missing?

Um, you are missing the concept of orbiting. There are far too many ways to phrase it, but one way is that the centrifugal force from your lateral speed counteracts the force of gravity so you don't fall out of the sky.

If your lateral speed is half of orbital velocity, then you only feel half normal gravity.

Additionally, your craft may experience some lift (depending on attitude) that would counter gravity, too. But the basic deal is that it does in fact take longer to fall if you have lateral velocity.

Edited May 14, 2016 by bewing

[Plusck]

Duna is very difficult to do without using engines at all.

The atmosphere is very thin, so you really want to go through as much of it as possible, especially during the very last stage as you approach the ground since the upper atmosphere does virtually nothing to slow you down at ordinary orbital speeds.

With lateral velocity, it is also possible to use body lift to slow your vertical acceleration. I must admit I've only had limited success with this since my landing craft have rarely been the sort of shape that would give any sort of lift rather than merely drag.

One thing that you could elevate to a general rule for all atmospheric bodies: ideally you want your trajectory to turn from orbital to suborbital shortly after passing Pe, so that you actually have a positive vertical component to your velocity, even though this is necessarly a very small amount. This maximises the time spent in the atmosphere.
The lower you can do this, the more certain you are to slow to terminal velocity before hitting the ground. However, to make this happen low down, you need to be going a lot faster to start with. For Duna, therefore, if you start with a 50x50km orbit, you'll never manage to get to zero vertical velocity low in the atmosphere. You have to start from much higher up, which means more heating, and all the additional difficulties that that entails.

All of which, obviously, plays havoc with your ability to target a particular landing site.

[Venusgate]

1 hour ago, davidpsummers said:

What I am I missing?

Lithobraking

I'm not sure what 1.1 has to do with this, tho. I know they've changed atmos recently, but Duna has always had an atmosphere.

There's nothing that says you CAN'T plummet like a rock with enough engines to slow you down, but at that point parachutes are kind of pointless.

[Streetwind]

2 hours ago, davidpsummers said:

Your downward acceleration due to gravity doesn't change with lateral speed and the time you take to cover the distance to the ground, by the same token, shouldn't change?

Your confusion likely stems from the difference between speed and velocity. This is not really your fault, because the aerospace community loves to use these terms interchangably, when they are in fact not.

"Speed" is an absolute value. "Your speed is 100 m/s" means simply that, and nothing else.
"Velocity" is a vector. "Your velocity is 100 m/s" makes no sense, strictly speaking, because there is information missing: 100 m/s in which direction? 100 m/s, parallel to the ground, northwards. Aha, now that makes sense! In other words, velocity is a combination of an absolute speed and a direction.

Your velocity vector can be split into constituent vectors, such as the rate at which you are falling towards the ground (vertical velocity, a downwards pointing vector) and the rate at which you are traveling parallel to the ground (horizontal velocity, a vector tangential to the curvature of the planet, in the direction of your orbit). You do this by essentially drawing a triangle.

Let's assume that you are reentering an atmosphere. At some point during your reentry, drag slows you down so that your capsule reports an absolute speed of 600 m/s. Let's also assume that this is not a terminal speed**, meaning that at this point, your spacecraft is still going faster than it should. Atmospheric drag is fighting to slow it down further.

Now, your direction becomes important. I will reference the two different cases I drew in the picture linked above.
- Example A: you are falling pretty much straight down at this point. Your velocity vector points very much towards the surface, and has only a small sideways component. As a result of this, almost all of the absolute speed is contained in your vertical component vector. You may be approaching the surface at 550 m/s here.
- Example B: you are coming in on an arc, from orbit. Most of your absolute speed is contained in the horizontal component vector. The rate at which you are actually approaching the surface here may only be something like 150 m/s.

Keep in mind: both examples have you at the same altitude above the ground. But in the second case, it will take you more than three times as long to traverse that distance and actually hit the ground. The atmosphere has that much extra time to continue slowing you down. This is why coming in sideways is helping you decelerate: because even though the speed is the same, and the altitude is the same, the direction of your speed is different.

But what about gravity? Well, your thought that gravity doesn't change is also correct. But gravity only becomes dominant once you have bled off all of your excess speed. The downwards acceleration due to gravity will push against atmospheric drag, and the two will find an equilibrium point: the terminal velocity. Once you have hit that point, it no longer matters if you have lateral movement or not. But! At that point, you have also already extracted all of the slowing potential that the atmosphere can give you. So either you are slow enough to deploy your chutes by now, or you better brought some engines to cushion your fall.

 

** See how easily these terms get mixed up? We're definitely talking about a speed here, not a velocity. But to be fair: "terminal velocity" is actually defined as going straight down in freefall, so calling it a vector is correct in that particular context.

Edited May 14, 2016 by Streetwind

[davidpsummers]

8 hours ago, Streetwind said:

Now, your direction becomes important. I will reference the two different cases I drew in the picture linked above.
- Example A: you are falling pretty much straight down at this point. Your velocity vector points very much towards the surface, and has only a small sideways component. As a result of this, almost all of the absolute speed is contained in your vertical component vector. You may be approaching the surface at 550 m/s here.
- Example B: you are coming in on an arc, from orbit. Most of your absolute speed is contained in the horizontal component vector. The rate at which you are actually approaching the surface here may only be something like 150 m/s.

 

 

 

I certainly understand speed vs. velocity.  Your example is that having the same speed straight down is worse than having that speed horizontally.  

But my issue isn't that I'm thinking that speed alone matters and not considering what component of it is vertical.  Lets say you kill your orbital velocity to zero and drop straight down.  Gravity will pull you down to the ground at a certain acceleration countered by the vertical component of drag.  If you only kill some of your orbital velocity, you gravity will still pull you down at the same acceleration, you just also have some sideways  velocity.  So the horizontal velocity doesn't change the acceleration toward the ground or the time to hit it.  But parachutes are limited by the total speed, so it seems to me that if you have significant horizontal velocity, that hasn't been killed by the horizontal component of drag, when you need to open your parachutes, you are worse off.

I think the difference may be what bewing pointed out, that planets are round and, as in an orbit, the ground is actually dropping way from you as you move laterally.

[Streetwind]

1 hour ago, davidpsummers said:

I think the difference may be what bewing pointed out, that planets are round and, as in an orbit, the ground is actually dropping way from you as you move laterally.

The lateral movement helps getting you to the correct part of the atmosphere, yes. If you are still at 95% of orbital speed, almost all of the acceleration imparted by gravity is nullified completely (after all, at 100% orbital speed, 100% of gravity is nullified. That's the definition of an orbit.).

If you're on Kerbin, you don't really start slowing all that well until maybe 35 to 30 km. If you get to that part of the atmosphere dropping straight down, you gained 100% of gravity as a bonus to your vertical speed. If you get there on a flat arc, you gained only 5% to 10% as much. So that helps for sure. But it's not the only thing that matters, because as you correctly observed, parachutes care about absolute speed. And coming in at 90% orbital speed is definitely still faster than what you can gather in a vertical drop, even if gravity adds almost nothing.

Once you're there, at the 30 km mark, then the thing with the vectors becomes important. You're starting to slow significantly below orbital speed, and your goal is to hit whatever terminal speed that part of the atmosphere offers you. If you hit terminal speed, then that means no matter what happens, you will always slow down faster than gravity can accelerate you. At that point, you have survived - even if you're still ten kilometers up going 800 m/s.

However, you are simply going to need a certain distance to reach terminal speed. It's like you're sitting in a car and need to brake really hard - your car has a minimum braking distance for whatever your current speed is, and if the obstacle you're trying to avoid is closer than that, you will hit it every time. And in case of Kerbin being the obstacle and a vertically dropping capsule being the car, the minimum braking distance (to terminal speed) is greater than 30 kilometers. You're going to hit Kerbin (and die violently) every time. If you were on a flat arc, however, you would have hundreds, if not thousands of kilometers of obstacle-free distance in front of you in which to brake. Even though your absolute speed on that arc is initially higher, it is not so high as to eat up all that distance.

Edited May 14, 2016 by Streetwind

[Plusck]

11 hours ago, Plusck said:

The atmosphere is very thin, so you really want to go through as much of it as possible, especially during the very last stage as you approach the ground since the upper atmosphere does virtually nothing to slow you down at ordinary orbital speeds.

... ideally you want your trajectory to ... have a positive vertical component to your velocity, even though this is necessarly a very small amount. This maximises the time spent in the atmosphere.

The lower you can do this, the more certain you are to slow to terminal velocity before hitting the ground. ..

Terminal velocity is the speed that you reach, in a given density of atmosphere, if you let gravity accelerate you.

If you start in vacuum, and accelerate into an atmosphere, that atmosphere has to brake you before you slow to terminal velocity. With high vertical velocity, and a thin atmosphere, there is simply not enough time to slow down to terminal velocity.

And on Duna, safe parachute speed is very close to terminal velocity at an altitude of only a couple of kilometres.

[davidpsummers]

4 hours ago, Streetwind said:

The lateral movement helps getting you to the correct part of the atmosphere, yes. If you are still at 95% of orbital speed, almost all of the acceleration imparted by gravity is nullified completely (after all, at 100% orbital speed, 100% of gravity is nullified. That's the definition of an orbit.).

Gravity really isn't nullified.  It pulls you down at the same acceleration as always. At orbital speed you are taking advantage of the fact the ground slopes down (since planets are round) so that so that you don't fall any faster than the grounds is sloping away and you "miss" the planet.  As I said above, this may be significant since you will have further to fall, though you sill fall just as fast.

4 hours ago, Streetwind said:

 

However, you are simply going to need a certain distance to reach terminal speed. It's like you're sitting in a car and need to brake really hard - your car has a minimum braking distance for whatever your current speed is, and if the obstacle you're trying to avoid is closer than that, you will hit it every time. And in case of Kerbin being the obstacle and a vertically dropping capsule being the car, the minimum braking distance (to terminal speed) is greater than 30 kilometers. You're going to hit Kerbin (and die violently) every time. If you were on a flat arc, however, you would have hundreds, if not thousands of kilometers of obstacle-free distance in front of you in which to brake. Even though your absolute speed on that arc is initially higher, it is not so high as to eat up all that distance.

If you drop a rock from 10 feet or roll a rock off a board 10 feet high, the vertical distance and time to hit the ground are the same.  So the vertical component of drag has to act over the same distance during the same time.  The Horizontal velocity only adds to horizontal drag.  It adds to distance but also makes you cover the distance faster and gives you more speed you have to loose.  I don't see how any of that helps you since none of this helps with the vertical components.

[Plusck]

(nvm)

Edited May 15, 2016 by Plusck

[Plusck]

Ah, sod it. I'll reply even if you aren't reading my posts.

Let's say that terminal velocity for your hypothetical rock is 200 m/s.

On Kerbin (or Earth) gravity's acceleration is 9.81 m/s^2. Therefore to reach 200 m/s (assuming no drag) you need 200/9.81 = 20.4 seconds.

From that standard equation S = ut+1/2at^2, which simplifies down to distance = 1/2 * a * t², we will reach terminal velocity after a total distance of 2041m.

So to get this hypothetical rock to fall at terminal velocity, on Earth (or on Kerbin) requires 2 kilometres of freefall.

If you are coming in from orbital speeds, you have a lot more than 2km to fall. Therefore your vertical velocity can exceed 200 m/s easily.

However, drag applies as the cube of velocity square of velocity.. It is therefore an order of magnitude higher than acceleration from gravity (which is only a square) (edit: wrong wrong wrong... and badly put... I'll have to come back to this in another post...), and you can assume zero acceleration once the limits of drag (i.e. terminal velocity) are reached.

Therefore if your hypothetical rock is dropped from 2 km up it will reach terminal velocity on impact. If instead your hypothetical rock is rolled off sideways at 200 m/s, it is already at terminal velocity and will not accelerate any more as it falls. Initially, that drag will only reduce the horizontal component of velocity but as gravity pulls on it, drag will reduce its vertical velocity too.

Therefore, it should be absolutely clear that this hypothetical rock, even if you assume that the planet has zero curvature (or is too big for 2km to make any real difference), if given a starting horizontal velocity of 200 m/s, will necessarily hit the ground later than the one that you simply dropped from a height of 2km.

Edited May 16, 2016 by Plusck

[AeroGav]

Fins / Winglets on the Duna re-entry vehicle would probably help a lot, at least enough so you can get to a safe chute opening speed.   Of course, they can also destabilize the launch rocket, unless you get creative with the layout (put the upper stage fuel tanks above the lander, use side by side boosters) to have the lander and it's wings lower down.

Ultimately this is why i prefer to go to Duna in a space plane.  

 

https://kerbalx.com/AeroGav/Astrojet-Citation  This one doesn't quite have the fuel to come back,  you either need to send some fuel ahead to Duna orbit, or if you think you might be visiting regularly, put an ISRU on the permanently sun facing side of Ike.

You could hop that space plane up some too, to give it the delta V it needs to go the whole way

• replace the Whiplash with a Rapier
• make a pair of jettisonable underwing pods with air intakes, some liquid fuel, and panther engines.  Use these to get to mach 2.5 and 17/18km before firing up the Rapier  (still a lot cheaper than disposable rocket, before you complain this is no longer SSTO)
• install  a fuel balancer like GPOSpeedFuelPump or TacFuelBalancer so you can get rid of the fuel ducts - less drag
• spike the engines - attach nose cones to the rear nodes of the engines for a large reduction in drag. offset them forwards so they are  hidden and don't block the thrust (gamey!)

....however,  this does mean landing on Duna with more fuel onboard, more weight equalling higher landing speed.

The other nice thing about bringing a plane is you won't need a rover.

[FullMetalMachinist]

@davidpsummers, I've been following this thread, but haven't said anything yet because you've gotten several good answers from @Streetwind and @Plusck. But something just isn't getting through, because you still don't seem to get it. That's ok, it's pretty complicated, and I don't completely understand it either. But I'll try and explain it a different way, maybe it'll help. 

I think where you're mostly getting confused is you're putting too much emphasis on acceleration due to gravity. And you're right, it's the same no matter if you come down straight or at an angle. The thing is, because it's the same in either situation, it doesn't make as much difference in reentry as "time spent in the atmosphere" does. 

Think about this. If you're in orbit and do a retrograde burn to stop all motion, you'll start falling straight down. The speed you fall at will increase due to gravity. If you were on an airless body, you'd have to burn your engines to stop, but with an atmosphere, that will slow you down for free (yay!).

But here's the problem: when the air is thin, terminal velocity is really, really fast. Like 1800 m/s fast. So that means that, even if you're in the air, you can still get going really fast when you fall. Then you finally hit denser air where the terminal velocity is slower, slow enough to open your chutes. But the tricky thing is that the dense air is just a thin layer right above the ground, and doesn't have enough time to slow you down. Just because you hit dense air doesn't mean you slow down immediately. 

Now look at the other scenario. You're in orbit and make just a small retrograde burn to lower your PE. This way your path is an arc through the air, and you spend more time in the denser air, partly because your horizontal speed means that the ground is still curving down and away from you a little. 

Really at the end of the day it's not about how much gravity pulls you down, it's about how much you can manipulate your flight path to spend as much time as possible in the air to shed your speed. 

[davidpsummers]

5 hours ago, Plusck said:

Therefore if your hypothetical rock is dropped from 2 km up it will reach terminal velocity on impact. If instead your hypothetical rock is rolled off sideways at 200 m/s, it is already at terminal velocity and will not accelerate any more as it falls. Initially, that drag will only reduce the horizontal component of velocity but as gravity pulls on it, drag will reduce its vertical velocity too.

As I read this, you are saying that once an object reaches terminal velocity, in any direction, it will not accelerate?  I don't believe this is true.  Terminal velocity is where the " the drag force (F_d) and buoyancy equals the downward force of gravity (F_G)".  Horizontal drag can't act to equal downward forces, any more than using rocket parallel to the ground will keep you from falling.  A rock rolled sideways at 200 m/s will fall and will accelerate until the _vertical drag_ equals the downward force.

[davidpsummers]

2 hours ago, FullMetalMachinist said:

@davidpsummersI think where you're mostly getting confused is you're putting too much emphasis on acceleration due to gravity. And you're right, it's the same no matter if you come down straight or at an angle. The thing is, because it's the same in either situation, it doesn't make as much difference in reentry as "time spent in the atmosphere" does. 

 

But does the time spend in the atmosphere change with horizontal velocity.  The standard claim is that a bullet fired horizontally from a gun 1 meter off the ground will hit the ground in the same amount of time (absent curvature of the planet) as a bullet dropped from 1 meter.

But as I was looking at a link for this, apparently it isn't quite true.  The Mythbusters did find a difference experimentally and, If you dig through math of aerodynamic drag you do come up with a difference...

http://www.wired.com/2009/10/mythbusters-bringing-on-the-physics-bullet-drop/

Though given the size of the difference and that fact that the differences in horizontal an vertical velocity are so large in this example, I am skeptical that this effect will be bigger than effects due the ground curving away.

[FullMetalMachinist]

7 hours ago, davidpsummers said:

But does the time spend in the atmosphere change with horizontal velocity.

Yes, because instead of going straight down, you flight path is longer. When you're going horizontal, you travel a longer distance versus dropping straight down, therefore you interact with more molecules of air that resist your travel (drag). 

The reason why is because this:

7 hours ago, davidpsummers said:

The standard claim is that a bullet fired horizontally from a gun 1 meter off the ground will hit the ground in the same amount of time (absent curvature of the planet) as a bullet dropped from 1 meter.

Isn't really what you're looking at for atmospheric reentry. It's more along the lines of a bullet fired horizontal versus a bullet fired straight down.

The reason is that the higher air is so thin that it might as well not be there (except for heat), so you fall and gain speed almost without restriction. So when you finally hit thicker air that will slow you down, it doesn't have enough time to slow you down because you're already going faster than terminal velocity. 

Think of this experiment. Imagine a hot air balloon is at 8km (about the altitude the bottom thickest layer of air starts). Fire (not drop) one bullet straight down, and fire one horizontal at the same time. The one going straight down will hit first, because it's going faster than terminal velocity when it starts, and the 8km of air acts fairly slowly to slow it down. The bullet fired horizontal is also going faster than terminal velocity, but it's vertical speed is very slow at the start. This makes the path for that bullet a long arc, meaning it travels a lot further than 8km through the air. Thus, the air has a longer amount of time to slow it down. 

[bewing]

20 hours ago, davidpsummers said:

Gravity really isn't nullified.  It pulls you down at the same acceleration as always. At orbital speed you are taking advantage of the fact the ground slopes down (since planets are round) so that so that you don't fall any faster than the grounds is sloping away and you "miss" the planet.  As I said above, this may be significant since you will have further to fall, though you sill fall just as fast.

You seem to be locked in to this way of thinking about it. The whole "ground slopes down" thing is only somewhat accurate, since "down" is not a well-defined concept in this case. It also makes it impossible to calculate the fall time for an orbital or suborbital craft, because you need to quantify how much the "ground slopes down" before you can calculate how much time things will take to hit the ground -- and it's not an easy thing to quantify. If, instead, you accept the concept of centripetal force, then you understand immediately that horizontal velocity DOES nullify gravity -- at a Vcraft / Vorbital ratio. Which allows you to calculate the fact that falling from 50 km on an airless planet takes twice as long if you have a 50% horizontal orbital velocity, than if you are falling straight down. The only reason a bullet fired horizontally takes about the same time to fall as one that is dropped, is that bullet speeds on Earth are not a significant fraction of orbital speed.

Plusck's comments about terminal velocity are somewhat wrong. Even at terminal velocity, the acceleration of gravity will modify your velocity vector to point more "down" all the time. So, technically, you are still accelerating.

Your basic calculation is correct. If you are slowing at .01g during your reentry, and you start at the same altitude every time, then in order to know how much you slow down, you have to know how long it will take to get to the ground. It's just that you are assuming that any craft always descends toward the center of Kerbin at 1g of acceleration no matter how fast they are going horizontally, and that is not correct.

 

 

[Plusck]

On 15/05/2016 at 9:58 PM, bewing said:

Plusck's comments about terminal velocity are somewhat wrong. Even at terminal velocity, the acceleration of gravity will modify your velocity vector to point more "down" all the time. So, technically, you are still accelerating.

I prefer "open to misintepretation" myself   In fact I was wrong to say drag is proportional to the cube of velocity... because it's actually proportional to the square of velocity. In fact, it is the power required to overcome drag which (because power is proportional to velocity) is proportional to velocity cubed. But I think that was the only actual error. The unclear bit was saying you "cannot accelerate" rather than "cannot accelerate to a higher total velocity".

I used that rock example simply as a thought experiment: to try to demonstrate what had to be true, by definition, without getting into numbers, but I messed it up a touch by being unclear.

If a rock is already going at terminal velocity, it cannot be accelerated by gravity to any higher absolute velocity (irrespective of vector) because by definition, drag is applying a constant force of (on Earth) 9.81m/s.

Gravity can, obviously, accelerate it downwards by 9.81m/s, while drag applies its slowing of 9.81m/s in its direction of travel.

However, gravity cannot magically cause the slowing by drag to apply only in the horizontal vector, or more to the horizontal vector than to the vertical. And because drag is proportional to velocity squared, you cannot separate the horizontal and vertical vectors and then apply the drag equation to each one before recombining them.

Therefore, it should be clear that gravity cannot cause downwards velocity to increase as quickly if it is applying to an object that is already going horizontal at terminal velocity. Therefore, necessarily, something going fast horizontally will take longer to hit the ground.

 

On 15/05/2016 at 7:28 AM, davidpsummers said:

As I read this, you are saying that once an object reaches terminal velocity, in any direction, it will not accelerate?  I don't believe this is true.  Terminal velocity is where the " the drag force (F_d) and buoyancy equals the downward force of gravity (F_G)".  Horizontal drag can't act to equal downward forces, any more than using rocket parallel to the ground will keep you from falling.  A rock rolled sideways at 200 m/s will fall and will accelerate until the _vertical drag_ equals the downward force.

So yes, I mean what I said but didn't put it very well ; )

It cannot accelerate to any higher speed and therefore the vertical acceleration has to be less than it would be if it were not already going so fast horizontally. You cannot separate drag into its horzontal and vertical components because it is proportional to the square of velocity. The size of the vertical component of drag therefore depends on the magnitude of horizontal velocity.

Edited May 17, 2016 by Plusck

[Plusck]

On 15/05/2016 at 7:51 AM, davidpsummers said:

But does the time spend in the atmosphere change with horizontal velocity.  The standard claim is that a bullet fired horizontally from a gun 1 meter off the ground will hit the ground in the same amount of time (absent curvature of the planet) as a bullet dropped from 1 meter.

But as I was looking at a link for this, apparently it isn't quite true.  The Mythbusters did find a difference experimentally and, If you dig through math of aerodynamic drag you do come up with a difference...

http://www.wired.com/2009/10/mythbusters-bringing-on-the-physics-bullet-drop/

Though given the size of the difference and that fact that the differences in horizontal an vertical velocity are so large in this example, I am skeptical that this effect will be bigger than effects due the ground curving away.

Actually, this difference is quite large. The "bullet fired three feet off the ground" has only a very minimal distance through which gravity has time to act and drag from the atmosphere has distance to slow it.

Unfortunately I'll need to use numbers to demonstrate this. For simplicity, I'll start with a 3:4:5 triangle.

Let's say we have an object coming in at an angle, total velocity (relative to the atmosphere) of 200 m/s. Horizontal velocity is 160 m/s and vertical is 120 m/s (our 3:4:5 triangle). And let's say that terminal velocity is, for this particular object at this particular altitude, 200 m/s.

Drag therefore applies a slowing force of 9.81 m/s opposite to the direction of travel (since this is the definition of terminal velocity - drag equals gravity). Over the next second, then, using pythagoras, we can split that into a horizontal slowing and a vertical slowing. Horizontally the velocity component is slowing by 7.848 m/s (=4/5 * 9.81) and vertically it is slowing by 5.886 m/s.

However, gravity is accelerating it by 9.81m/s vertically.

Therefore over the next second, it will accelerate by 9.81-5.886=3.924 m/s vertically, and by -7.848 m/s horizontally.

A second later, the horizontal velocity component will be 160-7.848=152.152m/s, and the vertical velocity component 120+3.924=123.924m/s.

If we recombine these components in our new triangle, we get total velocity = sq.rt(152.152² + 123.924²) = 196.233 m/s.

In other words, because there is a large horizontal component to our velocity, drag has slowed our craft to below terminal velocity.

And this demonstrates what probably any badminton player would be able to tell you intuitively: if you smash a shuttlecock horizontally over the net, there is a moment when it seems to just hang there in the air before dropping. Drag slows it to well below terminal velocity before gravity has time to accelerate it downwards again.

Edited May 18, 2016 by Plusck

[Zargg]

I think an easy way to visualize this is go in game and look at the orbit lines and look at how the velocity vector (speed AND direction) changes throughout this situation.

First start with a beautiful 100kmx100km orbit around the Mun. Your velocity is always 100% horizontal to the Mun, even though gravity is pulling on you, so you don't have to do any 'work' to fight gravity.

Then drop the PE down to just 10km and look at the velocity as you approach the Mun. At first, you're falling and gaining speed from gravity mostly in the vertical direction, right? But if you look specifically at the PE moment, what would you notice? You're velocity is 100% horizontal again without you doing any work! Your velocity vector has automatically had its vertical velocity nulled out just by the shape of the orbit! So you just dropped 90km in altitude, but didn't have to put any work into fighting gravity, so you got a free (as in vertical velocity) 90km drop in altitude.

Of course the energy you gained is now built up as horizontal velocity, but that doesn't put you in any danger, and the whole point would be that if there was an atmosphere, drag/chutes would have a much longer time to bleed that off since you are starting again with 0 vertical velocity.

As opposed to going straight down, where you would have gravity building up your vertical velocity for that whole 90km that you would have to cancel out in the few seconds you have before hitting the ground.

Edited May 17, 2016 by Zargg

[Warzouz]

On 14/5/2016 at 10:49 AM, Streetwind said:

"Speed" is an absolute value. "Your speed is 100 m/s" means simply that, and nothing else.

"Velocity" is a vector. "Your velocity is 100 m/s" makes no sense, strictly speaking, because there is information missing: 100 m/s in which direction? 100 m/s, parallel to the ground, northwards. Aha, now that makes sense! In other words, velocity is a combination of an absolute speed and a direction.

You're true, but beware that this may not apply in all languages. For example in French, the translation of "velocity" is not really used and is a strict synonym of "speed" (as celerity).

Considering speed as a measurement or a vector largely depends on the context. In orbital mechanics, we always consider speed as a vector (intensity and direction). There is not much sense to consider only the absolute intensity only.

On 14/5/2016 at 10:09 AM, Plusck said:

Duna is very difficult to do without using engines at all.

...

It's less difficult than in 1.0.5. Chutes have been fixed . In 1.0.5, regular chutes could only be open at 200m/s, now you can open them at 350 (I think). Drogues can be open at near orbital speed now (600m/s ?)

I've to check the precise value theis evening, I usually don't look at the speed'o'meter when do atmospheric reentry, but the chutes status.

[Kerbin vonKerbal]

On May 14, 2016 at 4:49 AM, Streetwind said:

Your confusion likely stems from the difference between speed and velocity. This is not really your fault, because the aerospace community loves to use these terms interchangably, when they are in fact not.

"Speed" is an absolute value. "Your speed is 100 m/s" means simply that, and nothing else.
"Velocity" is a vector. "Your velocity is 100 m/s" makes no sense, strictly speaking, because there is information missing: 100 m/s in which direction? 100 m/s, parallel to the ground, northwards. Aha, now that makes sense! In other words, velocity is a combination of an absolute speed and a direction.

I read this in Scott Manley's voice. 10/10 would be excited about the difference between speed and velocity again

Edited May 17, 2016 by Kerbin vonKerbal

[Streetwind]

Now that is a compliment I don't get often...!

[FancyMouse]

On 5/14/2016 at 0:32 AM, bewing said:

If your lateral speed is half of orbital velocity, then you only feel half normal gravity.

I have to correct this - centrifugal force is proportional to square of velocity (fixing r, which we can assume because the context here is near surface), so if you half the velocity, you feel 3/4 normal gravity.