Questions for my first build- Lightweight Cruiser

Hi, My name is Dominic and I have been reading in this forum for awhile. I am 16 and I am in a FRC robotics team. Also working a project car (MGA 1600). But all of that is irrelevant.

I have some questions for my first build. I am trying to make a lightweight cruiser that I can quickly take me a mile or two here and there when I miss the bus. I weigh less than 115 pounds so weight isn’t an issue. I also want to have a mixture of speed and torque, which is why I chose to run the motor at 8s instead of the usual 6s (I can gear down). I used Solidworks to CAD a motor mount. I know there are a lot out here I just like the pride of designing it yourself.

Parts of my build (some stuff I have already): -custom 29in cruiser that I made + hardware (have already) -Caliber Trucks 50 degree baseplate + riser pads (have already) -Abec 11 76mm flywheels + bearings (have already) -36T HTD 5mm pulley + pulley itself (have already) -SK3 280KV (have already)

I have some questions:

  1. Does my mount look good? I wanted everything to use locknuts cause why have it adjustable when you have to Loctite it?
  2. Does 36:16 sound good? I have some moderate hills that I need to climb but I am not looking to use this as my entire commute, maybe a last mile kind of tool.
  3. I want to use the vesc but I am scared that I am going to break it. I have some experience with electronics but after reading about the vesc on these forums I am worried I will waste my money by destroying it.
  4. How will the 8s1p battery work? I want to get maybe 4-5 miles of ride distance but not too worried about that, wanna have it be really light. I also want to use sleds for this so I can change out the batteries if they are not working so well. What battery is best? Sleds? Also, what kind of wire to wire it up? Best BMS?
  5. Suggestions for enclosures? I might be able to 3d print a single piece cause my battery is really small, idk.

I appreciate all the help. I will be posting updates and further questions in this post so I wont be obnoxious by making new forums for the same topic.


16 to 36 gear ratio will suit you just fine. From what I know 26 650 cells have a lower nominal voltage, so an 8s battery will be more like 7s or 6s (correct me if I’m wrong). is the best place to get a BMS in the states, but I think they’re having some pain issues at the moment. You can either spot while the pack with pure nickel strip or solder it with pure copper wire, but do not use sell holders because the tabs cannot handle the amount of current going through them. They are also not secure for the bumps when riding. A 3D printed enclosure is very possible if your battery is small. I find mine works great because I customize everything on it. I would not be afraid to use the VESC. You can just post all of your specs before you ride and have an expert check it on here.

This is based on cell chemistry not the size. a123 cells use LiFe are 3.3 nom 3.6V full li ion/lipo use LiMn (i think) 3.7 nom 4.2 full.

if they are liion like basen 26650 the motor might be a bit high kv. but since you are light, its not problem just limit max erpm on the vesc.

Also those battery holder can not handle the current, especially in a 1p pack. You need to spot weld, or solder.

I just did a post about range, with a simple calculator to estimate based on battery. Esk8 Range Calculator

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Ok, getting basen 26650 for 8s1p with a 60amp bms. Also getting switch from Problem is that i need something that can supply 32 volts to the bms so I can charge. All options that I found were super expensive for so little amps, and it would take 1/3 or the day to charge. Anyone know places to get cheap power supplys for 8s1p 26650 basen batteries?

Also, will 10 awg wire be sufficient for soldering the batteries together?

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Basen 26650 are overrated. They are 20~25A continuous. Better go with A123 cells.

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Oh, shoot. Already bought the basens. I could just limit the amps with the vesc for protection, right?

Yes, you can set 20A battery limit, but the voltage will still sag a lot in a 1P pack. Basen cells are good if used 3 in parallel or more. Only A123 cells or LiPos can do 1P.

What about limiting the damage done buying 8 more cells and making 8s2p? 40A should be enough, space cell is rated for that. It will weigh a bit more but should still be acceptable

Can i return the basen batteries from I dont think they shipped yet. They advertised that the battery could do 30 amps continuous? Why do i need to limit them to 20 amps if they can supposedly do 30 amps? Im still kind of confused on what voltage sag is. I am assuming its like our robot when too many motors run at once, it makes everything slightly slower?

They can do 25A continueous from the tests. But if you draw 25A, voltage will drop. Ideally, you want to draw < 10A from the cell. That’s why guys there make 4P packs of 20A cells and use 40A fuses / BMS instead of 80. Even this sometimes is still not enough.

Back to the voltage drop ( sag ). Less voltage = less POWEERRRRRRRRR. Now, seriously, in the end, you may end up not beign able to climb a steep hill and so on. Also, the battery will wear quickly and get quite hot.

Why hot? Because V ( drop ) = I * Ri Every single cell has it’s internal resistance. Imagine it as a resistor connected in series with the cell. Higher current = higher voltage drop. Energy won’t disappear. It transfers into heat.

Option 1: I limit the amps from the battery to 25, i get less heat but more voltage sag. Option 2: I limit the amps from the battery to 30, i get more heat and less voltage sag.

Am i doing this right? v=ir and the resistance is constant so the voltage is proportional to the amps?

Thanks for clarifying for me, I really appreciate it cause I just started physics his semester. Im taking AP physics 1&2, and we haven’t got to the 2 (electricity and magnetism) stuff yet.

Heat is directly proportional to the voltage drop. Voltage drop = heat. More current => higher voltage drop => more heat. Internal resistance is constant.

And yes, V = I*R. In this case, we are calculating the voltage drop itself. R = const, higher I = higher V ( voltagr drop ) = more heat.

With 1P pack, you ideally want to set your current limit to 15A.

More current/cell equals: -More voltage sag (bad) -More heat (bad) -Less range (bad) -Lower life cycles (bad) -Lower max power (bad)

I cant actually find any discharge curves for these cells so I can’t tell you how much is too much.

@IDVert3X, can you post some please if you know where to find them.

Either way, I recommend you either get a different cell or double your cell count.

Wrong. P = V^2/R. Or P= I^2*R P = Heat in watts

V=I*R in purely electrical systems. Batteries are not electrical systems. They are electrochemical systems, way more complex. Their internal resistance varies with temperature, state of charge, state of health, current draw, etc. This is why you want to rely on discharge curves and not V=IR

Correct. Both are correct. You just written the complete formula. From my formula, you get the voltage drop. And it is directly proportional to heat as higher drop = more heat. If you want the amount of heat in watts, you have to multiply it by current. And yes, that’s when I^2*Ri comes in place. Or step by step:

V(drop) = I * Ri [ Volts ] U = V(drop) * I [ Watts ] Which is equal to I * Ri * I which is equal to I^2*Ri.

This is all I have for the Basen discharge curves.

Drops are ridiculous.

That is not what directly proportional means… Directly proportional would be P= c*V(drop) Where c is some constant… Heat increases with voltage drop squared.

E: Read my edit above for the V=IR part of your comment…

Those curves are useless. They are with respect to time and not capacity, plus they are also resting the cell between draws. Very different from our application.

E2: You don’t learn this stuff in physics. You have to get to the Electrical engineering/ Chemical engineering courses to fully understand it

Well, I was taugh that directly proportional means when something increases by certain amount, something else increases too.

In my language, its called “priama úmera” which google translated as directly proportional.

And if voltage drop increases, watts increases too. Basically P=V(drop)*I and if drop is not directly proportional with heat ( watts ), how do you call it in english?

Directly proportional is this equation by definition. I’m pretty sure that’s a math thing not a language thing, but I dont know. X=cY where c is a constant. This is wrong because P=IV and I is NOT a constant.

This is the correct way to say it power is directly proportional to voltage squared. P = (1/R) * V^2 where 1/R is a constant

My whole topic deleted? Or someones comment? I kinda need this so i dont mess up anything.

He deleted his comment. Not thread.