# Power, Torque & Speed

First off, I’d like to point out, that I have no idea who you, dear reader, are. You may be a builder, a school teacher, or a particle physicist studying Bose-Einstein Condensate. You may be a maths nerd, or you may be terrified of numbers (if the latter then I’m sorry, there are a few below). I’m trying to make these post both informative and at the same time readable by as large a segment of the population as I can. So if things are too advanced, or too simple, you can try to comment, or else just try and push through. Of course there’s always the option of leaving and never returning, but I really hope you don’t take that option. Right, moving on …

With a form factor and performance criteria decided upon (trike, and 50kmph respectively) it was time to do some math and figure out how to achieve this. I’d already half decided to go with RC aircraft motors due to their abundance and low $/kW. A quick search on my favourite rubbish Chinese hobby site (www.hobbyking.com) turned up [these little gems]. While there exist higher power motors, what I’m looking for here is the low KV rating. KV is a measure of the motors speed for a given voltage. If I’m planning on running a 30V battery pack (a 6S lithium polymer for example) I can expect a top (unloaded) speed of 30 * 270 = 8100rpm. A higher KV will obviously lead to a higher rpm, as will a higher voltage.

The power in a motor is equal to its rotational velocity times its torque. Unfortunately it turns out that speed is way easier to achieve than torque for a given motor diameter. This results in some insane devices such as [this] which manages to produce 12kW (that’s 16hp!) in a package that weighs less than 1kg by turning at 50,000rpm. So what’s wrong with this you say? The problem, is a lack of torque. The rotor may turn at 50,000rpm with no load, but I’m going to have a load. I’m going to need to accelerate the bulk of me + trike, to over come air resistance (drag) and to push my way up some of the many hills around my home town. All these things need torque, not just speed. Now I could always connect the motor to an elaborate gearbox with a 200:1 speed ratio to obtain some serious torque at a reasonable speed, but it turns out that 200:1 gearboxes include some pretty major design and construction challenges I just wasn’t willing to tackle with my limited tool selection, hence the relatively lower power & rpm of the motor chosen.

So we know the maximum power, and the maximum speed of the motor, what about the torque? This is a fun one. I’m going to start by making a statement; Maximum power occurs at 50% of maximum speed. Using this and the knowledge that torque for a BLDC motor (like the one above) is at a maximum at 0rpm and decreased linearly to zero at maximum speed we can work out the maximum torque. I’d love to go into more detail about the math and physics behind this but there’s just not going to be time this post. If people want to know, message me and I’ll put it in another post.

So at maximum power, the rotational speed is 8100rpm/2 = 4050rpm. Power is equal to torque * rotational speed however we need the speed to be in the right unit, and rpm isn’t the right unit, we need rad/s. A rad is a measure of angle, like degrees, except that a circle has 360 degrees, but 2*pi rads. Thankfully it’s easy to convert. 4050rpm /60*2*pi = 424rad/s so at maximum power the torque is 2400W / 424rad/s=5Nm

To power this magnificent contraption I’m currently looking at using two 5800mAh 8S LiPo battery, like [this one] in parallel. These batteries can manage amazing power and energy densities while being reasonably priced and light weight. The only thing to keep in mind is the discharge current limit. The linked battery claims 25C, which means you can discharge at 25x the capacity, or in this case 25 * 5600mA = 145A per battery.

[For the math/physics geeks; try doing dimensional analysis on what a ‘C’ is. 25C * 5.8Ahr = 145A, so C mush have units of per hr, which is just per second with a scalar. Per second is also known as Hz and is a measure of frequency. Go figure, 25C is the same as 90kHz. If anyone can provide an intuitive insight to this I’m keen to hear it].

Back on topic. Since I have two batteries I could theoretically draw up to 290A. However given the slightly dubious source of these batteries I would recommend giving yourself some headroom. In my case the motors are only rated for 90A peak (of equally dubious amps) anyway, so I have plenty of operating margin.

The next decision made was around the drive wheel size. I wanted it to be as small as possible while still being a pneumatic tire. The reasoning behind this comes back to torque. My chosen motor spins fairly quickly. We know that torque, speed and power are all related, but none of these things directly result in forward motion. If we want to move forward, we need a force. The force is applied by the rear tires to the ground, and is directly related to the torque by the simple relation Force = Torque / Distance, where in this case the distance is the radius of the wheel. So, assuming that our motor gives a certain torque, the bigger the wheel, the smaller the force. Of course there is also a reciprocal relationship between Speed, Angular Velocity and Radius, so again assuming the motor is turning at a given angular velocity, reducing the wheel size reduces the speed. With the above in mind, I wanted to minimize the required gearing ratio between the motor and the wheel ergo, the smallest possible wheel.

There are a number of small pneumatic wheels available. From the very cheap hand truck sort, up to the much more expensive go kart wheels. In the end I settled on the back wheel of a pocket bike. If you’ve not come across a pocket bike before they’re small motor bikes, think clown bike small. They also have the huge advantage of being plentiful, reasonably cheap, and somewhat standardised. Browsing my favourite online chinese export company netted me [these beauts]. The major advantage of the pocket bike wheels is that they’re designed to be driven at speed, by a very small (though admittedly gasoline) engine. This means there’s also a selection of sprockets and chains designed to go with them available at low prices. Awesome!

Now that I have a wheel I can figure out how fast I need it to turn in order to get me to 50kmph. Our rim has a diameter of 6.5″ = 165mm. Most pneumatic wheels size the rim, not the entire wheel, the notable exception being bicycles. On this wheel we have a tire, which is 110mm wide, and has a profile that is 50% of the width, that’s what the 110/50 means. So we have a rim to road thickness of 55mm. Double that (the tire is on both sides of the wheel) and add it to the rim for a total diameter of 275mm. The circumference of the tire is the diameter * pi = 864mm/rev or 0.864m/rev or 0.000864km/rev so, for every revolution we will move forward 0.000864km. Next, 50km/hr divide by 0.000864km/rev = 57870 rev/hr. But we don’t want rev/hr we want rpm, or rev/minute so divide by 60 to get 965rpm.

Right, that was a fair bit of math, but it’s important. We now know how fast we need to turn our wheel to travel at 50kmph. Unfortunately here’s where things get tricky. We know the motor will spin at up to 8100 rpm so we could just use that number and calculate a required gear ratio of 1:8.4, however this is just an approximation. In reality the motor’s not going to spin at 8100rpm, there is going to be a load on the motor due to friction, drag, etc. It turns out that these things are pretty small, but I’m an engineer so I wanted to take them into account as much as possible. So not to bore you too much (if you’ve read down to here you’re a masacist anyway though) I’ll race through this next part really fast.

Drag due to air friction is given by Fd = 0.6 * A * Cd * v^2 where 0.8 is 0.5 * the density of air (~1.2kg/m^3), A is the cross sectional area of the object, Cd is the coefficient of drag, and v is the velocity in m/s. Here’s where I made some rough assumptions, A = 0.5m^2, Cd = 1. the coefficient of drag is especially pie in the sky stuff but I’d rather be slightly pessimistic than overly optimistic. Now that I can find my drag force at any velocity, and I can also plot the torque of the motor at any velocity, and I know the relationship between torque and force (remember the wheel radius?), I can plot both those things on the same graph, and where the torque from the motor = the torque from the load I have my maximum speed. At least I nearly can, I forgot about the gear ratio of the sprockets so now my torque is multiplied by the ratio, and speed is divided, but that’s easy enough to factor in.

I can actually work out a fair bit more than that too. I now have the difference in torque available and torque required to overcome drag at all speeds below maximum speed. The remaining torque can be used for acceleration.The simple relationship between force and acceleration is a = F/ m, where m is the mass of the vehicle and rider (we’ll say about 100kg) and F is the available acceleration torque referred back to a force through the gearbox and wheel, and obviously, a is the acceleration we’re trying to find. If we sub in the prvious equations relating back to torque where relevant we get one big equation, a = (Tm * Rg / rw – Fd) / m where, Tm = available motor torque, Fd = drag force, Rg = gear ratio, rw = wheel radius, and m is our 100kg lump. Cool, so now we’ve determined our acceleration at each speed, all we need to do is create a recursive loop which starts a 0kmph, determines our acceleration, calculates our new speed some small time later (lets say for arguments sake 0.1s), use our new speed to determine our acceleration for the next step, and repeat. The more math savvy readers out there will recognize this as an approximation to integration.

THE FINALE!

Rather than going though all this every time I want to change a parameter, I created a spreadsheet to do it all for me. I’m going to make the spreadsheet available under the proviso that it was built as a tool to be used by me. As a result it’s full of potential gotchas and caveats.

– Mike

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