I want to get tb110 wheels. Do any kegal style pulleys work on them?
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I think the new version comes to fit those, the first batch had some issue with flipped cores or something 
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Amount of energy is calculated in Watt hours or Wh,
Wh = V * Ah
So increasing either the Volt (Series) or Current/h (battery capacity) (Parallel) increases the range.
I might be wrong as Iâm no expert on the matter and people can correct me if Iâm wrong.
Doing some math on it, V = RI + E where E = Ke(Electro motive constant) * w(Speed).
Assuming constant current and same motor we get V = RI + Ke * w, moving these around we get RI0= V - wKe from this we get that in constant current if you increase the voltage the speed must also increase and with increased speed for the same duration means longer range.
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Important thing to remember about the tech were using in our electric Skateboard is that the energy coming out of your battery is not 1:1 to whatâs going into your motors, because your battery is DC and your motors are triple phase AC.
Why does this matter?
Because the output voltage to your motors is rarely at the maximum. Your battery will stay semi constant in output voltage, but the AC output voltage going to your motors is constantly varying as that is how the speed is controlled. Battery voltage only determines your maximum speed, youâre not pumping out 42-50v AC all the time- most the time youâre likely riding around 20-70% of that output.
So because of this conversion from DC to AC between battery and motors, think of your battery as a big fuel tank, and the best unit to measure overall fuel is in Wh (Watt-Hours, voltage x amp-hour).
To take it back a step and ELI5, think of your skateboard like a big juice blender machine. Your battery is a big vat of apples, the ESC is your blender, and the AC output to your motors is the juice. Now your juice output (speed) might be limited by how wide your Apple vat (voltage) is, and you can make more juice if you build a taller vat (more Ah/capacity). But ultimately the more apples you throw in, the more overall apple juice youâll end up with.
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@DerelictRobot Quick question. I have here a 10S8P 30Q (900 Wh) battery with 800 km on the clock and it has a delta voltage with exactly 63mV after use sitting on 37V. Is that drift normal? It was a 26,8 km trip btw.
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Is that a bug or a feature? Nothing mentioned about it on TB website
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It shouldnt be on the website, the first batch was sold awhile ago
It was a âfeatureâ, ala Bethesda
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Is this an average delta between cells? Or a maximum of one comparison set?
It wasnât a huge issue, honestly. Out of four wheels, two of mine were flipped and they went on the front. If I had gotten four bad ones, a little shrinkwrap would have sorted it. Amazeballs wheels for me.
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@Sn4Pz @BillGordon
Is that the only case of flipped kegel core wheels?
I was wondering what the Servisas reverese kegel pulleys are for 
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Love the analogy,
Though to be a pain in the ass and out of interest, what would be the mathematical proof of it?
Considering that voltage in the end is a form of pressure caused by the amount of electrons moving. There would be something in the conversion from DC -> AC that can use the increased pressure in order to draw less current from the battery.
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I donât know of others, but there could have been.
A lot of these analogies wonât hold up to a proof. One of the reasons EE is considered difficult is because electricity and how it interacts within a circuit is pretty difficult to visualize.
The analogies help explain some core concepts but the most common one of electricity vs water/pipes/pressure doesnât hold up particularly well beyond sound bytes.
On the topic of higher voltage/less amps, thatâs a lot of the core theory behind why full size EVs trend upwards in voltage. At higher voltage you can deliver the same watts at less amps, reduce operational ceilings, increase robustness of the system.
The takeaway is that itâs generally easier to just think of batteries in Watt-hr and motor output in Watts once the system has already been specified.
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I have had basic courses in EE, but they donât really go into anything practical. But they do show how much difficult it can be. Especially once AC is introduced.
The analogy I usually use for U = RI is that itâs a system of water flowing through a tube where I = how much water is going through, R = inverse to the width of the inner diameter of the tubing and U = the pressure the amount of water puts on the tubing. But once you start adding parts it becomes harder and harder to explain.
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@Krindor increasing the inductance in a circuit means increasing the time it takes for current to increase or decrease with applied or removed voltage.
so for example at 50% duty cycle, the battery could be supplying pulses of 60a, half the time, which averages to 30a. the battery side of the circuit has low inductance and so the current can ramp up to 60a and down to 0a very quickly.
on the motor side, the inductance is very high, so when the current from the battery shuts off, the circuit is rewired/switched so that the leads coming from the motor forms a loop, which allows the current to flow inductively (without applied voltage) for a short time.
the time scale of the on-off pulses from the battery are so short in duration that during the off times when there is no battery current, the current continues to flow in the motor entirely from the inductance.
so 60a from the battery half the time (30a average) can result in 60a in the motor the whole time, because the current still flows in the motor inductively while it is not flowing from the battery, but we say the 60a the motor sees the whole time is half the voltage of the battery if the duty cycle is 50%.
hope this helps.
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if you happen to be wondering what âpowersâ the current that continues to flow in the motor when you remove the battery current⊠itâs the collapsing magnetic field of the stator which stores energy in the magnetic field itself.
the inductance is directly proportional to the magnetic field produced by a given amount of current, and since the motor produces a large magnetic field for a given current, it has high inductance, and the current still flows for a while when you remove the battery current if you quickly short the motor to itself.
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That does help,
Correct me if Iâm wrong,
But if I were to make an analogy with a spring, if the inductor were to be a spring, the voltage the power used to push the spring back and the current the power/second when the spring retracts back into equilibrium. The retraction meter/second is constant. This would mean that the higher the voltage is, the more amps it can push back. Meaning that in the end, it would result in less amp usage with more voltage used instead.
Does this sound kind of sane?
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when you send current through the motor, because of the high inductance (created by the coil shape of the winding combined with the iron in the stator, which forms an elecromagnet) it takes extra time to reach a given amount of current, compared to the time it would take if you used a straight wire with no iron stator.
during this âextra timeâ you are building a magnetic field or âcompressing a springâ or âstoring energyâ⊠when you remove the battery voltage, the spring releases and powers the current that continues to flow for a short time without any battery voltage.
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That does make sense, thanks a lot for the help!
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