Decent rate question

[DHC6]

So I'm getting checked out at a Skydive place and they have some info that we are supposed to read before we are checkout out to fly and one of the items is about descents...

It says that the plane will not come down any faster in turning descent (Call it 45 degrees) than it would in a wings level decent. Now, I had some problems with this statement as it seems that the plane would absolutely come down faster in a turning decent than it would in a wings level descent.

So which is going to make the airplane come down faster (i.e. greater negative VSI reading) a 45 banked descent or a wings level descent?

To give you more info the on the plane (which I guess shouldn't matter),it is a twin otter and we do our descents at 140 KIAS to get down ASAP. My thought is that the plane will descend faster in a 140 KIAS decent with 45 degrees bank because for a given airspeed there is a "X" amount of lift, of which in a wings level decent the vertical component of lift is greater than it would be in turn where the some of the VCL becomes a HCL for the turn; Therefore less lift in the turn which would lead to higher descent rate.

Anyway, I could be way off and making this way more simplistic than it should be but what's the right answer?

Thanks in advance.

 

[tgrayson]

Therefore less lift in the turn which would lead to higher descent rate.

A descent rate is caused by a lack of power, not a lack of lift, and the lack of power is determined solely by your drag and velocity, D * V, assuming you have the throttle at idle. Since we're going to assume that velocity is a constant, then the missing power is determined only by drag. More drag will equal more descent rate. Because an aircraft in a 45 degree banked turn has a larger AoA than an aircraft at the same speed with level wings, then drag is going to be higher in a turning descent. Therefore, the descent rate will be higher.

 

[z987k]

And a slip will make it come down real fast. Speed brakes + spiraling slip = -2500fpm+ at about 100kias in the 206 and roughly the same in the 182 if you need to make if down real fast. CHT's stay up to.

 

[fish314]

A descent rate is caused by a lack of power, not a lack of lift, and the lack of power is determined solely by your drag and velocity, D * V, assuming you have the throttle at idle. Since we're going to assume that velocity is a constant, then the missing power is determined only by drag. More drag will equal more descent rate. Because an aircraft in a 45 degree banked turn has a larger AoA than an aircraft at the same speed with level wings, then drag is going to be higher in a turning descent. Therefore, the descent rate will be higher.

Hey Tgray,

The analysis above is only true if the airspeed is held constant, correct? Of course, at some point the airspeed must be held constant, because eventually even an airplane pointed straight at the ground with the engines at max will hit a terminal velocity due to parasite drag...

And of course, the pilot will artificially cap the airplanes maximum velocity so as not to exceed Vne...

But before this point, I would think that descent rate would be incredibly dependent on the rate at which you allowed the speed to build up. So an airplane that is accelerating slowly in the descent should have a lower descent rate than the same airplane in the same conditions, but allowed to accelerate faster. Right?

 

[tgrayson]

The analysis above is only true if the airspeed is held constant, correct?

As written, yes, since that's what the OP was doing. But allowing the airplane to accelerate would only increase the descent rate. V would go up and so would D as parasite drag increased with V^2. This is similar to when you have a student roll into a steep turn; if he doesn't increase the AoA, the airplane will accelerate and descend. Pulling back on the yoke can keep the airspeed at the entry value, but the aircraft will still be descending, although less than it was. If he pulls back further, he may stop the descent, but the airspeed will be below the entry speed. He really needs to add some power instead.

But before this point, I would think that descent rate would be incredibly dependent on the rate at which you allowed the speed to build up. So an airplane that is accelerating slowly in the descent should have a lower descent rate than the same airplane in the same conditions, but allowed to accelerate faster. Right?

If you're talking about accelerated climbs and descents, you'd have to introduce energy concepts to get accurate results, rather than just power curves, which assume non-acceleration. The power that's being used to accelerate is not being used to climb or maintain altitude, so the excess power would not be a good indicator of the climb/descent rate. But a quick and dirty way to look at the problem would be the alternate way of evaluating rate of climb, which is the vertical component of your velocity vector: ROC = V * sin(flight path angle). That becomes ROD when the flight path angle is below level flight, of course. But with a constant flight path angle, increased velocity is increased descent rate. An accelerated aircraft will, on average, have a higher velocity than an unaccelerated one, so it would have a higher average rate of descent. But a pilot could certainly adjust the flight path in order to keep the ROD the same, so in order to make the analysis comprehensible, you've got to hold some variables constant when you evaluate the consequences of adjusting another.

 

[DHC6]

And a slip will make it come down real fast. Speed brakes + spiraling slip = -2500fpm+ at about 100kias in the 206 and roughly the same in the 182 if you need to make if down real fast. CHT's stay up to.

That's all?!?! Haha, we get about 2500 to 3000 FPM descent at 140 KTS with the props back and power off in a straight and level descent. The otter is a parasitic drag monster. =)

 

[z987k]

That's all?!?! Haha, we get about 2500 to 3000 FPM descent at 140 KTS with the props back and power off in a straight and level descent. The otter is a parasitic drag monster. =)

wow, apparently.

 

[DHC6]

A descent rate is caused by a lack of power, not a lack of lift, and the lack of power is determined solely by your drag and velocity, D * V, assuming you have the throttle at idle. Since we're going to assume that velocity is a constant, then the missing power is determined only by drag. More drag will equal more descent rate. Because an aircraft in a 45 degree banked turn has a larger AoA than an aircraft at the same speed with level wings, then drag is going to be higher in a turning descent. Therefore, the descent rate will be higher.

Thanks for the explanation! I knew my solution wasn't correct in some form or another. I have another way of looking at it now that you mention this...

Power from the engine = Force

In my head I just did a thought experiment where I rotated the whole world of the plane descending to 20 degrees nose up so the plane is actually flying straight and level but the same forces exist in this straight and level world... essentially gravity is pulling the plane forward not the engines.

So say that the descent pitch to maintain 140 KIAS is 20 degrees nose down, and the world is rotated 20 degrees nose up.

Now if we say it doesn't matter what the force is that is pulling the airplane through the air whether it's engines producing thrust or some component of the force of gravity; if you were to turn 45 degrees in this world you would need more force from the engines or more force from gravity to keep the plane straight and level. Now you don't have anymore force from the engines ( because they are at idle ) so the plane will start a descent if you keep the same indicated airspeed. So this is why the airplane will descend faster in a descending turn than a straight and level turn...

Of course this could just be me totally destroying the world as we know it and not making any sense, but for some reason it makes sense in my world :rotfl:

 

[ryan1234]

A descent rate is caused by a lack of power, not a lack of lift, and the lack of power is determined solely by your drag and velocity, D * V, assuming you have the throttle at idle. Since we're going to assume that velocity is a constant, then the missing power is determined only by drag. More drag will equal more descent rate. Because an aircraft in a 45 degree banked turn has a larger AoA than an aircraft at the same speed with level wings, then drag is going to be higher in a turning descent. Therefore, the descent rate will be higher.

Sir,

I'm having a little trouble wrapping my head around this - I probably don't understand it that well - but from experience flying jumpers, I've gotten about the same descent rate turning or level wings at the same power setting.

could be wrong, but when descending I'm not sure that a 45 degree bank has to have a larger AoA than a level descent, or does it? On the g-meter they come up as 1 maybe even a little less at times. Now, if the turn involves loading to, say, limit the airspeed - it'll produce more drag, but also more lift - which translates to maybe the same descent rate as the level.

So the descent rate is a function of your flight relative angle to the ground and your airspeed. A descending turn usually increases the airspeed, but then you have to load the aircraft to limit the airspeed or reduce the bank angle. Steep turns, you're talking about a level, loaded turn - which is a different animal, right?

I don't know much...so I'm sure if any of that makes any sense.

 

[tgrayson]

ould be wrong, but when descending I'm not sure that a 45 degree bank has to have a larger AoA than a level descent, or does it?

It does if the airspeed is to remain constant. The vertical component of lift must equal the weight of the airplane; when you're banked, you need a larger quantity of total lift, which must either come from an increase in AoA or an increase in airspeed. If you don't adjust your AoA, then the airspeed will increase by about 19 percent in a 45 degree banked turn.

Note that all steady turns are loaded, climbing, descending, level, doesn't matter. Think back to the steep spirals for the Commercial/Flight Instructor...there's a good 2 g load during the steepest portion, even though your descending rapidly.

 

[ryan1234]

It does if the airspeed is to remain constant. The vertical component of lift must equal the weight of the airplane; when you're banked, you need a larger quantity of total lift, which must either come from an increase in AoA or an increase in airspeed. If you don't adjust your AoA, then the airspeed will increase by about 19 percent in a 45 degree banked turn.

Note that all steady turns are loaded, climbing, descending, level, doesn't matter. Think back to the steep spirals for the Commercial/Flight Instructor...there's a good 2 g load during the steepest portion, even though your descending rapidly.

Tryin' not to sound stupid here... but if a descending turn produces an increasing vertical component of lift, it would then limit the descent rate right? ....By limiting the relative angle to the ground and airspeed respectively.

 

[tgrayson]

but if a descending turn produces an increasing vertical component of lift, it would then limit the descent rate right? ....By limiting the relative angle to the ground and airspeed respectively.

No, no, the vertical component doesn't increase, the total aerodynamic force generated by the wing increases in order that the vertical component of that force (lift) remains equal to the weight of the airplane. The steeper the bank, the greater the total aerodynamic force needs to be in order to generate lift = weight. This is what produces the load factor.

If the vertical component of lift increased over the weight of the airplane, then the airplane would be accelerating towards outer space. Likewise, if lift were less than weight, the airplane would accelerate downwards like a brick. In any steady flight regime, all forces are in equilibrium. There is no net force acting on the airplane.

 

[periksmoen]

I've never found a detectable difference in descent rates between a level descent vs. a banked descent. The exception to this is in the initial descent from level flight where I do find that a banked turn does allow you to accelerate quicker. Once I have established my descent airspeed, I find that in a level descent it's easier to scan for traffic without the earth spinning round and round below me. Additionally, when you're flying 25+ plane loads per day, you'll find that you're a whole lot less fatigued at the end of the day if you didn't spend an hour of your day spiraling at the ground.

After the last jumpers leave, I initially start a 1G descending turn to quickly accelerate to descent speed. I find this is better than just pushing the nose over to start a steep descent, which is putting zero and/or negative G's on the airframe which is was never really designed to do repeatedly. Just be aware of where the jumpers are that you just released because you will pass them in freefall as your initial descent rate in the Otter will be in the vicinity of 10,000+fpm. As you level out to maintain your descent airspeed, those jumpers will eventually pass your level again, so it is important to be well clear of where you just dropped them. I level out on a heading that is 90 degrees off from the jump run that I just flew. In the Twin Otter, I use 150KIAS in a level descent with the props at 90%. This will give you a stabilized 6000fpm descent rate. If you are descending over a populated area, then you will probably want to pull the props back to 75-80% in order to keep the noise down, but this will drop the descent rate to 4000-5000fpm.

To the pilot that suggested slipping the whole way down, I would NOT recommend doing this. I have spoken with multiple operators who have had stretched and/or broken rudder cables and bent rudders from their jump pilots doing prolonged high speed slips. It's one thing to slip an aircraft for a few seconds at 70 knots while on final, but doing that 20 times a day at 100+ knots for minutes at a time will do damage. The Twin Otter had a massive rudder and replacement parts and maintenance is getting more and more expensive, so please treat it with care. If you take good care of it you will find the the Twotter is one of the most dependable, predictable airplanes to fly jumpers in.

Good luck and have fun!

 

[shdw]

I've never found a detectable difference in descent rates between a level descent vs. a banked descent. The exception to this is in the initial descent from level flight where I do find that a banked turn ...your initial descent rate in the Otter will be in the vicinity of 10,000+fpm. As you level out to maintain your descent airspeed...This will give you a stabilized 6000fpm descent rate.

Sound like you see it each time you fly. You just leave the bank before it can stabilize there. Would you be doing about a 60 degree bank to initiate your descent?

 

[ryan1234]

I've never found a detectable difference in descent rates between a level descent vs. a banked descent. The exception to this is in the initial descent from level flight where I do find that a banked turn does allow you to accelerate quicker. Once I have established my descent airspeed, I find that in a level descent it's easier to scan for traffic without the earth spinning round and round below me. Additionally, when you're flying 25+ plane loads per day, you'll find that you're a whole lot less fatigued at the end of the day if you didn't spend an hour of your day spiraling at the ground.

After the last jumpers leave, I initially start a 1G descending turn to quickly accelerate to descent speed. I find this is better than just pushing the nose over to start a steep descent, which is putting zero and/or negative G's on the airframe which is was never really designed to do repeatedly. Just be aware of where the jumpers are that you just released because you will pass them in freefall as your initial descent rate in the Otter will be in the vicinity of 10,000+fpm. As you level out to maintain your descent airspeed, those jumpers will eventually pass your level again, so it is important to be well clear of where you just dropped them. I level out on a heading that is 90 degrees off from the jump run that I just flew. In the Twin Otter, I use 150KIAS in a level descent with the props at 90%. This will give you a stabilized 6000fpm descent rate. If you are descending over a populated area, then you will probably want to pull the props back to 75-80% in order to keep the noise down, but this will drop the descent rate to 4000-5000fpm.

To the pilot that suggested slipping the whole way down, I would NOT recommend doing this. I have spoken with multiple operators who have had stretched and/or broken rudder cables and bent rudders from their jump pilots doing prolonged high speed slips. It's one thing to slip an aircraft for a few seconds at 70 knots while on final, but doing that 20 times a day at 100+ knots for minutes at a time will do damage. The Twin Otter had a massive rudder and replacement parts and maintenance is getting more and more expensive, so please treat it with care. If you take good care of it you will find the the Twotter is one of the most dependable, predictable airplanes to fly jumpers in.

Good luck and have fun!

I agree - I can't seem to find a better descent rate in a turn versus level..... can't explain it -

 

[z987k]

I've never found a detectable difference in descent rates between a level descent vs. a banked descent. The exception to this is in the initial descent from level flight where I do find that a banked turn does allow you to accelerate quicker. Once I have established my descent airspeed, I find that in a level descent it's easier to scan for traffic without the earth spinning round and round below me. Additionally, when you're flying 25+ plane loads per day, you'll find that you're a whole lot less fatigued at the end of the day if you didn't spend an hour of your day spiraling at the ground.

After the last jumpers leave, I initially start a 1G descending turn to quickly accelerate to descent speed. I find this is better than just pushing the nose over to start a steep descent, which is putting zero and/or negative G's on the airframe which is was never really designed to do repeatedly. Just be aware of where the jumpers are that you just released because you will pass them in freefall as your initial descent rate in the Otter will be in the vicinity of 10,000+fpm. As you level out to maintain your descent airspeed, those jumpers will eventually pass your level again, so it is important to be well clear of where you just dropped them. I level out on a heading that is 90 degrees off from the jump run that I just flew. In the Twin Otter, I use 150KIAS in a level descent with the props at 90%. This will give you a stabilized 6000fpm descent rate. If you are descending over a populated area, then you will probably want to pull the props back to 75-80% in order to keep the noise down, but this will drop the descent rate to 4000-5000fpm.

To the pilot that suggested slipping the whole way down, I would NOT recommend doing this. I have spoken with multiple operators who have had stretched and/or broken rudder cables and bent rudders from their jump pilots doing prolonged high speed slips. It's one thing to slip an aircraft for a few seconds at 70 knots while on final, but doing that 20 times a day at 100+ knots for minutes at a time will do damage. The Twin Otter had a massive rudder and replacement parts and maintenance is getting more and more expensive, so please treat it with care. If you take good care of it you will find the the Twotter is one of the most dependable, predictable airplanes to fly jumpers in.

Good luck and have fun!

Who do you fly for around st louis?

 

[periksmoen]

Sound like you see it each time you fly. You just leave the bank before it can stabilize there. Would you be doing about a 60 degree bank to initiate your descent?

The reason for the initial high descent rate has little to do with bank angle and more to do with pitch angle. Even if you started your descent with zero degrees of bank, you can still reach a 10,000+fpm initial descent rate in the Otter as it accelerates in a steep nose-down attitude. Turning while beginning your descent simply allows you to keep positive G's on the aircraft while you do this. My initial bank angle is only about 45 degrees but the pitch angle starts rather steep as the aircraft accelerates, giving me a 10,000+fpm descent rate at only about 100KIAS. If you held that pitch angle indefinitely though, the aircraft would eventually accelerate beyond VNE and you soon find that your descent rate would increase even further as the wings part company with the aircraft.

 

[ryan1234]

Would you be doing about a 60 degree bank to initiate your descent?

Our practice is to let the guys out - get everything all situated, roll and kick the rudder over into a 80-90 (...uh...I..mean...60...degree) bank and get nose pointed down, level the wings and start a gentle pull out to whatever airspeed/angle you're descending at.... we generally roll out about on a 90 degree course to the jumpers (our jump run), let the dive stabilize and then either make a sweeping, gentle turn back in or continue a level descent.

The approach to landing is probably the most fun passing of time. Downwind, base, final is usually just one steep, descending turn. Usually use the last part of the descent to look for traffic around the airfield, get the airplane ready for landing - trim it up, fly down wind, tight about 1/8-1/4mi at the very most... and depending on where you want to land, where your taxi way is,trees, obstructions, etc start the turn right before 'abeaming the numbers. Usually you have so much energy from the descent, you can pull the throttle to idle, do a 180 degree descending steep turn, loading/unloading the aircraft as needed... everything is just one constant turn... wings level a second before touch down.

....'course if probably varies based on person, airplane, airport.

 

[shdw]

total aerodynamic force generated by the wing increases in order that the vertical component of that force (lift) remains equal to the weight of the airplane.

Because an aircraft in a 45 degree banked turn has a larger AoA than an aircraft at the same speed with level wings, then drag is going to be higher in a turning descent.

I wonder if this combination is causing confusion with drawing this mental picture for some of you. Specifically the underlined segment. In the right picture, lift is equal to weight to hold a stable flight condition (climb, descent, or level flight). Here is a picture:

Warning: The picture refers to the flight conditions as "descending" and "level." Replace those words with "unstable" and "stable" for a more clear picture. Where stable refers to a constant rate of climb/descent (this is what you experience in level flight as well) and unstable refers to a changing rate of climb/descent.

Do you guys think that the forces acting in a banked level turn (stable) differ from those in a constant (stable) descent? They don't. A 45 degree constant airspeed/descent rate (stable) descending turn will have ~1.4G, same as a level 45 degree turn. The small difference in the lift/weight relationship in a descent or climb are being neglected in this statement.

Notice tgray says that there is more drag in the turn because of the increased AOA. You know this to be true because you know the procedure for a steep turn is to increase back pressure to increase total lift (by increasing AOA) and subsequently add power to overcome the increase in drag (caused by the increased AOA) that results.



Said differently:

For any given flight condition (airspeed) we find the AOA, via pitch, that will achieve a stable flight condition where lift and weight are equal. If lift does not equal weight then we are increasing our climb or descent rate and our speed is changing relative to the condition we're in (unstable flight).

When we want to maintain a stable condition we need to set lift equal to weight. In a turn we accomplish this by increasing our g-forces to 1 / cos (bank angle) to achieve this lift (vertical lift when we speak about banked flight - see diagram) equal to weight relationship. So, where 1.4 g's at wings level equates to a increase in climb rate (unstable), in a 45 degree turn it only provides a lift equals weight condition (stable) where the climb/descent rate would remain constant and thus a stable flight condition.

This condition can happen whether you are level, climbing, or descending. In other words, the fact that we are climbing or descending has little to do with this discussion; really we are speaking in terms of static rate of climb (stable) versus changing rate of climb (unstable) flight conditions.

 

[Capt Obvious]

What about just idle power, and dumping the flaps in, and pushing the nose over to just below Vfe. Not sure in an Otter if this will make a differnce, but got a higher decent in a c172 doing this then a decending steep turn in a clean configuration.