# Could 1.9 Nm be enough to accelerate 130kg electric go-kart?

I'll apologize in advance for a fuzzy question. My problem consists of two parts.

For starters, I'm trying to buy a DC engine for learning purposes and most of them only specifies the watt (which I can use in order to convert and get the horse power that I'm normally used to) or torque. And secondly, I've never built something like this before, usually I build bikes or cars. Normally engines are pre-designed to haul a car or two but when it comes to DC engines I could accidentally buy something that's meant to drive a overhead crane.

What torque would be required to match your run-of-the-mill go-kart that you usually find at theme-parks or amusement parks (I understand that a 70 bhp go-kart doing 120mph is another ballpark)?

Would 1.9 Nm be enough? Sounds weak when my bike weighs 196 kg with 96 Nm does 0-60 in roughly half the time a normal go-cart reaches it's top speed.

TL;DR:

Could 1.9 Nm be enough to accelerate 130kg as fast as a "normal" go-kart?

I have no place to ask these silly questions and no friends to ask, that's why I'm asking these vague questions here. Again I'm sorry!

• I think this is a great place for this type of question, but I think we still need to flesh it out a little bit. You talk about acceleration and fast. To me these are two different things. If geared correctly, you can take any motor and make it accelerate any mass to as fast as the motor will propel you ... but the trade off is the amount of speed which can be derived. To answer your question, we need to know what "normal" is: what is the expected top speed? Also, what are you using for gearing, or is gearing going to be dependent? Using a transmission or straight drive (chain or otherwise)? Mar 9, 2015 at 15:56
• @Paulster2 I haven't really considered gearing yet more than the fact that I understand that I'll need to work that out later in order to get the proper proportions between acceleration and top speed. To answer your question, roughly 20km/h (~12.4mph) would be enough for me. Would the gear on the drive-wheel need to be so huge that it's not practical (meaning the ground clearance would have to be unreasonable high or the engine would need to be way up reducing the low gravity of the kart? Assuming 1.9 Nm of torque on the engine. Chain drive with two sprockets btw is my thought. Mar 10, 2015 at 8:57
• Also i forgot to mention that I prefer high acceleration, so around 3-5 seconds for full on speed is my goal :) Mar 10, 2015 at 10:39

This is a great vehicle dynamics question that essentially has two parts to it:

1. Is the motor able to hit top speed, 120 mph?
2. Is the torque enough to accelerate it to top speed within 5 seconds?

# The motor in question

`````` Power : 1000 W (~ 1.36 hp)
Speed : 3200 RPM
Torque : 1.91 Nm
``````

Something interesting to note here is the apparent discrepancy between the three quantities. Since the three quantities are related, one could back-calculate how much power is delivered by the quoted torque at 3200 RPM:

``````Power = Torque x Angular Speed
= 1.91 Nm x ( 2π / 60 x 3200 RPM ) 1/s
= 640 W
``````

So clearly, something is up with the manufacturer's claimed figures. For 1000 W output at 3200 RPM, the required torque should be around 3 Nm, not 1.91 Nm.

However, errare humanum est, so we'll stick with the generous 1000 W for the rest of this answer.

# So what does 1000 W buy?

In short, not much.

The aerodynamic drag force alone, in the absence of other parasitic factors such as rolling resistance, inclinations and headwind, is enough to severely limit the top speed of the go-kart.

At top speed, the motive forces on the go-kart equal the drag forces acting on it (resulting in zero net force, zero net acceleration).

According to physics (and the Bosch Automotive Handbook):

``````Aero Drag = 0.5 x Air Density x Drag Coeff. x Cross-Sectional Area x Speed²
``````

With some reasonable assumptions for a go-kart:

``````Air Density = 1.225 kg/m³ at sea level, 15 °C
Drag Coeff = 0.4 (generous for a go-kart)
Area = 0.5 m² (rough, based on a 1.2 m wide chassis)
``````

The aerodynamic drag is a function of the square of the vehicle's speed:

``````Aero Drag = 0.1225 x Speed²
``````

How does this relate to power required?

``````Power = Resistive Forces x Speed
``````

Assuming that aerodynamic drag is the only resistance that the vehicle needs to overcome, the power required to overcome this force can now be estimated:

``````Power = 0.1225 x Speed³
``````

What this means is that if you double the top speed requirement, you need eight times the power.

So at best, 1000 W at the wheels would let you get up to 20 m/s, a hardly-earth-quaking 45 mph.

To hit 120 mph, one needs around 19 kW at the wheels, or roughly 25 hp, which is the equivalent of nineteen 1000 W motors.

• I'll try to answer the acceleration question later.
– Zaid
Mar 10, 2015 at 19:50
• Dear lord! Would theoretically this engine be able to hit 45mph? In my question (via the comments) I was hoping to get out ~12mph.. Can't wait to read about the acceleration! Mar 11, 2015 at 7:57
• @Zaid ... I don't see you estimating anything for gearing, though. If you use a 2:1 gear reduction, your torque is doubled and top speed halved. There are a lot of factors involved for estimating the theoretical top speed (speed excluding wind resistance), to include tire size and gearing. Great answer thus far, btw. I love mathematics (though I've forgotten most since I left HS). Mar 11, 2015 at 23:43
• Here is a good tire to use as an example because it has a 5" radius. Should make for easier computations. This should equate to a 31.415" rolling distance per rev. Mar 11, 2015 at 23:47
• 31.415" x 3500 = 109952.5"/minute = 9162.7'/min = 1.7895 miles / minute = 107.375mph ... that's the theoretical maximum with the 10" tire, 1:1 gearing, and 3500rpm. Make the gearing about 9:1 to get to ~12mph, and have about 17.19Nm of torque at the wheel. Mar 11, 2015 at 23:56