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dc to dc step up converter for 22v 6 cell lipo to get to 44 volt , need advice Answered

Im building a power harness for my hang glider, the project is very near completion, my brushless out runners require 44 to 48 volts. I am using two sets of 22v 6s in series, but the weight of two large 6s batteries banks is significant , what would be the potential to create this step up converter and what are the down side of designing dc to dc step up converter into my power system.

My hope to eliminate the in series battery ( double battery  weight ) set up. 

I suspect loss of amps or total run time ? what about problems with electronics  heat in the circuit, issues with ticky speed controllers, or the out runners , how will they be effected , or will these components work normally in a step up voltage / current  as compared  batteries in series. ( which is working beautifully ) I guess volts are volts, it just sounds to good to be true , double volts and less battery weight

Any and all qualified opinion and advice is appreciated. 

Thanks

Jeff

Discussions

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rickharris

3 years ago

1.5 Kg is about right for 6S lipo. - As my shoes weigh about that you must be able to find something you can abandon to save the weight.

I really think your running too close to your max allupweight Not a good idea for flying.

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Quadrifogliorickharris

Answer 3 years ago

I am curious. What happens when the threshold is crossed that the hang glider is mostly aloft due to the provided thrust and the thrust stops? I know fighter jets make lousy gliders, but what about a hang glider that is designed to glide? Put the nose down and go for a fast horizontal landing and a big flair at the end?

The gear seems very awkward with unintentional vectored thrust, including the tarmac. Except for the additional weight, it seems that variable pitch and folding propellers would be helpful. Also. it is a shame that there is already a substantial spar (keel?) that can't be used. And just a afterthought, the rig with the legs looks just like a grasshopper leaping.

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JeffR135Quadrifoglio

Answer 3 years ago

not sure how to answer all the questions, but maintaining the simplicity of free flight of the glider after power is off , is what were after, the electric propulsion harness being prototyped today is very limited in flight duration as compared to gas harness of the 90s ( even still currently available for mosquito NRG ). The grass hopper you saw. These can fly up to an hour on a gallon of gas.
The electric harness built with brushless out runner motors and lipo battery honest duration at max throttle is enough to get up to 2k AGL and seek thermals, or extend the flight time off mountain launch to seek thermal. Designs are focusing of course to limit drag and the additional weight. So free flight is not compromised in power off. Other means to get aloft traditionally are aero tow or winch tow. As we design an electric power harness small foot print that does not affect the glider performance during power off fee flight, we can now fly in many many places not previously an option. currently limited numbers of mountain launch, aero tow sits are the only way to participate. Its like trying to snow skip in th south. We could simply foot launch off flat land and enjoy free flight once at altitude. vector issues are there, weigh shift is the means of controlling pitch and roll, so with the push of the power harness large turns will climbing would make the most sense, the skill set for controlling the hang glider requires pitch and roll input where you lead from the hips, avoid cross control of your body while adding input to control the glider, so the thrust does present added dimension to controlling you weight shift and keeping the glider at correct angle of attack.
Some have HG pilots have experimented with motor / thrust attached directly to the main keel, but apparently this presented a vector problem primarily in pulling out of a dive.
It appears that the most successful power supply on wing without control surfaces ( gravity controlled pitch control ) is when it is attached to the pilot.

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JeffR135

3 years ago

Thank you all for the input. The 6s lips I have are about 3.5 lb each if you could eliminate 4, that would be fantastic weight reduction. but is not a deal breaker. Looking for every way to decrease weight, drag, increase thrust. Ill post the harness and results soon. Im getting about 50# combined thrust at peak output constant for two minutes on the bench. Using 2, 220 kv Gmax 5355 helicopter motors. 250 amp turnigy ESCs and 22x10 props. All Aluminum tube frame. total eight without batteries is about 10# , search " e help hang glider " on youtube if you want a good visual of what Im doing, expect Im using twin 220 kv motors instead of one 100kv .

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Downunder35mJeffR135

Answer 3 years ago

You could squueze a few pound out by using carbon fibre rods for frame.
Quite cheap to optain in kite shops these days.

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Downunder35m

3 years ago

If these battery packs are already too heavy you will need a bigger glider.
Using just one battery means you will have less than half the runtime and most likely not enough power to run the motors properly.

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Jack A Lopez

3 years ago

If you understand the law of conservation of energy, you can prove mathematically that which you suspect about voltage, current (i.e. "amps"), and run time.

The first part of your proof requires a model for a DC-to-DC converter

An ideal DC-to-DC converter converts power without loss. That is to say, the power output, Pout = Vout*Iout, is exactly equal to the power input, Pin = Vin*Iin

Pout = Vout*Iout = Vin*Iin = Pin [ideal, with eff=1]

For a practical DC-to-DC converter, there is some loss of power, as heat, and you can account for that by adding a factor called "eff", for "efficiency".

Pout = Vout*Iout = eff*Vin*Iin = eff*Pin [practical, with 0<eff<1]


The second part of your proof requires a believable model for an electric battery.

It is not far-fetched to say a battery stores, can deliver, only a finite quantity of energy (where energy is the time integral of power), and if you want a number, an estimate, for how much energy that is, it is a safe bet that quantity of energy is not more than the nominal battery voltage multiplied by the current*time capacity.

By the way, if you want your number for energy, delivered by the battery, to be in units of joules or kilojoules, notice that battery manufacturers usually give their numbers for battery capacity in ampere*hours, and 1 hour is equal to 3600 seconds. In other words:

(1 V)*(1 A*h) = (1 V)*(1 A)*(3600 s) = 3600 J = 3.6 kJ


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rickharris

3 years ago

Any electronic way to increase the voltage will proportionally decrease the available current - This may not be what you want.

As a 6S battery pack isn't all that heavy it suggests you may be very close to the limit of your rig anyway.