At 2 amps that means I can get 31 amphours - the full charge into the Chafon generator - in 15.5 hours. That's not too shabby for just pedaling a bicycle - no need for sun!! The solar panels will get 5 to 6 amps and the car charger will get 5 to 6 amps also.
So is this new motor, even though it has more horsepower than the last motor, really just a much higher RPM motor and therefore a much lower torque power at lower rpm?
https://elixirfield.blogspot.com/2019/01/the-joy-of-finally-liberating-dc-motor.html
That is the first Treadmill motor.
5075 RPM at 15 amps and the voltage rating is 130 voltsSure enough - it is probably 1000 RPM less then the motor I know have! Wow more like 1625 less rpm:
6700 RPM 18 AMP MAGNET
130 volts x 15.7 amps = 2041 watt motorSo it's about 2.75 horsepower motor - but that is what my current motor claims. The big difference is the RPMs! So my current motor does not list the RPM but the same mounting bracket lists the RPM - based on the same motor model.
It was a higher amp also though - 18 amps.
So he's claiming at 60 volts he gets 3000 RPM!!
This is because if electrical resistance in the motor is what limits your # of amps, then when you divide the voltage by 2, you have to divide the power you get by 2 squared (3, 3 squared, etc).So what was that equation again?
OK so we are dealing with OHMs as Volts (130) divided by Amps (15.8) = 8.2 Ohms as resistance. Then take the volts (17) divided by Ohms (8) to get the amps again as 2 amps.
Wow - so I get the same result.
Only that was at 17 volts - and I have to now pedal up to 60 volts! Why? because the Ohms are less - less torque.
How much less?
130/18=
7 Ohm.
Oh wait - it might be 100 because it says 100 VDC yes and my first motor was 130 VDC. So a higher voltage with less RPM is definitely a way better motor for operating power.
yeah so 5.5 Ohm on the motor I have now.
V is the voltage drop of the resistor, measured in Volts (V). In some cases Ohm's law uses the letter E to represent voltage. E denotes electromotive force.
I is the electrical current flowing through the resistor, measured in Amperes (A)
R is the resistance of the resistor, measured in Ohms (Ω)
in the case of a typical cheap treadmill motor, results in the HP falling by the square of the speed- as in, at half RPM, HP is a quarter rating, at a quarter speed it's a sixteenth, etc.
If you lower the voltage the current will be lower.So If I am getting up to 3000 RPM at 60 volts - that is less than half speed which means it's less than one quarter the horse power. So it says 2 horse power continuous or 1492 watts. So less than 370 watts. I calculated 60 volts x 2 amps = 120 watts... why? because the RPM value is a lot lower.
So in order to calculate the actual power we need to find the R or resistance of the motor. If we use ohms law V = I x R and rearrange it to V/I = R then 95Volts/21.4amps= 4.44 ohms.
The formula for Power in watts is Vsqrd/R or
48v x 48v / 4.44ohms = 519 watts
At full voltage 95v x 95v / 4.44 = 2033 watts
V = volts, I = current, R = resistance in ohms.
So we have 100 volts = 18 x Ohm, as I figured about at 5.5 Ohm.
So then we rearrange volts for power so at 60 voltes we get 60 x 60/5.5=
Actually based on the rating of 100 volts and the watts as 1492 then the Ohms is 6.7
So then we get how many amps?
60x60/6.7=
but then I need to add the 2 ohm of the battery - assuming that is the accurate, as that is the typical resistance of a 12 volt battery.
413 watts
So the amps is
still too high - 6 amps.
So what is the back current ohms of the DC-DC buck converter?
it says 94% efficiency - someone tested it as 85%
But what about the Chafon generator?
https://elixirfield.blogspot.com/2019/02/the-skinny-on-chafon-lithium-battery.html
OH - so I need to reset the voltage level on the new DC-DC Buck Converter!! I have it set too HIGH - I have it set at 12.7 volts or so....So then the Solar port DC charging states that with a solar panel at 18 volts (to maximize amps) you lose 20% of the amps to convert to the DC battery charging.Also the car DC charging through the Solar port states if the input charge voltage is over 12.4 volts - then the battery voltage regulator flashes on the green light.This is what happened to me BEFORE I got the DC/DC buck converter. I always saw that green light flash on the solar charger before it turned to Red. I had no idea that meant the voltage regulator was having to then lower the voltage.So when I set the DC/DC buck converter to 12.2 volts that was the IDEAL most optimum voltage - even though each "pack" of cells is actually at over 13 volts to charge it. But the advanced battery management is not based on the pack (wired in a series) nor is it based on the 9 packs (wired in parallel) but it's based on EACH cell at 3.7 volts.
OK I reset it to 12.2 volts but then the amps dropped to only less than half of 1 amp at highest gear speed on the bicycle!! So the reason is because this new DC-DC Buck Converter draws 400 millivolts - so at 12.2 volts then the actually volts are below 12 and so the Chafon is not charging. Quite the drop!!
So then to experiment I increased the volts to just under 14 - and the amps went back up to just under 2 - sometimes 2 - but not STEADY at 2 amps. So that proves that it definitely was the voltage change that caused the amps to drop to less than one half amp!
So now I'll set the volts back to 12.7 - which seems to be the magic number. Because if I go any higher than I go over 12.4 volts. At 12.7 then I am drawing .4 for the DC-DC buck converter - so or a little less than .4 so I should be above 12.2 volts but under 12.4 volts. Maybe if I try 12.6 then it should be ok. I think it was set at 12.6 before.
OK so now I plugged in the 1.5 amp DVD player that has a 3.5 amp lithium battery charging up. I think the increase load on the DC Motor then enabled to get more amps at a lower speed!
So I can pedal in the 2nd highest gear while still maintaining a 2 amp charge into the Chafon 31 amphour lithium battery inverter generator. Pretty awesome!!
So once the DVD player battery charges up then it will stop drawing on the generator - and I'll see if the load goes down or I mean if the resistance goes up - so that the amps go down too much while in the 2nd highest gear.
On this device - it has to be powered first to change to amp reading and I can't reach it while riding the bicycle.
OK you do have the best vid on youtube for a treadmill motor generator. So getting 8 amps is very impressive! I am just really wondering though about what kind of amp draw do you get for charging a 12 volt battery? Because obviously the resistance is a lot lower on a battery - right? I mean if two light bulbs are rated to need about 8 amps to power them at 12 volts - then you're gonna get 8 amps through your DC-DC Buck converter! I am charging a 12 volt lithium battery - in series - 31amphours - and I can only get a 2 amp draw even though I also have 55 volts on my treadmill motor. So it's just because the 12 volt battery has a lower ohm resistance? 3.3 ohms I think. Let me look up the ohm resistance on two light bulbs. So that is why you can get higher amps - right? I would like to get higher amps but I have not hooked up a higher draw directly to my DC/DC buck converter (and yes I think it's a 12 amp limit). I can go up to 75 volt input. Yeah it says up to 12 amps output. It's the Drok: D3806.
Actually Mr. Electron is only getting a 2 amp draw average while charging the 12 volt battery. But he gets a 5 amp draw while charging the 60 watt lightbulb. Why? Because the internal resistance of a 12 volt battery is much lower and so the amp draw is much lower - even though the total amphours of the 12 volt batteries are 48 Ah. The internal resistance of a 12 volt car battery is 0.02 ohms. The internal ohms of a 60 watt 12 volt light bulb is 12/5 so 2.4 ohms. So Mr. Electron - you have ANSWERED a question I posed to another youtuber - but my lithium inverter generator has a 120 volt input for charging. So I think if I invert my DC motor amps (stepped down in voltage via a DC-buck converter) - into AC then I should get a higher load draw from the 12 volt battery since it would be higher volts so higher ohm resistance. But then it still has to convert it back to 12 volts inside the Chafon generator. Oops I just blew my car inverter experimenting on this since I could not tell which wire was positive or negative. I got smoke and a loud pop. I did not see any blown fuse inside but lots of little twine fraying. Strange - that stuff will really take a spark well! thanks for the vid. Fascinating.
thanks for the advice. Yes I am using a DC-DC buck converter. So I get the RPMs up to about 3000 based on about 60 volts - on the bicycle treadmill motor. So that is then stepped down to 12.6 volts for charging the battery. I will now go buy a halogen light that is 60 watts. So that will have an ohm resistance of 12/5 (12 volts/5 amps) so over 2 ohms. A typical 12 volt battery only has an ohm resistance of .02 ohms. So that is why the light bulb should charge at 5 amps compared to the battery charging at only 2 amps. So that will verify that my DC-DC buck converter is producing the larger amps. Then if I switch from DC input on the Chafon 12 volt battery - to AC input then that should increase the resistance to increase the amps. So yes I can see why that would be more dangerous at the connection that I wire to the inverter. But if my experiment works - it would save about 10 hours of pedaling on the bicycle. haha.
Ok so the Chafon generator says it charges faster off the 12 volt car charger. So I checked my 12 volt charger for the drill - it's actually 20 volts! The Chafon can also take up to 20 volts. So I then increased the DC-DC buck converter to 20 volt output because I had not realized before - that will increase my watts. So based on the comment I had mentioned earlier - the volt input changes the Ohm resistance significantly.
"One thing that really bothers me is your use of the battery as a load. If you have a battery of 12V and a resistance of 0.1 ohms at a point where the circuit is at a potential of 14.4V, you only have a voltage of 2.4V across that resistor."So that is a difference of 2.4 volts - so that means if you have - he said 24 amps so then the ohm is .1 as the division. But if I have only 2 amps then the ohm resistance for the draw is 1.2 ohm. But if I am now using 20 volts so then I have a voltage of 6 across the resistor then the ohm is now 3 ohm - so twice the resistance. So I get more watts but it takes more pedaling resistance - TWICE as much - to maintain the 2 amps. I can't maintain 2 amps at that resistance.
So if I charge at 120 volts - I'm not sure what the charging resistance is inside the Chafon. First I have to see if I can increase my amps - so I'll go buy two 50 watt bulbs and first do one - as 12 volts - so 4 amps. Then I'll try 2 bulbs wired together - as is shown in one of the vids - that will be 8 amps. So then - ok here's the deal - if my resistance does not increase on the load side of the generator then I don't see how the draw will increase.
If the draw doesn't increase - oh wait at higher volts will the resistance increase on the Chafon side? And even if it does that means I would be pedaling with that much more resistance. So it doesn't look like it will work. In fact the person lighting up the 100 watts of bulbs could barely do so since the pedaling resistance was so strong. So if I can barely pedal at 40 watts - then to get 100 watts would be difficult for very long.
Yes it says the 120/110 volt charge takes 5 to 6 hours (the last hour is just battery management). So we say 5 hours of charging which is 31 amphours - so that means it's 6 amps!! So for ohm resistance is volts/amps = an ohm level of 18.
So there's absolutely NO WAY I can pedal at 18 ohms resistance.
A solar panel is a current source, not a voltage source. The spec on the [100 watt] panel is 18 volt Vmp and 5.55 Imp. When connected to a discharge battery the voltage of the panel is pulled down from 18 Vmp to match the battery voltage of around 12.1 volts and is supplying its constant current of 5.55 amps. Do the math 12.1 volts x 5.55 amps = 67 watts. You loose 33% of the panel powerSo the Chafon is stepping down the voltage and lose the WATTS anyway!! So no point in pedaling twice as hard!
Best solution is to find a charger for 6V NiMH 400mAh. But that will be hard, maybe impossible. So you need another solution.
What I would do: use a simple 12V DC adapter; very cheap. Connect it to the battery pack: - of adapter to - of battery; + of adapter to a resistor; other side of the resistor to + of battery pack. So the resistor is in series with the battery pack. Resistor value is 120 ohm, 0,5W. Very cheap too. Plug the adapter in. Start counting 14 hours. Disconnect everything. Battery should be full!https://batteryuniversity.com/learn/article/charging_with_a_power_supply
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