climb

[Guest Marty]

anyone knows how to get over high mountains?

[Nate--IRL--]

Fly at 130km/h IAS, this gives optimum climb.

 

Nate

[Arne Anka]

Loose weight, so either, join Weight Watchers, or perhaps loose some unnecessary payload/fuel... :music_whistling: Either through jettison or perhaps easier on the ground; Maintenance -> Change payload -> Weight reduced payload -> have your pick. Same goes for Refuel...

Edited August 21, 2009 by Arne Anka

[159th_Viper]

http://forums.eagle.ru/showthread.php?t=39411

 

http://forums.eagle.ru/showthread.php?p=691172#post691172

[EtherealN]

NOTE: This post contains errors. Read on to be better educated and see the maths of it in action.

 

To explain the 130km/h IAS thing a bit:

 

Marty, do note that the wingstubs that the weapons are hanging from are actual aerofoils - that is, they have the same properties as regular airplane wings in that they create lift when air flows around them. For this reason there is a quite substantial difference in the amount of lift you get while at a hover (~0 airspeed, thus no lift from the wings) and when at speed, since in the latter case you are getting lift from both the rotors and the wings. With the Ka-50, this adds up to be most effective at 130km/h.

 

Note though that there are limits to how high you can get, and that your weight matters. You'll be able to get quite a bit higher when low on ammunition and fuel than when fully armed and with a full gas tank. Do not expect to fly over really high mountain peaks, but with an airspeed of 130km/h you should have no issues with mountain passes a few km up - just make sure to maintain airspeed. The most common mistake at altitude is that when you note you won't make it over the pass you raise your nose to climb (like a fixed-wing would), but since this slows you down it decreases the lift given by your wings and you might even find yourself starting to fall. Instead - gently turn in a circle while maintaining airspeed until you have the desired altitude.

Edited August 21, 2009 by EtherealN
Correcting myself

[osok]

I only load two pylons instead of four if I know I'm going to have to do some mountain flying.

[Panzertard]

anyone knows how to get over high mountains?

Lots of rope, patience and a icepick?

[GGTharos]

This isn't entirely correct. It's not about the wings - they might add something, but the most important thing is that translational lift from the rotors is more efficient than lift in a hover (ie. you get more lift per unit power than in a hover) and this occurs for the rotors at 130kph IAS (well, it occurs at different airspeeds, but is most efficient - ie you get the most lift - at 130kph IAS)

 

That is, once more, IAS (STEAM GAUGE!) not what you see on the HUD, which is DNS measured ground speed.

 

To explain the 130km/h IAS thing a bit:

 

Marty, do note that the wingstubs that the weapons are hanging from are actual aerofoils - that is, they have the same properties as regular airplane wings in that they create lift when air flows around them. For this reason there is a quite substantial difference in the amount of lift you get while at a hover (~0 airspeed, thus no lift from the wings) and when at speed, since in the latter case you are getting lift from both the rotors and the wings. With the Ka-50, this adds up to be most effective at 130km/h.

 

Note though that there are limits to how high you can get, and that your weight matters. You'll be able to get quite a bit higher when low on ammunition and fuel than when fully armed and with a full gas tank. Do not expect to fly over really high mountain peaks, but with an airspeed of 130km/h you should have no issues with mountain passes a few km up - just make sure to maintain airspeed. The most common mistake at altitude is that when you note you won't make it over the pass you raise your nose to climb (like a fixed-wing would), but since this slows you down it decreases the lift given by your wings and you might even find yourself starting to fall. Instead - gently turn in a circle while maintaining airspeed until you have the desired altitude.

Edited August 21, 2009 by GGTharos

[EtherealN]

Hm, interesting point GG. I'll have to find the maths of it somewhere to see if I can figure it out. Thanks.

 

Being a fixed-wing pilot IRL I guess I'm biased in where I look for explanations to aircraft behaviour. But I guess sometimes you can come to the right conclusion for the wrong reasons. :P

[GGTharos]

That's okay - you just weren't thinking that the wings on the heli move faster than the fuselage. Perfectly understandable! :D

[EtherealN]

I'm still going to screenshot you someday. :P

 

But since it's somewhat on the topic of this thread, any short bytesize way to explain it or link to explanation elsewhere? Basically, it was my understanding that (simplified) the lift from the rotors while at speed decrease and increase relatively uniformly (that is, cancelling each other) to to decreasing airspeed on one side and increasing airspeed on the other, cancelling each other out. Admittedly I never ran maths on that, it's a pure gut calculation. Am I correct in understanding that this is where my error lies?

[SkidMark]

The "Coln Kurtz" (i think) once mentioned it 2 me,....along these lines......"Try turning off ur heaters"......then giving more power.....but then (of course) once over. ON they come again. (check with him on time you have 2doit).:detective_2:

[Yo-Yo]

I'm still going to screenshot you someday. :P

 

But since it's somewhat on the topic of this thread, any short bytesize way to explain it or link to explanation elsewhere? Basically, it was my understanding that (simplified) the lift from the rotors while at speed decrease and increase relatively uniformly (that is, cancelling each other) to to decreasing airspeed on one side and increasing airspeed on the other, cancelling each other out. Admittedly I never ran maths on that, it's a pure gut calculation. Am I correct in understanding that this is where my error lies?

 

 

The explanation is simple and correlating with the facts you know about fixed wings.

 

Basicly there is a physical rule that the more air (or any media) you push down the less amount of speed you must accelerate this air to. Of course having the same lift. That's why the wings with higher aspect ratio is more effective. The amount of accelerated air increases with the airspeed, so inductive part of drag decreases. This is common for fixed wings and helos. The only different is that helos can produce lift without airspeed...

The total drag is a sum of inductive and profile drag.

As the airspeed increases profile drag increases so the total drag has a minimum. It is valid either for fixed wings or helos.

That's the point.

[EtherealN]

I'm slightly confused about it, but suspect that it may be down to language. (Swedish being my native language and the one I had my pilot training in, and russian being yours, and us communicating in english.)

 

Is there some math to describe this that doesn't take too much time from you?

[GGTharos]

Think of the rotor as a non-moving wing surface in forward motion; forget the details (ie. part of the rotor retreating etc).

 

If you assume that the rotor (magically) produces lift while stationary, you can say its airspeed is zero (it isn't, we add rotor speed to this datum) ... once you start moving forward, the rotor, like a wing, begins to produce lift as airspeed builds up. Add this translational lift to the lift you had while hovering.

 

It's an extremely simplified way of thinking about it I suppose.

Edited August 21, 2009 by GGTharos

[EtherealN]

Hm.

 

So simplified I guess I could say that I base my "gut calculation" on this:

 

L=R+F, where L is lift, and R is the lift of the retreating blade and F is the lift of the advancing blade. When I go by this (rediculously simple) model a change in airspeed of the helicopter as a whole seems to linearly change the lifts of the two opposites, so in a hover we could have:

 

10=5+5

 

Increase airspeed by 20% and it would be (obvious problem: multiplying 0 by 0.2 = 0. Something is broken in my gut calculator. :P )

 

10=4+6

 

Obvious caveats: this extremely simple model does not take into account such things as alpha modification as could be used on a single-rotor to avoid retreating blade stall. But obviously my knowledge of how to calculate all of this properly is lacking, and I realize that the mathematics might be too complex to conveniently type out.

 

Your post there does however reinforce the notion that I am probably thinking in a completely wrong manner, since all my thinking has been extremely focused on the retreating blade.

 

So instead we are looking at not calculating lift as an artificial linear number, which is obviously dangerous, but rather looking at the mass of air being displaced? Seems rather obvious when I spell it out, but in the case of ignoring the retreating blade to understand it it seems weird to remove the "detail" of the retreating blade.

 

I feel that I am forgetting the more specific details of aerofoil interaction, and focusing too much on the airspeed of the aircraft as a whole. So I'll go get some maths to describe just the rotor blade and forget about the helicopter, I guess.

 

EDIT: reading the edit done to the before post and re-examining.

Edited August 21, 2009 by EtherealN

[GGTharos]

I told you, FORGET the details (... yes, that means the retreating blade :P )

 

If you really want to go into it - you could simplistically assume you lose some amount of lift equivalent to the airspeed you're gaining for that half of the rotor. The other half of the rotor gains that much lift - but this lift is gained with respect to a rotor moving through the air.

 

In other words you have to think of the rotor as a non-moving airfoil, like the wing of a cessna. It's over-simplified, but it works.

Edited August 21, 2009 by GGTharos

[Panzertard]

I told you, FORGET the details (... yes, that means the retreating blade :P )

Impossible - he's Ethereal! Have you seen him post less than 3 paragraphs? :D

[EtherealN]

Maths rule. I believe I understand it now, explained by the following (some numbers lifted arbitrarily from completely random airfoils):

 

L=C1*½*D*V2*A

L=Lift

C1=Lift Coefficient

D=Air Density

V=Velocity

A=Area

 

Lh=Lift for blade in hover

Ldh=Lift of disc in hover

La=Lift of advancing blade

Lr=Lift of retreating blade

Lm=Lift of moving helo

 

Lh = 0,55 * 0,5 * 0,0023769 * 300 * 300 * 160

Lh = 9 412,5

Ldh = Lh+Lh = 18 825

 

La = 0,55 * 0,5 * 0,0023769 * 400 * 400 * 160

La = 16 733,37

 

Lr = 0,55 * 0,5 * 0,0023769 * 200 * 200 * 160

Lr = 4 183,34

 

Lm = La+Lr

Lm = 20 196,71

 

Lift when hovering with blade velocity of 300: 18 825

Lift when moving with total aircraft velocity of 100: 20 196,71

 

---

 

As we can see, the thing my gut got wrong was the fact that we need to square velocity, and this has profound effects. To illustrate:

 

L=C1*½*D*V*A

L=Lift

C1=Lift Coefficient

D=Air Density

V=Velocity

A=Area

 

Lh = 0,55 * 0,5 * 0,0023769 * 300 * 160

Lh = 31,37508

Ldh = Lh+Lh = 62,75016

 

La = 0,55 * 0,5 * 0,0023769 * 400 * 160

La = 41,83344

 

Lr = 0,55 * 0,5 * 0,0023769 * 200 * 160

Lr = 20,91672

 

Lm = La+Lr

Lm = 62,75016

 

Lift when hovering with blade velocity of 300: 62,75016

Lift when moving with total aircraft velocity of 100: 62,75016

 

---

 

As we can see, with the proper order of squaring the influence of airspeed, we get an increase in total lift of the "wings", even allowing for the retreating blade being slower, than if we make the mistake (like I did) of going "gut" and assuming a linear relationship. Silly mistake by me in making the gut reaction of positive 100 with negative 100 making 0, which obviously is wrong when you look at an equation that requires you to square it. Still some work in figuring out why it tops at a certain airspeed, but I am at least satisfied that the maths prove that a linear increase/decrease in airspeed between the two blades does cause a net increase in lift.

 

Thanks for your patience, GG and Yo-Yo. :)

 

------

 

EDIT: And dammit! I've made several posts shorter than 3 paragraphs! Hell, when I broke 1000 posts someone said he was sure at least a third of them had some kind of point! :P

Edited August 21, 2009 by EtherealN

[GGTharos]

Kudos on going out there and doing your own math :D

 

At least someone does, instead of waiting for someone else to tell them everything!

 

Refreshing :)

[EtherealN]

Pointers on where to look to figure out why the lift doesn't increase constantly with airspeed would be awesome. I'd like to do the maths myself (best way to understand it) but hints on the best way to go about it would be gratefully received.

[Jrod]

For some reason I thought I could cerebrally keep up with this thread....failed! :doh:

[Panzertard]

For some reason I thought I could cerebrally keep up with this thread....failed! :doh:

lol .. you're not alone :D

Ethereals calculations goes WAY past my understanding ;)

[topol-m]

For some reason I thought I could cerebrally keep up with this thread....failed! :doh:

 

Lol :laugh: me too. Though when i look at these calculations so detailed and well-arranged they look correct to me :lol:

[xxvintersorgxx]

For some reason I thought I could cerebrally keep up with this thread....failed! :doh:

lol, well keeping up with this thread is something you don't need to do... they are just trying to explain the answer giving in the second post. If you cant follow the math, just accept the answer. :joystick:

 

Fly at 130km/h IAS, this gives optimum climb.

 

Nate