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rp181

Sat Apr 10 2010, 06:47PM

Registered Member #1062 Joined: Tue Oct 16 2007, 02:01AM
Location:
Posts: 1529

I am looking to make a ground vehicle for a project. For this project, I have chosen to use 4x 350W motors.
The chassis will be steel, dimensions around 12"x18"x6" without wheels (I plan on 8" wheels).

I was wondering how I could calculate the motor RPM with a given load. I know the motor is 350W, or .469HP (each). The RPM is 2200 no load, so torque is 1.12 pound-foot.

With this, I am trying to see if I can run the motor direct drive to the 8" tire, without excessive speed (~15mph goal). The vehicle needs to have some terrain capability, so that needs to be factored in. How would I go about this?

Sulaiman

Sat Apr 10 2010, 07:14PM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140

For a dc motor, roughly speaking, speed is proportional to voltage and current is proportional to torque, pwm will give torque (current) control due to motor inductance if the switching frequency is high enough, usually a few kHz.

Industrial dc motor controllers usually have two feedback loops, an 'inner' current (torque) control and an 'outer' (overall) voltage (speed) control.

radiotech

Sat Apr 10 2010, 07:35PM

Registered Member #2463 Joined: Wed Nov 11 2009, 03:49AM
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Posts: 1546

For any motor :

Torque in pound-feet = (5250* Power in HP) / Speed in RPM

rp181

Sat Apr 10 2010, 07:43PM

Registered Member #1062 Joined: Tue Oct 16 2007, 02:01AM
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Posts: 1529

Let me clairfy, my post wasn't very clear.

I am trying to figure out the gear reduction, if any, I need. For example, with no load, the motor spins 2200RPM, way too fast. As A load is applied, the RPM lowers and the torque increases as the motor draws more current. How can I estimate the RPM drop with a load? The vehicle will be 50-100 lbs.

I may implement voltage and current control.

Sulaiman

Sat Apr 10 2010, 08:50PM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140

Quick calculation;
15 mph = 15 x 63360 / 60 = 15840 inches / minute
15840 / (PI x 8) = 630 rpm
2200 / 630 = 3.49 gear ratio (no-load 15 mph, 8" dia wheel)
So about 3:1 gear ratio required for full power at 15 mph.

For dc motors the back-emf is truly proportional to rpm
The applied voltage = (back emf) + (motor equivalent series resistance x current)
for medium size motors I'd expect full-load rpm to be about 90% of no-load rpm.
Many theoretical analyses of dc motors give
(maximum power output) at (1/2 x max rpm) x (1/2 x stall-torque)
This is rubbish (same as for transformers, batteries etc) (max power theorem)
Motors (transformers, batteries etc.) are manufactured for much less than this because such operation would result in < 50% efficiency.

Dr. Slack

Sat Apr 10 2010, 09:11PM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

How can I estimate the RPM drop with a load?

You shouldn't have to. The motor supplier should give you a speed/load graph which allows you to read it off directly. If you need to estimate, and if they are halfway-decent permanent magnet motors, then you should expect around 75% efficiency for the rated loading, so a speed drop of no more than 25% below no load speed. It will be less than 25% because the "no load" speed is already loaded by friction, windage, eddy currents, and as the current rises when delivering a load, efficiency suffers with copper losses. So guess around 15% speed drop for rated load and you won't be far wrong.

However for an all-terrain capability, the motors will deliver much more than their rated load for short transients, all the way up to the stall torque, for a 100% reduction in speed!

Hint, when doing motor calculations, use SI units consistently, it's much easier. Watts in == watts out = Nm torque * radians/s angular speed. Working around conversion units through measuing time in minutes rather than seconds, angle in cycles rather than radians, distance in feet rather than metres, furlongs, hundredweight, barely corns, carats, slugs, is all so much extra work. Once you are all in SI, then speed = wheel radius * angluar speed.

Another hint. It's generally possible to deliver more voltage to a motor to get it to spin a bit faster. The robot I built with 24v motors used a nominal 36v supply. Actually a few of these volts went in the controller losses and the battery output impedance. You lose motor lifetime through increased armature arcing, but hey, overrunning anything usually carries a lifetime cost, but you get more power and a larger speed range. The only thing you must observe if you over-volt a PM motor is to limit the current to below that which would be taken by the stalled motor at the rated voltage. This is so you don't demagnetise the field magnets. A motor has strong enough field magnets to withstand the demagnetisation H field when starting from stationary at the rated voltage, but not much more. Oh, and don't over-do it, 50% extra is about the limit without doing tests for the armature bursting speed.

rp181

Sat Apr 10 2010, 09:38PM

Registered Member #1062 Joined: Tue Oct 16 2007, 02:01AM
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Posts: 1529

Thanks for the wealth of information!
The motors are from surplus, so there isn't much information about them:

There is also a 450W 36V version for 4 more dollars, I decided to o with the 350W one for the power supply.
Also, I said 2200 RPM, its actually 2600 RPM (I think Ile go 3:1 or 4:1).

Another question:
How are brushed motor ESC's designed? What design should I go with?

How come H bridges are used? I don't understand how a H bridge would work as it supplies AC power.

Thanks.

ScotchTapeLord

Sat Apr 10 2010, 10:16PM

Registered Member #1875 Joined: Sun Dec 21 2008, 06:36PM
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Posts: 635

rp181 wrote ...

How come H bridges are used? I don't understand how a H bridge would work as it supplies AC power.

Super-duper low frequency, variable duty cycle AC can be very useful for a motor that you don't want to run in only one direction. :P

Arcstarter

Sat Apr 10 2010, 11:05PM

Registered Member #1225 Joined: Sat Jan 12 2008, 01:24AM
Location: Beaumont, Texas, USA
Posts: 2253

How come H bridges are used? I don't understand how a H bridge would work as it supplies AC power

Well, you can use an H bridge to control the direction the motor spins. The H bridge does not switch at any real 'frequency', it is either forward, reverse, and off.

If Q drive is high, the motor spins one way, if inverted Q drive is high, the motor spins the opposite way. That is how an H bridge in Tesla coils work, but they are repetitively turned on then reversed for the AC output.

~2hp of electric motors could do over 15mph, something like 25 would be easy, so terrain other than perfectly flat streets at 15mph should be a breeze, depending on the load weight, tires etc...

rp181

Sun Apr 11 2010, 02:40AM

Registered Member #1062 Joined: Tue Oct 16 2007, 02:01AM
Location:
Posts: 1529

Well, thats obvious. I was stuck on using full bridges with high frequency.

How should I implement speed control? I want to use PWM, but is their a better way than just connecting it to a IGBT or FET?

One last question: Does anyone know where I can good priced 8" wheels, wide rubber? not those skinny plastic spoked ones, as I said, its for decent terrain (loose sand, packed dirt, grass, steep grades).

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