Climb angle for best rate of climb?

[mnowell129]

1) Assume a powered aircraft with sustained vertical capability, i.e. thrust >> weight for some finite period of time.
What is the climb angle for the fastest rate of climb?
I'm looking for some unbiased independent verification of what I already believe.

2) Is this the correct angle for a DLG sailplane launch to achieve maximum height, and if not, what is? and why?

Let the games begin :)

[Brandano]

perhaps the angle that allows you to achieve the best LD ratio? Since thrust here is more than weight, probably an angle much closer to vertical, since the plane doesn't need the additional lift, and using less wing lift will reduce the induced drag. I'd just try to achieve the best LD ratio speed, and then pitch up as much as the plane will allow while maintaining the same speed.
A DLG glider has a different problem. It starts to lose speed as soon as it's launched. Probably the best option would be to launch it almost vertically and have it coast on a parabolic trajectory to the glide speed at the top, while keeping the wing unloaded to minimize drag. You have a finite amount of energy to work with, and any maneuvering will waste some of it.

[richard hanson]

1) Assume a powered aircraft with sustained vertical capability, i.e. thrust >> weight for some finite period of time.
What is the climb angle for the fastest rate of climb?
I'm looking for some unbiased independent verification of what I already believe.

2) Is this the correct angle for a DLG sailplane launch to achieve maximum height, and if not, what is? and why?

Let the games begin :)
speed and climb angle at speed
look at time to altitude
Just because it has sustained vertical climb capability does not mean that it has fastest time to altitude in vertical flite
combining lift with power is USUALLY the best --even tho lift adds drag.
playing with small 3D foamplanes will show this
what angle is best?
who knows - you need a fixed setup to proove it - for that particular setup.

[eflightray]

Now I'm a complete dummy when it comes to the 'technical side' of lift drag, but reading through the posts made me think, (must go and have a lie down soon).

A DLG launch - if the wind was blowing, (when is it never), a vertical launch is still cutting across the air flow, (wind), so would there be less drag if throwing slightly with the wind, (slightly down wind), but still near to vertical?. But, would it actually lose out in the climb because it is losing some of the lift energy of the wind?.

Luckily I fly electric, but still wondered.

[vintage1]

I THINK that you will find if the rate of climb goes over the best L/D cruise speed then its vertical, but plenty of fully vertical planes climb faster at less than 90 degrees.

[vintage1]

Now I'm a complete dummy when it comes to the 'technical side' of lift drag, but reading through the posts made me think, (must go and have a lie down soon).

A DLG launch - if the wind was blowing, (when is it never), a vertical launch is still cutting across the air flow, (wind), so would there be less drag if throwing slightly with the wind, (slightly down wind), but still near to vertical?. But, would it actually lose out in the climb because it is losing some of the lift energy of the wind?.

Luckily I fly electric, but still wondered.


The maximum kinetic energy with respect to the wind for a given launch speed with respect to the ground, is achieved by bunging it dead flat into the wind, and then using up elevator to gain height..if there is wind shear as well, it will do even better.

[Tom Harper]

54 degrees

[jfrickmann]

The way you throw a DLG is almost horizontal when it leaves your hand. Then as it pulls up to a vertical climb, it should do that with the maximal L/D and continue vertical until it is time to level out.

[Brandano]

Jesper, I think the reason why a DLG is thrown horizontally is for safety in the case of a bad throw and for the fact that it would be really hard to throw it any other way, since you need to put the discus thrower movement into it. Eflighttray, throwing it into the wind would get you a little "free" energy (emphasis added to avoid flames, we all know there's no free energy) right at the start, but once the plane is flying you are better off getting it to climb as vertical as possible even if it's looking like it's flying backward to you. That said it might be hard to keep it in sight. The same consideration goes for a powered model ifit's initially thrown. If it takes off on it's own power, the only help the wind will give is a shorter takeoff run.

[mnowell129]

54 degrees
Yea, so why?

[Brandano]

Hmm drawing vectors of the thrust and generated lift the angle that gives you the longest vector. May well ne 54, degrees (why not 45?) but even then , how is this obtained? Or is just an empirical measurement?

[kcaldwel]

There are good equations for best climb angle, and best rate of climb (different things) for all power levels, and all types of aircraft. Perkins and Hage "Theory of Flight" has them, if you have some idea of thrust vs. speed, weight, Cd, etc. Since L/D ratio might be 10:1 or larger, you can get a lot of lift from the wing for little drag. The best angle depends on the airplane, power vs. speed, propeller efficiency at different speeds, etc.

DLG launches have been simulated in every detail by Dr. Drela. I've attached some of his plots for different climb angles, wind speeds, pull-up Cl, etc.

You can't really throw them ballistically, because the wing will hit the ground on the back of the spin. Keeping the rotation plane horizontal, with a pretty flat throw is recommended, with pre-sets on the Tx for a pull-up after release.

Kevin

[jfrickmann]

The almost horizontal throw is only because it's the way you can give the plane the most speed. Human physiology, not aerodynamics! As for the 54 degrees, it is obviously wrong. We all know from the Hitchhiker's Guide to the Galaxy the answer to everything is 42 ;)

[L1z4rd]

I seem to remember from uni 7 years ago that if your power to weight ratio is greater than 1.5 then a vertical climb gives the quickest climb rate...

[Brandano]

L1z4rd, that could be right, but to climb vertically the plane needs to keep its nose just shy of the 90 degrees datum, or it will be flying backward. Unless it's been rigged completely neutral.

[vintage1]

I seem to remember from uni 7 years ago that if your power to weight ratio is greater than 1.5 then a vertical climb gives the quickest climb rate...

1.5 what?

power is not the same units as weight..

[Brandano]

oh, nit-picking... I assume he meant thrust instead :)

[vintage1]

No that doesn't work either.

I think there is a way to approach this that works.

For a given prop amd motor effeicoency, the rate of climb (max) is the power to weight ratio, making allowance for power losses due to drag.

Obviously in the limt, that will be all climbing on the prop with the airframe at its lowest drag position. Approximately vertical, but also a machine that can only just hover and climb a little vertically is able to climb on the wing a lot faster.

So just because a model can climb vertically under power doesn't mean its the fastest way.

I therefore think that the condition where vertical flight becomes optimal, is where the climb speed equals the speed at which the L/D ratio is at a maximum..

Say the L/D(max) is 20:1 at say 20mph..drag losses will be weight/10 x 20mph.

power needed will be that, plus whatever is needed to gain potential energy at 20mph upwards.

So the power point is 110% of 20mph x weight. 20mph is 29.33 feet per second or 39.77 watts per pound. Add in 10% for drag, and that is 43 watts per pound.

Assuming the model is propped well for climbing, so the prop is say 70% efficient and so is the motor, that's around a 50% battery-airframe efficiency, so the point at which the best climb rate becomes essentially vertical is 86W/lb.

There are a lot of assumptions in there of course : if the machine is propped for speed, the prop efficiency at that 20mph may be very low, and it may be better to optimise for more power out at higher airspeeds, even if that means the model goes very fast at say 45 degrees. But 86W/lb would seem to be the minimum power needed to get a light model that will climb vertically faster than any other way.

[richard hanson]

No that doesn't work either.

I think there is a way to approach this that works.

For a given prop amd motor effeicoency, the rate of climb (max) is the power to weight ratio, making allowance for power losses due to drag.

Obviously in the limt, that will be all climbing on the prop with the airframe at its lowest drag position. Approximately vertical, but also a machine that can only just hover and climb a little vertically is able to climb on the wing a lot faster.

So just because a model can climb vertically under power doesn't mean its the fastest way.

I therefore think that the condition where vertical flight becomes optimal, is where the climb speed equals the speed at which the L/D ratio is at a maximum..

Say the L/D(max) is 20:1 at say 20mph..drag losses will be weight/10 x 20mph.

power needed will be that, plus whatever is needed to gain potential energy at 20mph upwards.

So the power point is 110% of 20mph x weight. 20mph is 29.33 feet per second or 39.77 watts per pound. Add in 10% for drag, and that is 43 watts per pound.

Assuming the model is propped well for climbing, so the prop is say 70% efficient and so is the motor, that's around a 50% battery-airframe efficiency, so the point at which the best climb rate becomes essentially vertical is 86W/lb.

There are a lot of assumptions in there of course : if the machine is propped for speed, the prop efficiency at that 20mph may be very low, and it may be better to optimise for more power out at higher airspeeds, even if that means the model goes very fast at say 45 degrees. But 86W/lb would seem to be the minimum power needed to get a light model that will climb vertically faster than any other way. As I noted early on you simply can NOT calculate best angle - you can predict the possible range of the best angle but that's it- period.

[Tom Harper]

Richard,

Perhaps it is a range but the range is very narrow and is centered on 55 degrees.

The optimum angle of climb utilizes both thrust and lift to accelerate vertically. The thrust is always parallel to the line of flight so is felt equally at any angle. Lift, however, is felt at right angles to the direction of motion. When the model climbs vertically, the wing makes no contribution to the climb. At 55 degrees they optimally share the load.

"Model Aircraft Aerodynamics" Martin Simons 1978, P35

I believe Donald K. Foote came to the same conclusion in 1952 but I can't find that book.

Tom

[richard hanson]

Richard,

Perhaps it is a range but the range is very narrow and is centered on 55 degrees.

The optimum angle of climb utilizes both thrust and lift to accelerate vertically. The thrust is always parallel to the line of flight so is felt equally at any angle. Lift, however, is felt at right angles to the direction of motion. When the model climbs vertically, the wing makes no contribution to the climb. At 55 degrees they optimally share the load.

"Model Aircraft Aerodynamics" Martin Simons 1978, P35

I believe Donald K. Foote came to the same conclusion in 1952 but I can't find that book.

Tom
That info MAY have been true then but the times they have changed
Now, we have motor thrust setups unimaginable then
I have models with thrust, 3 times the weight - these appear to do best with a vertical powered take off.
The new batteries and motor combos are dramatically superior
There is no drag induced by providing lift
lift is drag and drag is lift - no matter what you call it.

[Tom Harper]

Perhaps you are correct, but I dont think the laws of flight dynamics have changed dramatically in 30 years.

How are you measurig your rate of climb. I don't doubt that you are able to climb vertically, but I believe the same aircraft climbing at 55 degrees would climb faster. It would be interesting to test this with a data logger.

"There is no drag induced by providing lift" - not true - If there is no cost to generating lift then you have an infinite glide - I'd like to get me some of that.

"lift is drag and drag is lift" - I don't follow that one.

[richard hanson]

Perhaps you are correct, but I dont think the laws of flight dynamics have changed dramatically in 30 years.

How are you measurig your rate of climb. I don't doubt that you are able to climb vertically, but I believe the same aircraft climbing at 55 degrees would climb faster. It would be interesting to test this with a data logger.

"There is no drag induced by providing lift" - not true - If there is no cost to generating lift then you have an infinite glide.

"lift is drag and drag is lift" - I don't follow that one.
Not in the Bible terminology but consider this
In order to produce lift - you must have a difference in pressure on the wing/fuselage -any part of the craft
IF- you remove that difference in pressure --(zero lift )
you also removed the drag caused by producing lift.

You can call it whatever term suits you but as you increase lift ( increase pressure differences)- more power is required.
The fastest speed craft use tiny wings -just enough lift to do the job

[Brandano]

"There is no drag induced by providing lift" - not true - If there is no cost to generating lift then you have an infinite glide - I'd like to get me some of that.

Ah, I am afraid this is one of the occasions where your wonderful language fails you. What I think he meant is that, as the wing is producing no lift in a vertical climb, it isn't also subject to induced drag, but only to friction drag. "Drag is lift and lift is drag" is probably along the same lines. Lift and induced drag are almost directly proportional through a wide AOA range until you reach the stall angle.

[Tom Harper]

Kind of.

In a vertical climb the wing produces lift and drag. The wing will move the airplane horizontally so the climb will not be vertical. If you move the elevator so that the wing produces no lift it still has weight and profile drag.

Vintage covered L/D ratio above.

At high speed the cost of lift is very low due to the nature of induced drag and the L/D ratio. So, you will reach the greatest rate of climb at an angle of around 55 degrees.

"You can call it whatever term suits you but as you increase lift ( increase pressure differences)- more power is required."

That statement is not true. If you increase lift by increasing power, thus speed, drag will decrease. That phenomena was reoported by Langley. It would be a free ride except that parasitic drag eventually becomes dominant.

The drag curve, with velocity, is U shaped. Some birds take advantage of this by flapping their wings until parasitic drag begins to increase then coasting until induced drag increases. That way they operate in the bottom of the U.

Check out the curves here:

http://classicairshows.com/Education/Aerodynamics/SimpleAerodynamicsPartFive.htm

Scroll down the the bottom of the page.

Some fast aircraft do not have 'tiny' wings.

[kcaldwel]

High power aircraft have been around since WWII, so high power electric RCs aren't anything new. Here's a movie of an Me163 climbing out - looks much steeper than 55 degrees to me:

http://luftwaffefighters.co.uk/me163video.htm

Perkins and Hage "Airplane Performance and Control": big long equation for rate of climb for aircraft capable of accelerating during climb, that needs successive approximations to solve. The most interesting point is: "The effect of angle of climb on rate of climb is very small. In equation (4-56) gamma (angle of climb) affects only the induced drag term, which is very small at the high climbing speeds associated with high performance aircraft."

So it doesn't make much difference what angle you climb at if you have lots of power, the rate of climb is about the same. And it ain't easy to solve for, and involves a whole pile of aircraft parameters.

But since propeller driven RCs will run into their top speed limit pretty quickly, once the aircraft has accelerated to somewere near top speed, vertical will be the way to go.

Kevin

[richard hanson]

Kind of.

In a vertical climb the wing produces lift and drag. The wing will move the airplane horizontally so the climb will not be vertical. If you move the elevator so that the wing produces no lift it still has weight and profile drag.

Vintage covered L/D ratio above.

At high speed the cost of lift is very low due to the nature of induced drag and the L/D ratio. So, you will reach the greatest rate of climb at an angle of around 55 degrees.

"You can call it whatever term suits you but as you increase lift ( increase pressure differences)- more power is required."

That statement is not true. If you increase lift by increasing power, thus speed, drag will decrease. That phenomena was reoported by Langley. It would be a free ride except that parasitic drag eventually becomes dominant.

The drag curve, with velocity, is U shaped. Some birds take advantage of this by flapping their wings until parasitic drag begins to increase then coasting until induced drag increases. That way they operate in the bottom of the U.

Check out the curves here:

http://classicairshows.com/Education/Aerodynamics/SimpleAerodynamicsPartFive.htm

Scroll down the the bottom of the page.

Some fast aircraft do not have 'tiny' wings.
You are not reading what I wrote--I said IF you increase lift (change pressure differences) yo haveto increase power - . Ifyou don't do this you will slow down.
IFyou increase lift by increasing power -that is another matter.
also in vertical accelerated flight NO effective AOA (no lift )is desired or required and no vector is involved
We use symmetrical shaped on all of our models as we use power a LOT and simply change ele to produce lift and OR drag as desired
aerobatic setups --thrust available is 2-3 times weight

[Tom Harper]

But, your original question was "what is the climb angle for the fastest rate of climb?".

Although your model, or a 163, may be capable of vertical climb, you will find that arround 55 degrees you are climbing faster.

To use your scenario, assume your model is in level flight with full power to the prop. To increase lift you raise the nose angle. The model will increase lift and rate of climb and will decrease speed. This process will continue until the climb angle is 55 degrees. At that point the forward speed will slow a greater amount for each degree of increased climb angle. This is because the lift previously contributed by the wing must be replaced by thrust. The only way the propeller can provide more thrust (since RPM is already at maximum) is to decrease velocity. So the rate of climb will decrease after the climb angle of 55 degrees is passed.

Vertical may look great but it is slower.

[N9DP]

1) Assume a powered aircraft with sustained vertical capability, i.e. thrust >> weight for some finite period of time.
What is the climb angle for the fastest rate of climb?
I'm looking for some unbiased independent verification of what I already believe.

Let the games begin :)

Problem 1) is under defined and does not alow a quantitative answer. Most technical writers would assume >> to be of the order of a factor of twenty or greater. In that case, I believe the answer is vertical, based on single-stage rocket formulae. I know of no model airplanes with thrust >> weight.

Good landings,

Dennis

[richard hanson]

Problem 1) is under defined and does not alow a quantitative answer. Most technical writers would assume >> to be of the order of a factor of twenty or greater. In that case, I believe the answer is vertical, based on single-stage rocket formulae. I know of no model airplanes with thrust >> weight.

Good landings,

Dennis
2-3 time thrust is presently ,easily done we do it all the time now - even my 23lb EDGE with piped 80cc is almost twice the power to weight (the static thrust with a fairly high pitched prop is 60 pounds and it has far less power to weight than my electrics -- we check static thrust on a stand with a scale and watts consumed - - full speed on these comes within a very few seconds --under 5 --
the old formulae assume less power
air speed may be greater at 55%-maybe not
but time to altitude is not improved
The last 10 years has produced huge improvements in this
A friends full scale Pitts 12 has true1-1 power to weight - very unusual except for military fighter stuff .

[kcaldwel]

Although your model, or a 163, may be capable of vertical climb, you will find that arround 55 degrees you are climbing faster.

It will really depend on the thrust/speed curve, the L/D polar of the aircraft, weight, etc. There is absolutely no reason 55 degrees would be magic.

What if you have a very light, very high drag model, with a lot of power and a low pitch prop? There is no reason you couldn't build something that would have 95% of it's level flight speed in a vertical climb. What's the best angle then?

The Me163 accelerated in a vertical climb, as do many of today's jets. They don't slow down when you raise the nose past 55 degrees, they still accelerate. The Me163 accelerated from 200 kph to nearly 1,000 kph in a vertical climb to 6 km high. What is it's best climb rate angle?

There isn't anywhere near enough information to calculate the best climb rate angle for this hypothetical aircraft, but there is no reason it would be 55 degrees.

Kevin

[vintage1]

Only a handful of full size military jets are capable of vertical climbs, and then not with a full tank and weapons load.

It's not that much a useful feature.

[biber]

I'd say, for any given model, if you are free to chose a fitting power combo and have enough power to go reasonably vertical, you will end up with a big, slow turning prop and upright vertical as optimum climb angle.
It is because you will go the shortest way to your destination with the least mechanical work done on the model.
Any other (more flat) angle will require to go a longer trajectory and at a higher absolute value of aerodynamic drag,
both of which increases the energy required to go there.
So to get there at all, the most little energy is required with going vertical.
The goal to get there as fast as possible will give the same result, as the aerodynamic power consumption goes roughly cubic with the speed.
I. e. you will always try to go the shortest way, to get there the fastest way.

Of course all that does not account for any efficiency issues of the power combo, but you should always optimise that part anyway and it's likely to reach similar values of overall efficiency in any case.
(e.g. eta_motor * eta_prop = eta_combo will typically max out at approximately 0.8 * 0.8 = 0.64 if it is optimised for the particular application)

biber

[Tom Harper]

Acceleration in vertical flight just depends on the starting point. An Me163 could accelerate in a steep climb from take off. My LT-40 accelerates in climb from take off. No aircraft will accelerate if it enters vertical flight at max speed.

Vintage #18 lays out a good test case. The illustration of one extreme is a model that can hover at max power. The model is still capable of climbing at the same power but at a climb angle less than 90 degrees.

The other extreme is a model that can accelerate, to some limit, in vertical flight. As Caldwel's reference points out the difference between max climb rate and the vertical rate will decrease as thrust becomes large. With infinite thrust the difference will be zero.

So for any model with less than infinite thrust there is a climb angle of less than 90 degrees that yields the max climb rate.

Martin Simons provides the calculations to show that the optimum angle is 55 degrees.

[richard hanson]

- the ability to accelerate in absolute vertical climb, was not seen until recently-
On our own models we developed this first on small foam model weiging 12 ounces and powered by electricmotors and LiPo batts
Now we use A123 cells with extreme instant amperage capabilities,which make for "rocket" like vertical acceleration, if desired.
My large gas models with piped two strokes can not approch the electric setups -in power to weight but 2-1 thrust to weight is possible now - even at 1.5 to 1, - the performance is exciting as you can hear the engine gain speed from a vertical take off in 10 ft.
We were never into "calculating" , any of this as models are a hobby.instead , hands on experimentation showed me that much of the text book stuf is simply outdated - not false -just out dated.

[biber]

No, Tom, you certainly don't need infinite thrust for an optimum of 90°.

My thesis is, that it will be 90° whenever your power suffices to go vertical with a speed you can fairly control your model at.
(If your combo is optimised for the climb task, that is)

That just needs the thrust to be as much as the wheight is, for that certain minimum speed of control.
That speed might even be slower than stall speed for horizontal flight.

For all setups, that lack the power to maintain that minimum control speed in vertical climb, you should rather aim for the Cl of the lowest sinking rate and adjust the climb angle to that and the available power.

biber

[Tom Harper]

Sounds great!

With an obscene amount of power, and a correct CG, no math is needed.

[biber]

You still need some math to optimise the combo, anyway.

And the required power is related quite rigidly (more or less linear) to the desired climb rate, anyway, if the whole system is fairly well optimised.

biber

[richard hanson]

You still need some math to optimise the combo, anyway.

And the required power is related quite rigidly (more or less linear) to the desired climb rate, anyway, if the whole system is fairly well optimised.

biber
desired?
I thot you wanted the max!
How many of you guys fly performance models ?

[subtleguru]

No, Tom, you certainly don't need infinite thrust for an optimum of 90°.

My thesis is, that it will be 90° whenever your power suffices to go vertical with a speed you can fairly control your model at.

biber

Am I missing something here ?

Perhaps you need to be clearer on what vertical climbing means.

Pointing the nose directly vertical will result in a climb angle less than vertical due the horizontal "lift" generated during the climb. Thus you are travelling a greater distance, and at a slower speed, than a truely vertical climb in which the nose was pointed skyward at something just less the vertical.

[richard hanson]

Am I missing something here ?

Perhaps you need to be clearer on what vertical climbing means.

Pointing the nose directly vertical will result in a climb angle less than vertical due the horizontal "lift" generated during the climb. Thus you are travelling a greater distance, and at a slower speed, than a truely vertical climb in which the nose was pointed skyward at something just less the vertical.
what horizontal lift - if you use a zero/zero/zero setup trimmed for zero ou go straight up
Personally ,I don't use any thing except that on aerobatic stuff

[Tom Harper]

Richard,

If all you want to do is bore holes in the sky, you don't need math. That seemed to be the case in your post #35.

If you want to optimize performance you need math and a way to measure results. A data logger like Eagletree is a must.

[richard hanson]

Richard,

If all you want to do is bore holes in the sky, you don't need math. That seemed to be the case in your post #35.

If you want to optimize performance you need math and a way to measure results. A data logger like Eagletree is a must.
It's a hobby to me - everyone has their own thoughts on what is fun-
Dataloggers etc.,
I have the data stuf for my Spektrum radios - to optomize radio antenna etc..
My point was that the angle take off is NOT optimal if you design with max power for openers and know power /thrust parameters
I guess I have been at this too long ----
Obviously I am in the wrong group here - I am looking for real time data not text book exerpts

[Tom Harper]

"My point was that the angle take off is NOT optimal if you design with max power for openers and know power /thrust parameters "

Your statement does not make sense and is not related to your original question. I suspect we have been had by a troll.

[Brandano]

the pogo VTOL prototype could maintain control while accelerating vertically. Nevertheless its best climb angle was most probably not near 90 degrees. The same goes for the Harrier and pretty much any other "time to altitude" aircraft record. The F15 can climb vertically, but in order to make a record I think the pilot had to follow a precise path.

[mnowell129]

-
Obviously I am in the wrong group here - I am looking for real time data not text book exerpts
You may not be in the wrong group, but you may have missed the point of this thread. I started this thread to see if someone had some theoretical analysis showing that even if a model with enough power to climb vertically (T >> W) was this in fact the maximum climb rate?
I'm familiar with M. Simons analysis of 55°, but curiously this analysis is missing from the latest edition of his book??
Clearly with infinite thrust, vertical is the way to go since weight doesn't matter, but somewhere between 1:1 thrust to weight and infinite thrust to weight are some optimum climb angles.
Simon's analysis must assume 1:1 since the climb rate goes to 0 at 90 degrees, indicative of 1:1 thrust to weight.
So assume some thrust to weight that is in the range 2:1 or 3:1, say in a high performance sailplane climb, is vertical the fastest time to altitude or is it 85°, 55°. I'm still looking for some plausible mathematical argument.
I've run some real numbers based on the fact that there is
some Cdzero, that is some minimum drag while making no lift. The steady state vertical climb speed is then calculable based on available thrust. Taking a non vertical line and only a tiny increase in drag due to induced Cl yields a much faster speed up the climb angle, since the wing is carrying the weight (thrust to weight 3:1), and this projected back on vertical yields a higher climb rate than vertical. But I've no closed tidy form of this equation ala Simon's.
If you have an idea please, let's hear it.
My second question was about launch/climb angle for DLG's which turns out to be largely irrelevant (according to Drela, thanks to whomever posted that, and where did it come from by the way?), since the launch height is basically dictated by the launch velocity, which was my basic assumption anyway.

[richard hanson]

You may not be in the wrong group, but you may have missed the point of this thread. I started this thread to see if someone had some theoretical analysis showing that even if a model with enough power to climb vertically (T >> W) was this in fact the maximum climb rate?
I'm familiar with M. Simons analysis of 55°, but curiously this analysis is missing from the latest edition of his book??
Clearly with infinite thrust, vertical is the way to go since weight doesn't matter, but somewhere between 1:1 thrust to weight and infinite thrust to weight are some optimum climb angles.
Simon's analysis must assume 1:1 since the climb rate goes to 0 at 90 degrees, indicative of 1:1 thrust to weight.
So assume some thrust to weight that is in the range 2:1 or 3:1, say in a high performance sailplane climb, is vertical the fastest time to altitude or is it 85°, 55°. I'm still looking for some plausible mathematical argument.
I've run some real numbers based on the fact that there is
some Cdzero, that is some minimum drag while making no lift. The steady state vertical climb speed is then calculable based on available thrust. Taking a non vertical line and only a tiny increase in drag due to induced Cl yields a much faster speed up the climb angle, since the wing is carrying the weight (thrust to weight 3:1), and this projected back on vertical yields a higher climb rate than vertical. But I've no closed tidy form of this equation ala Simon's.
If you have an idea please, let's hear it.
My second question was about launch/climb angle for DLG's which turns out to be largely irrelevant (according to Drela, thanks to whomever posted that, and where did it come from by the way?), since the launch height is basically dictated by the launch velocity, which was my basic assumption anyway.
I never doubted that speed will likely increase as the angle is lowered
but distance to goal INCREASES --obviously.
This isn't the same as firing a bullet horizontally and dropping one at the same time (both hit the ground at the same time )
trying to calculate an unknown thrust with an unknown l/d -tho seems a bit iffy-
If the 3-1 thrust to weight model hits max velocity in very short time - say 3 second it has covered almost double the vertical distance as a 55 degree setup as initial acceleration won't vary much
and after that if speed is the same -
what is there to calculate?
Also IF the subject was a jet -(as a example) the RATE of climb would remain constant not taper off as a prop job may do as it reached max pitch speed.

[Tom Harper]

mnowell,

The point of Simmons analysis is that the drag resulting from lift at a climb angle of 55 deg is less than the weight vector as the angle is increased from 55 deg.

The case where velocity goes to zero is the worst case. As thrust increases the effect is less. But the optimum climb angle is still less than vertical.

Another way to handle excessive lift in a steep climb is a banked spiral. I am sure you are familiar with:

"Circular Airflow" Frank Zaic et al 1964

Tom

[richard hanson]

mnowell,

The point of Simmons analysis is that the drag resulting from lift at a climb angle of 55 deg is less than the weight vector as the angle is increased from 55 deg.

The case where velocity goes to zero is the worst case. As thrust increases the effect is less. But the optimum climb angle is still less than vertical.

Another way to handle excessive lift in a steep climb is a banked spiral. I am sure you are familiar with:

"Circular Airflow" Frank Zaic et al 1964

Tom
No - - as you may have guessed I spend little time perusing those studies .
I take the direct approach - DO IT-

[kcaldwel]

My second question was about launch/climb angle for DLG's which turns out to be largely irrelevant (according to Drela, thanks to whomever posted that, and where did it come from by the way?), since the launch height is basically dictated by the launch velocity, which was my basic assumption anyway.

Dr.Drela has posted his DLG launch analysis to the Hand Launch forum, and maybe the Yahoo group data file. He has done some amazing analysis of all aspects of DLGs. I just save his stuff, and posted it here. If you search the Hand Launch forum you can probably find the original posts.

For low thrust cases it is pretty simple in theory, but still requires a thrust versus speed plot, and an aircraft polar:

Rate of Climb (ft/min) = (T -D)*V/W

T = Thrust
D = Drag
V = speed

Basically, all excess thrust above the drag goes into rate of climb. It follows that maximum rate of climb then occurs near min sink speed.

For high power cases, the formula is much more complicated. I can type it in, but no one is going to have the required data for an RC model.

Glad someone clarified the 55 degree optimum is for a 1:1 thrust to weigh ratio, but I would think that will vary with the aircraft polar, and also makes the assumption that the thrust to weight is constant no mater what the speed is, which isn't true either.

Kevin

[mnowell129]

Dr.Drela has posted his DLG launch analysis to the Hand Launch forum, and maybe the Yahoo group data file. He has done some amazing analysis of all aspects of DLGs. I just save his stuff, and posted it here. If you search the Hand Launch forum you can probably find the original posts.

For low thrust cases it is pretty simple in theory, but still requires a thrust versus speed plot, and an aircraft polar:

Rate of Climb (ft/min) = (T -D)*V/W

T = Thrust
D = Drag
V = speed

Basically, all excess thrust above the drag goes into rate of climb. It follows that maximum rate of climb then occurs near min sink speed.

For high power cases, the formula is much more complicated. I can type it in, but no one is going to have the required data for an RC model.

Glad someone clarified the 55 degree optimum is for a 1:1 thrust to weigh ratio, but I would think that will vary with the aircraft polar, and also makes the assumption that the thrust to weight is constant no mater what the speed is, which isn't true either.

Kevin
Thanks, sure type it in, and a reference would be handy. Most of
my aero books are in storage.

[mnowell129]

I take the direct approach - DO IT-
Do what?
I want to know the best climb angle to get the highest rate of climb. Are you saying it's always vertical? If it's not vertical then at what angle?

[richard hanson]

Do what?
I want to know the best climb angle to get the highest rate of climb. Are you saying it's always vertical? If it's not vertical then at what angle?
I did not say that --

[JetPlaneFlyer]

Tom,
Regarding the notion that best climb angle is 55 deg... A lot of low powered models, and most full size light aircraft would simply not be capable of a 55 degree climb, so to claim this is optimum for all aircraft is clearly false.

Also regarding high power aircraft... simple trig tells us that an aircraft climbing at 55 degrees must have an airspeed af 122% of a vertically climbing aircraft in order to have the same rate of climb. Because drag is equal to velocity squared the drag of the 55 deg climb will be 150% of the vertical climber for the same climb rate, this is excluding the added induced drag of the 55 degree climbing model.
For fast flying models a 50% increase in drag could outweigh any advantage that wing lift gave, so a vertical climb, or a climb somewhere between 55 and vertical, could prove better.

There is no one best climb angle that 'fits all' aircraft.

Steve

[MarkusN]

There is no one best climb angle that 'fits all' aircraft.
Yep. The problem is not a trivial one, but it can be solved with the usual optimisation methods. Precision of the result heavily depends on accuracy of the model (the mathematical one) you are using. I am pretty sure, though (plain gut feeling) that for T/W > 2 the optimum angle quickly approaches 90°.

Things like starting conditions and propeller efficiency at low speeds (during the acceleration phase) will also heavily affect the result for a "real world" problem. I am pretty sure that the optimum trajectory for best climb starts at a shallow angle to accelerate, then turns vertical, or close to it.

[Tom Harper]

With such high thrust values, and the demonstrated ability to climb vertically, the optimum point is just a matter for discussion. It would be interesting the see data logs for different climb angles.

[richard hanson]

With such high thrust values, and the demonstrated ability to climb vertically, the optimum point is just a matter for discussion. It would be interesting the see data logs for different climb angles.
Yeh the data would be nice I guess--- but from a purely practical standpoint it is obvious that an increase in thrust will result in a higher climb angle.
I simply made different electric powered foam model and started changing power to weight (thrust ) my thrust stand (I designed it years back) quickly showed me watts consumed to thrust relationships .
This easily converted to model performance setups
and it worked . frankly my math skills are limited - I let the jr engineers do that stuff for me when I was doing machine design.

[kcaldwel]

For high angles of climb, L no longer can be assumed to equal W:

Rate of climb in ft/min =
(88*V/W)*[T-sigma*V^2/391-124.5*(W^2*cos^2gamma/sigma*e*b^2*V^2)-W*a/g]

This must be solved by successive iteration after assuming an approximate acceleration. If we assume it is high angle unaccelerated climb, then the last term just reduces to zero.

V = velocity
W = weight
T = Thrust
a = acceleration
g = acceleration of gravity = 32.1740 ft/sec^2

sigma = ro/rozero = density ratio air density/ air density sea level: assume =1 for a short climb at sea level.

Assuming the aircraft has a parabolic performance curve:
e = span efficiency
f = off "speed solution chart" which is aircraft speciic, or can be derived from:

T at L/D max = 1.132* (W/b)*(f/e)^1/2
b = wing span
gamma = climb angle

So, I need to think about which of these goes constant for short, unaccelerated climbs...

Kevin

[mnowell129]

Thanks, got a reference?

[kcaldwel]

Edit: My simplification was all wrong so I removed it.

The above correct equation is from Perkins and Hage, "Aircraft Performance Stability and Control", 1949.

[eflightray]

Now math and me parted company when I left school, Ok, not quite true but I'm a try it and see not a calculate and see.

Now this question is sort of related to the 'best rate of climb' part of the question. - What prop?
Is there any data as to what type of prop would give the best rate of climb?

For horizontal speed a smaller diameter coarse pitch prop would be recommended.
For vertical climb based on thrust, then a larger diameter prop of ? pitch would be recommended.

So for a fixed power input, (obviously you could just keep increasing power theoretically to go faster until.....?), is it possible to 'calculate' the right prop for maximum climb speed? (at whatever angle).

[biber]

Like wingspan on gliders, for max efficiency, you want the biggest possible prop in diameter most of the time
which often limited by quite practical points(hight of landing gear/ground clearance or length of fuselage in front of the wing with swinging blade props).
Javaprop is a very useful tool helping to design and analyse props (find it on www.mh-aerotools.de).

biber

[richard hanson]

well son -if'n it's a jet --- thrust is constant (sorta)
If'n it's a prop job , thrust available diminishes as speed increases .
On my model at 3 to 1 power to weight - acceleration is -- ? but rate of acceleration will diminish as prop pitch speed max is reached--if pitch is a constant
--I can start at zero airspeed in a hover and accelerate out
no one says what the vertical distance (altitude to be reached) - would be ----
I see lots of numbers & formula- but what is the altitude goal?
This here question is akin to "how long is a piece of string."

it is fer sere that in today's power to weight world - a 55 degree setup -is for low power and a heavy airframe

[Tom Harper]

That's arguable.

[mnowell129]

I see lots of numbers & formula- but what is the altitude goal?

The goal is not the accelerating case.
I want the estimate of the angle for best rate of climb, in the steady state, non accelerating case.
Repeat : I don't want the accelerating case, I want the steady state case.
To repeat. Once the vertical speed is established, not accelerating and constant is it still better to maintain 90° or is some slight angle actually faster rate of climb?

[kcaldwel]

Let's try this again!

W * sin gamma = T - D - W*a/g

If we reach a steady state climb at the given angle, a goes to zero, and the last term drops out. So:

sin gamma = (T-D)/W

If thrust is assumed constant, I believe that means you want to climb at the angle where V = best L/D speed where induced drag = parasitic, and you are at the drag minimum. This is complicated by the fact L no longer equals weight because of the steep climb angle.

If the plane can climb vertically at over best L/D speed, then vertical is going to definitely be the best angle.

If the vertical climb speed is below best L/D speed, someone with better math skills than me should be able to differentiate the big equation above and find the minimum, assuming a parabolic drag equation, and thrust = a constant for simplification...

K.

[biber]

Better speak in Cl of best L/D rather, than in speed of best L/D. ;)

biber

[kcaldwel]

Nope, not Cl of best L/D, since Cl should be zero in a vertical climb.

I meant that you should definitely climb vertically if the vertical speed is higher than the aircraft level flight best L/D speed.

In fact, I wouldn't be surprised if climbing vertically was optimum down to quite a bit below best L/D speed.

Kevin

[biber]

Strictly speaking the speed of best L/D is a function of the lift which is again a function of climb angle.
Not so the Cl of best L/D, it's constant for a given model (except for reynoldsnumber effects).

But if you go vertical you have (per definitionem of lift) no lift at all and thus L/D in this case is zero, anyway.
So what relevance can the terminus 'speed of best L/D' have in this case?
Is it applicable at all?

Edit: Ok, i think I understand your point now.
But I would still use the Cl instead of the speed.
Once your engine has as enough power to force you to fly a Cl below that one of the best L/D, you certainly have to maintain vertical for optimum climb rate.

biber

[richard hanson]

Nope, not Cl of best L/D, since Cl should be zero in a vertical climb.

I meant that you should definitely climb vertically if the vertical speed is higher than the aircraft best L/D speed, as I said.

In fact, I wouldn't be surprised if climbing vertically was optimum down to quite a bit below best L/D speed.

Kevin
darn -I don't see how anone can establish best rate of climb with just a "formula .
Having done this -- it always was a matter of watching an altimeter --then playing with power and trim. but these were pokey little puppies with a hundred or so HP.
IF you had telemetry - you could do it but calculate it? NFW
close ? yeh- maybe but that's it.
so unless I am wrong - the higher the power (all else considered equal) - the steeper the angle and back to my current toys -vertical of course.-

[kcaldwel]

Richard,

I assure you they have been calculating best rates of climb since the 30s at least. They generally have a lot more test information of course, but have to deal with way more altitude and load variations.

Given today's fuel prices, the airlines will be flying very carefully calculated minimum fuel trajectories, which come from the same equations.

25 years ago I could have figured out this angle of climb thing easily, but I'm pretty rusty.

Kevin

[kcaldwel]

Edit: Ok, i think I understand your point now.
But I would still use the Cl instead of the speed.
Once your engine has as enough power to force you to fly a Cl below that one of the best L/D, you certainly have to maintain vertical for optimum climb rate.

biber

It gets more complicated because lift does not equal weight in a steep climb, so the induced drag drops. Minimum drag is when induced drag equals parasitic, and as the induced drag goes down with the lower lift needed at a steep climb angle, the minimum drag speed will go down as well...

Hmm, the key is there somewhere, but I'm not seeing it.

K.

[biber]

Strictly speaking the speed of best L/D is a function of the lift which is again a function of climb angle.


;)

biber

[richard hanson]

Richard,

I assure you they have been calculating best rates of climb since the 30s at least. They generally have a lot more test information of course, but have to deal with way more altitude and load variations.

Given today's fuel prices, the airlines will be flying very carefully calculated minimum fuel trajectories, which come from the same equations.

25 years ago I could have figured out this angle of climb thing easily, but I'm pretty rusty.

Kevin
yes -but not using a model with only observed data -the formulae for high reynolds numbers are not very good for these little things
remember the data the airliners will use was derived from actual use and those corrections added to the original formulae.
Back in th '30s the expurts were still trying to figure out if a painted B17 was faster /used less fuel than a polished one .

[kcaldwel]

For high angles of climb, L no longer can be assumed to equal W:

Kevin

Yep, I said that before I typed the original equation, which takes that into account.

I'm just not seeing the simple solution that is there, with of course assumptions of constant thrust at all speeds, and a parabolic performance curve.

Kevin

[kcaldwel]

Ah. I knew somebody smarter than me must have figured this out before! I don't quite understand his units to be honest but he has nice graphs:

http://www.rcgroups.com/forums/showpost.php?p=7123347&postcount=31

Seems to confirm my intuition that you fly best L/D speed, and just keep pointing the nose up to the maximum angle that it can maintain that speed, or is vertical, whichever comes first.

And at the bottom of this page it has plots, if you understand German or use Babelfish:

http://www.czepa.at/geschwindig.html

Kevin

[richard hanson]

Ah. I knew somebody smarter than me must have figured this out before! I don't quite understand his units to be honest but he has nice graphs:

http://www.rcgroups.com/forums/showpost.php?p=7123347&postcount=31

Seems to confirm my intuition that you fly best L/D speed, and just keep pointing the nose up to the maximum angle that it can maintain that speed, or is vertical, whichever comes first.

And at the bottom of this page it has plots, if you understand German or use Babelfish:

http://www.czepa.at/geschwindig.html

Kevin
If you take max speed ain level flight --then point the nose up and can continue to maintain AND/OR exceed that speed -you end up -vertical (90) as best angle and THAT is possible with some of the models we fly--- these are little foamies with built in drag to hold level flight speed down -such that maneuvering is at a CONSTANT speed no matter what the maneuver -for example a vertical snap can be exited and almost instantly regain selected maneuvering speed .
-----------Just an example ------------------
All of this type power to weight was non existant 10-20 -30 etc years ago.
The formula shown is an example of establishing max climb as any old pilot will tell you

[Tom Harper]

Richard,

'Just do it' is a viable approach. That's how I design my AP models.

But when you ask a question about 'what's best?' you enter a new realm. To know 'what's best' there are two paths - theory or measurement. You can't eyeball it.

The power control issues on Free Flight are far greater than just hovering with an electric foamy. The power to weight available 30 years ago is similar to what you are dealing with. VTO was a common technique in Free Flight competition - unassisted and they accelerated vertically.

Yep - there is some neat stuff at the hobby store today. But, don't knock the pioneers.

[richard hanson]

Richard,

'Just do it' is a viable approach. That's how I design my AP models.

But when you ask a question about 'what's best?' you enter a new realm. To know 'what's best' there are two paths - theory or measurement. You can't eyeball it.

The power control issues on Free Flight are far greater than just hovering with an electric foamy. The power to weight available 30 years ago is similar to what you are dealing with. VTO was a common technique in Free Flight competition - unassisted and they accelerated vertically.

Yep - there is some neat stuff at the hobby store today. But, don't knock the pioneers.
I was flying free flight in 1950-- I ain't no teeny bopper . Iam very familiar with the basics -I used a Wasp .049/ K&B.049-what engines were you using?
Theory?-- well it is still just theory till it is applied and proven- I had the patents to prove THAT point.

[Tom Harper]

In 1950 I was flying an Anderson .049 and a KB .02. Had a McCoy .60 and some British diesels, but I got busy with school and then the USAF.

Tom

[Cory]

I've seen a graphic of the Streak Eagle time to climb record from back in the 70's. I wish I could find it now. It's definitely not straight up. Here's a description of the climb profile:

The profile for the 30000 meter run was something like this.
Gear up and rotate at first indication of airspeed (about 70 kts). Accel to about 420kts. Rotate verticaly into and Immelamnn and hold about 2.6 Gs. Arrive level and upside down at 32000 FT and Mach 1.1. Roll 180 to right side up and accel to 600 kts. and then up to Mach 2.25. Pull 4 Gs to 55 degrees. Hold 4 degrees AOA untill mission control tells you to recover (after passing 98000 FT). Engines would be expected to quit at about 80000 FT. Passing through 55000 FT on the way down look for 4 green lights to indicate boost pump is on and attempt engine re-start once above 350 kts.

[kcaldwel]

Interesting!

That is a very different case though. It is time to climb, which is from a dead stop on a runway, it has huge altitude changes and the accompanying aerodynamic and engine performance changes, supersonic flight, and acceleration into a ballistic profile for engine out coasting to max altitude.

Somehow I don't think this is translatable to most RC models!

Kevin

[Cory]

I think the main difference is that the altitude change experienced by RC models does not involve a noticeable difference in air density. Although the previously posted account doesn't mention it, I remember there was a slight descent in the thinner air at the top of the Immelman in order to build airspeed as quickly as possible for the final climb.

[kcaldwel]

The Streak Eagle records were quickly broken by the Soviets with an Su-27. All these aircraft have a maximum thrust to weight ratio (stripped) of just over 2:1.

And there is considerably more than just the altitude differences:

"Zoom Climbs
An aircraft can operate briefly outside its level-flight envelope. This can be done either by diving (so that, as in a glide, a component of weight acts opposite the drag) to reach airspeeds above its maximum level-flight speed, or by performing a zoom climb. A zoom climb occurs when an aircraft climbs so as to convert airspeed into altitude. If an aircraft is flown to the edge of its operating envelope (so that T = D ) and then forced to climb, it will move along a constant energy height line on the Ps diagram. It will decelerate as it climbs, but (at least initially, because T = D ) its total energy will remain constant. As the aircraft slows down, it may deviate from the He = constant line as drag changes and no longer equals thrust. If the aircraft whose Ps diagram is shown in Figure 5.32 where flown to its absolute ceiling, h = 57,000 ft and V = 630 knots, this corresponds to an energy height of 74,000 ft. If it entered a zoom climb from this condition, and thrust remained equal to drag, it would move along the He = 74,000 ft line decelerating until it reached zero velocity at an altitude of 74,000 ft.

Minimum Time to Climb
The Ps diagram can be used to determine a strategy for climbing to a given altitude in absolute minimum time, as when the F-15A Streak Eagle set minimum-time-to-climb records in 1975. The maximum rate of climb at any given altitude is achieved at the speed where Ps is maximum. However, because the aircraft can be zoomed at the end of its climb to get to a particular altitude faster, the minimum time to climb is achieved by changing energy height, not just height, as fast as possible. The aircraft increases energy height the fastest when it moves perpendicular to He = constant lines at the point where Ps is maximum on each line. A trajectory is shown on Figure 5.32 which satisfies this requirement to cross each energy height line where Ps is maximum. At one point along the trajectory the aircraft descends and accelerates following an He = constant line, then continues on the climb profile. The constant-energy-height descent/acceleration moves the aircraft quickly through the transonic regime to an altitude and supersonic speed where Ps is maximum along the higher He = constant lines."

Ref: INTRODUCTION TO AERONAUTICS: A DESIGN PERSPECTIVE

CHAPTER 5: PERFORMANCE AND CONSTRAINT ANALYSIS


Kevin

[Jurgen Heilig]

...
The Me163 accelerated in a vertical climb, as do many of today's jets. They don't slow down when you raise the nose past 55 degrees, they still accelerate. The Me163 accelerated from 200 kph to nearly 1,000 kph in a vertical climb to 6 km high. What is it's best climb rate angle?
...

Vertical climb? Were did that information come from?

From a Me163 pilot:

“The engines were completely throttleable. Acceleration at first seemed surprisingly slow. The aircraft was poised on small wheel blocks, and when you went from ground idle to full power, the dolly wheels would roll over the blocks, and it took a while to build up flight speed. Once airborne, you accelerated to speed for best climb."

“We’d take off and try to keep the plane low, say 15 feet up, and then start to trim it for speed and then very gradually climb as we accelerated to the best climbing speed, which was about 420 miles per hour. The airplane was very short-coupled, so you didn’t want to over-control it in pitch as you took off. When we reached the best climbing speed, we’d pull back and climb at approximately 70 degrees."

:) Jürgen

[mnowell129]

Kevin, thanks for the f-15 climb info.
It took a couple readings to get it, but it all makes sense now.
I never knew why the dive was there, but getting through transonic
quickly makes sense.

This and the soviet experiment both seem to indicate that something other than vertical works even at T:W ~= 2:1.
I've ordered an altitude tracker (ram3), may have to perform some real experiments with high powered models.

[Jurgen Heilig]

More info about the Me163:

http://www.historyofwar.org/articles/weapons_me_163.html

With an empty weight > max. thrust the Me163 for sure did not have true vertical performance.

:) Jürgen

[Tom Harper]

Jurgen, thanks for the link - interesting stoff!

Tom

[nmasters]

The maximum takeoff weight of the 163B was 3950kg and it landed at 1900kg. The Walter HWK 109-509 motor developed 1500kg of thrust so there couldn't have been any sustained vertical flight much less acceleration. Why did this thread drift off into reaction motor driven aircraft anyway? The performance curves of reaction motors don't compare at all, you guys know that. Rate of climb of propellor driven aircraft is solely Dependant on excess thrust horsepower. I. e. it takes so much power to overcome the drag and any HP beyond that is available to lift the airplane. Since 1hp is 550 ft lbs/second if your plane is 55 lbs and there's 1hp still available at cruising speed then the plane can climb at 550/55=10 ft/sec just by throttling up without changing trim. If you want the fastest climb possible trim the plane for its best glide and gun the engine. If you want the steepest climb trim for minimum sink. Don't worry about angles or speed the plane will find its own equilibrium.

--Norm

[Jurgen Heilig]

...
Since 1hp is 550 ft lbs/second if your plane is 55 lbs and there's 1hp still available at cruising speed then the plane can climb at 550/55=10 ft/sec just by throttling up without changing trim. If you want the fastest climb possible trim the plane for its best glide and gun the engine. If you want the steepest climb trim for minimum sink.
...

Your calculation is based on a 100% effcient prop and an airplane without any drag. ;)

If you have a thrust to weight (&drag) ratio >> 1, the other two recommendations also are no longer valid.

For fastest climbs - just watch the F5B pilots: Accelerate to max. climb speed and pull into an almost vertical climb (minimum drag climbout).

:) Jürgen

[nmasters]

Your calculation is based on a 100% effcient prop and an airplane without any drag. ;)

No it isn't. I said "thrust horsepower". THP=BHP x prop efficiency. And the THP required curve definitely shows drag which is dependent on both AoA and speed

[kcaldwel]

I. e. it takes so much power to overcome the drag and any HP beyond that is available to lift the airplane. Since 1hp is 550 ft lbs/second if your plane is 55 lbs and there's 1hp still available at cruising speed then the plane can climb at 550/55=10 ft/sec just by throttling up without changing trim. If you want the fastest climb possible trim the plane for its best glide and gun the engine. If you want the steepest climb trim for minimum sink. Don't worry about angles or speed the plane will find its own equilibrium.

--Norm

That clearly doesn't work. If the aircraft has a low static margin, it will just accelerate with little change in attitude. If it has a high static margin, it will loop.

With excess power, you will need to actively fly an angle or airspeed.

I hadn't looked at the details of the Me163 for a long time, sorry. I just remember it going almost vertical in the movies at the Canadian Aerospace museum in Ottawa. The Me163 does a zoom climb, exchanging speed for altitude, as in the F-15 example.

Kevin

[nmasters]

That clearly doesn't work.
Climbing on the engine doesn't work either. Remember propelor thrust drops off as airspeed increases. At max level speed the prop is making just enough thrust to counter the drag of the airplane which is only very slightly higher that the zero lift drag (for highly cambered airfoils minimum drag is actually at some positive lift). And don't forget that the pitching moment doesn't disapear when lift is zero. The wing (even a crappy foamy wing) is a more efficient lifting surface than a propelor. It just doesn't make sense to expect a plane with a fixed pitch propelor to have it's best sustained rate of climb vertical no matter how powerful the engine. Jets and rockets with
thrust > 1.5 x weight may be a different story as are variable pitch props.

If the aircraft has a low static margin, it will just accelerate with little change in attitude.Higher speed=higher CL plane gos up


If it has a high static margin, it will loop.
Power duration free flight models solve that problem by turning or using a mechanism that adjusts elevator trim, andor wing camber, for the power. Of cource a spiral climb just wasts the excess power that would have made the plane loop in a strait ahead climb and an auto trim system requires a sensor on the plane. If people insist on putting more power in an airframe than it can handle then you're absolutely right. You can't trim it for a maximum performance climb because the plane can't keep up with the engine

--Norm

[Jurgen Heilig]

No it isn't. I said "thrust horsepower". THP=BHP x prop efficiency. And the THP required curve definitely shows drag which is dependent on both AoA and speed

Sorry, but you did not use "thrust horsepower" but "mechanical horsepower" in your calculation:
Since 1hp is 550 ft lbs/second if your plane is 55 lbs and there's 1hp still available at cruising speed then the plane can climb at 550/55=10 ft/sec just by throttling up without changing trim.

:) Jürgen

[richard hanson]

nice charts and they make sense
The question as i stands tho -is "best ROC"--but for how long?
from a practical standpoint, there has to be a goal somewhere
My 23 lb EDGE had a static thrust of way over 55 pounds - really good power to weight --it accelerated - vertically - absolutely vertically - rolling as it went to clarify attitude so vertical position easily demonstrated .
of course it changed (slowed ROC as pitch speed caught up) - but in a drag race to usable altitude - it was VERY hard to beat .
My foamies have higher power to weight - but obviously a lower practical altitude
but in a short drag race -it takes a rocket to beat one of these things . the original question is still a "how long is a piece of string " question. for an infinite climb (?) I have no idea what angle is best.

[kcaldwel]

Higher speed=higher CL plane gos up

Power duration free flight models solve that problem by turning or using a mechanism that adjusts elevator trim, andor wing camber, for the power.
--Norm

My point was, just trimming an airplane for best L/D speed in level flight, and adding power does not result in the best climb angle. Higher speed does not = higher Cl for a neutrally trimmed aerobatics model for example. Higher power just = more speed, without adding up elevator.

Yes, free flight comp models are trimmed for a climb under power, hopefully near the best rate for that model.

The graphs I linked to previously look to be right to me, for climb angle versus power. Of course propeller driven aircraft have a very narrow speed range where the engine/propeller combination efficiently makes thrust.

For maximum climb rate, looks like you should prop the engine for best L/D speed. You raise the nose of the aircraft to the angle to maintain near best L/D speed. If the aircraft ends up climbing vertically at faster than best L/D speed, then the engine should be re-propped for a higher speed. Iterate. That will give the best climb rate for that airplane power combination.

There is a continuous curve of angles for best climb rate that depends on the thrust to weight ratio. The angle gets steeper with increasing power to weight, and there is no magic angle.

The actual numerical solution isn't that complicated (see previous formula), but depends on too many factors that aren't available for most RC models.

Kevin

[Jurgen Heilig]

Again: Watch how the F5B guys climb to altitude.

As they get penalized for motor-on time, you bet that they want to climb as fast as possible. ;)

http://www.youtube.com/watch?v=wP9srodkvEM&NR=1

:) Jürgen

[mnowell129]

Again: Watch how the F5B guys climb to altitude.

As they get penalized for motor-on time, you bet that they want to climb as fast as possible. ;)

:) Jürgen
This doesn't necessarily prove anything. Lots of guys do things in competition just because everyone else is doing it. There is no proof in this video that he couldn't have climbed higher at some other angle. If this showed two back to back runs with the same plane and a recording altimeter at two different angles then it might show something.

The question still is for one of these models what is the best angle for rate of climb with the power available. For pure vertical thrust - weight - drag at Cd0 gives you a vertical velocity. It still is has not been proven to me that there is not some other climb angle near Cd0 where the lift contributes to the climb so the combination of lift + thrust - weight - drag produces a higher net climb rate projected across the climb angle.

[mnowell129]

nice charts and they make sense
The question as i stands tho -is "best ROC"--but for how long?
from a practical standpoint, there has to be a goal somewhere

2000 feet.

[biber]

With the questioned value being connected very closely to the drives characteristics (thrust as a function of airspeed)
you can either take a particular example of which you know the characteristics/function,
or you can just assume an optimised drive and calcuiate with a prospected optimum eta of .65 for the motor prop combo.
But either way you have to decide for one of both scenarios and then go from there.

biber