Voll gas oder?

Mikko

User
Hello !
I've a Volcano Valenta mit Polytec 600-24 6,7:1 and 4SLipo.

I've not yet telemetrie for spannung. ( vario and speed from Jeti)
A question : make sense to have two motor position, for example 50-60% and voll gas?

I fly ONLY on alpine slopes where I live, so motor is used to come back in spring/summer time and to fly like in the ebene in winter time.

I'm not able to find a formula to understand if 60% motor ( enough to gain quote) it's also spare in mhA or not due to motor optimisirung make no sense.
Thanks. Write also in deutsch eventually I'll use a translator.

Ciao from Lago Maggiore
 
Some people use two motor settings for launching (if full throttle is too much power to hold the glider or if it is even dangerous)

I do not understand your question. Are you asking if half throttle is sufficient for maintaining or gaining height? What propeller do you use? What is the current draw at full and half throttle?

300W should be sufficient for slow climbing (Volcano with 3.2kg)

Stefan
 

GC

User
Why don´t you use a switch for

- butterfy or

- continues gas?
Why only full gas or no gas?
 

Wombat

User
Hello,

for my Surprise 15 F5F-Model with 4S and about 3000 W Power I have programmed an three-position-switch with 0-50-100 % throttle.

I use the 50% position only for about 1 second for the first climb to reduce risk of initial hand-launch.

In all other climbs I push the switch directly from the 0% to the 100% position.

To reduce the peak-current in the moment of motor-start I have programmed a "ramp" for the throttle-function of about 0,7 seconds to get am smother run-up. Of course the "ramp" is active at the first climb but this is no problem at all.

If there a questions regarding the programming of the "ramp" I can only give advisements for the Futaba FX32 Transmitter I use.

Guido
 

Wombat

User
No! No long discussions! The weak point of nearly any really high powered system with more than 150 A Current is the lower-partial-load and because of this it's not recommandable to use a proportinal-function to control power. You have to select a switch which takes sure, that the throttle is set immedeatly to 100% and not underneath of this position!

Every Modeler who fly eletrical-powered systems with low-power can of course use a proportional function for throttle...
 

Mikko

User
My Volcano is 4,3 Kilos. Lipo 4S 3000 mhA., propeller 18x9 Aeronaut. With 5 sec. motor at 50% I launch when wind it's not enough and it's risky.
After I look for thermic, for sure.
When I'm deep in the valley - for example - I use 100% motor to come back home and I've measured with Jeti telemetrie in 9 sec. I gain 100 meter.
More or less 14-15 sec. use of motor in both position -as told before - I drain 279 mhA ( last measure this week).

The question is :
- ok for launch some sec. at 50% for safety reasons.
- when flying in the valley make sense to gain altitude using 50% gas to save mhA??

OR using 50% the motor runs not optimized and so no save or probably more use of power than using 100% ?

In other words : if I gain 100 meters in 9 sec or if I gain 100 meters in 14 sec. there is a difference in the drain ?

I know that with telemetry it's simple ( and cheap) but for the moment I've not.

I prefer to read your interesting answers to take advantage of your experiences ( is not this the sense of a forum?)

Miko
 
There is no big difference in my opinion.

The exact calculation is difficult. I recommend an experiment. Compare the total discharge of a flight with same number of climbs at 100% and 50%.
 

Mikko

User
Thanks !

I will order a MUI from Jeti ( MUI150?) and I'll make test.
Probably there are so many factors that there is no a "law" to judge if at 100% I use 55 Ah at 50% the drain is ....... 27,5Ah? 35Ah?

But when you use ( I try with some difficult for the moment) a E calculator you note important differences in the % optimisirung of the motor system.

And I suppose that calculation is always done at 100% so if you use proportional set probably you "waste" energy with no sense.

I've to do experiment on this because the konzept it's not clear for me.
 
There is definitely no (easy to apply) law. 50% throttle may mean anything in terms of power.
The motor may have higher efficiency at lower power levels. The ESC typically has lower efficiency at partial loading.
The glider has higher drag at higher speed. I don't believe in calculation here. The experiment is much easier.

Watch the temperature of the ESC at 50%.

You do not need a current sensor. You can check the accumulated current draw during charging.
 

Mikko

User
Looking for the next motor for next 5 mt Discus I've seen on the Hacker' e Calc an answer to our argument "Voll gas oder?"

It's clear that in case you use only 50% of throttle power is less than 50% and naturally flight time is longer........ok any case must be compared
the clim rates but reading this part of tabelle seems to be pretty good to have also a reduced position of throttle ( in my case via a switch). Schermata 2015-08-19 alle 18.49.13.jpg
 
Hello Miko,

if you look for best climb efficiency (highest altitude gain from battery energy),
then the steepest climb rate is probably best, because the glider will climb along the shortest path.

Therefore full throttle (vollgas) is probably more efficient than half and recommendable anyway in order to
keep your ESC alive.

However, it is very important to operate the prop at it's optimum point.
The optimum point is usually at a flight speed of 80% of prop pitch speed.
Variating the climb angle such that this speed is effectively reached during climbing will greatly improve prop efficiency.

You can calculate an overall efficiency of your drive system as follows:
Eff = (weight in N * altitude gain in m) / used battery energy in Ws

An overall efficiency of 45% is acceptable, 50% is pretty good, 55% is near maximum.
Sources of loss are prop first (<80% eff), motor second (~80%), glider third (glide rate 20 means 95% eff),
battery system fourth and esc and electrical connectors fifth.

Therefore, the choice of the right prop that fits both your motor regarding rpm and the plane regarding weight/thrust is more important than electrical efficiency (however the second should not be neglected because of heat!). Putting it in other words: it is better to choose the prop first according to plane weight and desired climb rate and then choose the motor according to required rpm and power.

With best regards,

Georg
 
Hello Miko,

calculating overall efficiency from yor data results in:
Energy from altitude gain: 43N * 100 m = 4300 Ws.
Energy invested: 0.28Ah * 15V * 3600s/h = 15120 Ws.
Overall eff = 28%.
While electrical efficiency of your drive is good, it looks like things could be improved
via using more power and a higher climb rate - half throttle is not an option.

Of course, I forgot to mention : glider efficiency of 95% mentioned is only valid with 90° climb angle -
if you climb at less than 30° then the glider drag is the highest source of climbing inefficiency.
Half throttle means longer climbing path and higher drag losses then.

Hotliner pilots know that more power in the system can actually save energy ;)

Wtih best regards,

Georg
 

Mikko

User
Dear Georg

let me study your lesson !

First I understand that my method to chose a motor is completely wrong. I use weight and W to see if I'm in the range of 100-120 w/kg and then I consider as perfect the propeller that the motor producer indicates.

Second it's clear that some test regarding climbing rate must be done in relation with consumption : another parameter it's the temperature of the ESC at "middle throttle"....

Third : I am supposing that lot of people is using in average the motor set in pretty bad conditions.

And let me say that "re-seller" propose you motor that normally is in stock.... and not the technical best for your case ( strange i'nt it? )

I repeat I want to study your messages to come out with a solution for my next project ( 5 mt 8 kg Discus, where I would like to use the Topmodel system : only the motor axe come out from fuselage, no spinner at all).

Thanks, again this forum is great and populated by really experts !! I enjoy because at my age ( old) I've lot of thing to learn and this keep my mind open and young !
 

Mikko

User
Another point : using a mixer - like me - throttle to down elevator to avoid to reach too steep climb it's no sense.

But I judge it's better to keep a constant climbing rate not too steep to be able to switch off motor and find safely and immediately a stable flight position.

In other words probably I use too much power and I spend power also with down elevator.
But the option is to set full throttle at 80% and delete the mixer ..........or change propeller? instead of 18x9 a 17x9 ?
 
Hello Miko,

some more hints.

The short version is:
What you actually need to be more efficient is probably just more power and maybe another prop.

Or even shorter: Voll Gas = Yes!!


The long version is as follows:
( I am using a lot of rules of thumb here because there are many variables in the equation.)

Let me first elaborate a bit more on climbing and role of the climb angle with efficiency.

Let's look at glider efficiency in two extreme situations and then in a third more normal situation:
1) level flight (no climbing)
2) vertical climbing
3) 30° climbing

1)
At level flight, you need to compensate the glider drag with the propeller,
but climb efficiency is zero, as all the power is lost with compensating and the climb path is therefore infinitely long.
The power needed to compensate glider drag can be calculated from the glider weight and the glide ratio.
Let's e.g. assume a weight of 8kg, a ratio of 20 (20 should be pretty good for a model glider), a glide rate of 10m/s and a sink rate of 0,5 m/s.
Ratio of 20 means that the drag forces are 1/20 of the weight forces.
That would be 4N of drag at 10m/s, requiring 40W of power.
You could also calculate the power needed to lift 8kg at 0,5m/s - it is the same 40W.
With an optimum efficiency of the electrical drive system including motor and prop of ~55% (will write more on this later), that would be 73W of input power.

2)
At vertical climb with 10m/s, the 4N of drag add on top of the 80N of weight and therefore climb efficiency is 95%.
However you will need much more power - with the same calculation as above, it is 1527W input power.

3)
At a climb angle of 30°, you need 2m of climb path to reach 1m of altitude.
Therefore, the efficiency will be 0,95*0,95 which is roughly 0,9.
Smaller climb angles will result in much longer climb paths and therefore lower efficiency.
However, you also need only half of the force to overcome the weight.
Therefore, you would need 480W of effective power or 872W of input power for a 30° climb of 5m/s.


Now let's look at the magic 55% of drive and prop efficiency - where are they coming from (this is all rules of thumb):
- Battery efficiency 95%
- Losses in cables, plugs and ESC 5% = 95% Efficiency
- Motor/Gearbox efficiency 80% at operating point
- Propeller efficiency 75%

Then let's have a look a the kind of prop you need to get 4800g of thrust at 10m/s at an efficiency of 75%.

This is the actual problem, because you would need a very large prop spinning at a comparatively low speed.
Let's assume a 23x12 operating at 1000W mechanical power (~1300W of input power at the ESC) - it will deliver 8kg of static thrust with a pitch speed of 90km/h which is equivalent to 25m/s. At a speed of half of the pitch speed - 12,5m/s - it will deliver roughly half of the thrust at an efficiency of roundabout 70%. At 10m/s, the thrust of 4800g is probably right, but the efficiency drops to 65%.

Alternatively, you can go for more power and a higher climbing speed.
E.g. at a mechanical power of 1500W (~2000W of input power at the ESC),
the pitch speed of a 23x12 is 103 km/h - equivalent to 28m/s.
The static thrust is 10,6kg. 4800g of thrust are probably reached at 16m/s with an efficiency of roundabout 75%.
Therefore, at higher power and climb speed, the efficiency goal can be reached with this prop.

What do we learn from this ?
- Go for the largest prop possible
- More power means more efficiency

Look at hotliners - they have to produce maximum (and fast) climbing from limited energy:
- Prop is as large as possible, reaching from the nose to the wing.
- Prop is "quadratic" (e.g. 17x18) which increases efficiency.
- Power is > 1kW / 1kg.

Of course this cannot be realized for a normal glider.
However, the same principle applies.

Therefore I am used to "overpowering" of my gliders for better efficiency -
using a drive train that theoretically would allow vertical climbing for something like a 30° - 60° climb angle.
The result is quite a few more climbs from the same battery (which of course must allow for a high C rate).
The efficiency is much dependent from the actual climb angle chosen -
therefore it probably pays off making measurements at different angles.
As a substitute for measurements, I use to watch my glider when climbing.
The glider should not climb at maximum angle and comparatively low speed
but rather give some "dynamic" feeling when climbing.


With best regards,

Georg
 
One addition:
It is very easy to find the most efficient climb angle if you have a vario:
Climbing is most efficient if the climb rate is highest.

Best regards,

Georg
 

Mikko

User
George
I am not a Mathe or fisic like you ! I've tried to remake tour calculation with difficulties.
But if I ve under stand something tre answer to the question voll gas oder?
It s here
Level flight 73W climb = drag /eff climb 0 meter

Vertical flight 153 w Every meter climb 1 meter

30'. Flight 87 w Every meter climb 0,5 meter
To climb 1 meter needs 174w

So at 30' flight I need 13% more power to climb 1 meter!!

Voll gas oder?
50% power to launch and tiene ALLWAYS 100% gas to avoid to loose power
And: When chosing a motor better to exceed power 150-200 w per kilo......if possibile
(I m writing from a mobile device from a kraotian island....)
 
Hello Miko,

just change W to Ws (Wattseconds) and you got it right.

However, this was assuming that your prop efficiency is 75% in all cases.

With a 30° angle you have a real chance to achieve this efficiency,
but you rather need 200W/kg to get there.

With vertical climbing you would need an unrealistic
400W/kg to achieve the same efficiency with the same prop,
but this would be more energy efficient.

With 100W/kg you are more likely to climb at 15° and loose more energy along the way...
so your conclusion is also right.

With best regards,

Georg
 
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