I recently discovered electric vehicles and I've started researching a conversion. I have a ton of questions though, most related to the engine and batteries. It's been too long since I took a physics class, I hope I've still got everything straight. I'd appreciate any help people could give me.
The minimum specs I'd consider to make a conversion worth it are:
~90mph top speed
30+ mile range
I'm not going to offer any opinions at all, pure checkable facts only. I'm not even going to calculate anything for you. Well maybe some examples, but not specific to your situation. Please feel free to do the calculations and post. I'm happy to check your calculations, but not do them. If you wonder why, please check the archives...
Basically I need to know how to calculate these (variables, formulas) so I can choose components that meet my needs.
No problem, it's quite straight forward. You need to find out two things about your donor bike. Its maximum horsepower and its top speed. You'll also need to estimate the final weight. Everything from there is a simple conversion.
Power required is simply related to the speed in a cubic relationship. Go twice as fast, need 8 times the power. Go half as fast, need one eighth of the power. Go third as fast, need one twenty seventh of the power.
So the first calculation you need to do is find the ratio between the top speed of the donor bike and the speed you want: 90 mph. Say for example the donor bike can do 180 mph. Simple, the final bike goes half the speed of the donor, it needs one eighth of the power. Look up the power of the donor bike, divide by 8 and that's the power you need.
You will use that info twice. Once to work out the size of the motor, controller, wiring. Secondly as *part* of the calculations to work out the battery size you need.
So you have worked out the power in watts. (if you've done it in hp then multiply by 745) Now work out how long it will take to cover the distance at that speed. Express that figure in hours. Mulitply the number of watts by the number of hours. The result is in watt hours. Divide by 1000 to get kWh.
This assumes that you ride on the flat and you never stop and start. If you don't ride on the flat figure out your elevation change that you may need. The energy of change of height is at a first approximation, Metres X Kilograms X 10. The answer is in Joules. Battery capacity is calculated in kWh. To convert divide by 3.6 million. Say your bike weighs 250 kg, you weigh 100 kg and you want to ride up a total of 1500 metres (all the hills added together). 1500 X (250+100) X 10 / 3 600 000 = 1.45 kWh. Notice that the steepness of the hills is *not* part of the calculation.
If you want to stop and start the energy will go into the brakes and be lost as heat unless you have regen braking, forget that unless you can figure out how to do it on the front wheel. If you're mad for regen then google Swigz and see what's involved in doing that. To figure out the energy lost it's the half the total mass in kg X velocity in m/s. Again the answer is in Joules. Again, divide by 3.6 million to convert to kWh.
Add up the energy needed for riding, climbing and restarting. The answer will be in kWh. That's the energy that needs to arrive at the back wheel to do what you want. Of more importance to you is how many kWh you need to have in the battery.
Take the result from the above operation and divide by the efficiency of the motor (generally about 0.
and the efficiency of the controller (generally better than 0.95). For example, say you calculated you need 12 kWh at the back wheel. Divide 12 by 0.8 and divide again by 0.95. That's 15.8 kWh
Now you've got the amount of energy that you need to be able to draw from the battery pack. You're nearly there!
There are three main choices for battery packs, Lithium (LiFePo is the right one for Electric Vehicles), NiMh/NiCd and Lead Acid.
NiMh/NiCd batteries are happy to be completely discharged. Lithiums shouldn't be drawn down below 20% remaining and Lead Acid shouldn't be drawn down below 50% remaining. If you draw faster than the test rate then the efficiency of the batteries will be less. You're talking about going 30 miles at 90 mph. That means full discharge in 1/3 of an hour. (sometimes expressed as 3C, see below for a discussion of what that really means. Almost all Lead Acid batteries are rated by discharging them for 20 hours (0.05C, often expressed as 20/C). Draining them at 3C they'll produce about 60% of their rated energy. Lithiums and NiMh/NiCd are a bit better in that regard, but say 80% of their rated capacity.
So depending on the chemistry chosen you should divide by those figures to get the size of the pack.
So for instance, Lead Acid, 12 kWh to the rear wheel: 12 X 0.8 (the motor efficiency) X 0.95 (the controller efficiency) X 0.5 (the maximum discharge) X 0.6 (the loss of capacity due to high discharge rate). So the battery pack you need (for example only) is 52.6 kWh. To work out the number of amp hours you need, convert to wh rather than kWh (mulitply by 1000) divide by the voltage that you are using. Say for example only, 72 volts. So you'd need to divide 52600 watts by 72 to get 730 Amp Hours. To work out how many actual cells you need, divide the voltage you want by the voltage of each cell. Lead Acid is normally a nominal 12 volts, however that's actually six 2 volt cells. So you can use 36, 2 volt cells, or 18, 4 volt batteries or 12, 6 volt batteries or 6, 12 volt batteries. They will be connected in “seriesâ€. That is to say, positive connected to negative to make a string.
To get the amp hours you need you will have to either buy 730 ah batteries, or put two 365 ah batteries in parallel, or three 240 ah batteries in parallel or... you get the idea. Parallel is when you connect the positive to positive and negative to negative.
So to get both the voltage and the amp hours you need you have to have both parallel and series together. You often see this expressed as s and p. 18s4p would be 18 cells in series with 4 cells in parallel. Using say 20 Ah A123 brand cells (which produce about 3.2 volts) that would get you 4 X 20 Ah (80 Ah) at 18 X 3.2 volts (57 volts). Multiply those figures gets you Wh. 80 X 57 = 4560. Divide by 1000 gives you 4.5 kWh.
Motor:
I've done some research into DC vs. AC motors and the consensus seemed to be "go with what fits your needs/budget" which is kinda vague. It looks like AC motors are more expensive (with more expensive controllers), but can have more hp and higher rpms. For the speed I want with decent acceleration I'm assuming I'd need at least 72V. Are there benefits to going higher? The way I understand it, horsepower is a function of wattage and torque is a function of current. Is this correct? How can you estimate the top speed of the bike? Is it strictly a function of max rpms of the motor, sprocket ratios, and tire size? Do other things like weight of the bike/rider and wind resistance come into play or are those negligible? Does this mean I should choose a motor primarily based on max sustained rpms?
There's a lot of questions there, but I think most of them are answered above. It's really wind resistance that counts above all. Generally a lower rpm motor is more suitable as the back wheel turns comparatively slowly. Lower rpm motors need less reduction in the gearing (smaller rear sprockets). Have a look around and you will see the giant sprockets that end up on the back of some electric bikes. The benefit of going to higher voltages is smaller wiring. The amperages going through even low powered bikes are huge by normal electrical standards. The largest household wiring copes with about 20 amps. The Zero S/DS has 500 amps at 48 volts flowing through it (and is much less powerful than you would need)
What is the difference between these two motors?
http://www.electricmotorsport.com/store/ems_ev_parts_kits_emc-rt_7234.php
http://www.electricmotorsport.com/store/ems_ev_parts_kits_perm_7234.php
The cheaper one looks like it has better specs.
Assuming 17" rear wheel I calculate: 17" x 3.14 x 3700rpm x 60min/h / 12"/' / 5280'/mi = 187mph with even gearing
Is that right?
Without a torque curve (which they don't give) you can't calculate how long it would take you to reach that speed, but it should be reached eventually right?
No, that's not right. It would never reach 187 mph. As you go faster more energy is expended by disturbing the air. When the rate of energy used disturbing air equals the maximum rate of energy conversion of the system (battery, controller, motor) then you have reached top speed and can't go any faster.
Batteries:
It looks like Lithium Iron Phosphate is the way to go. The cheapest I've seen is ~$450/kWh vs. ~$220/kWh for SLA.
Is it safe to assume I should ignore "maximum" rates (current, discharge, torque, rpms, etc.) and base everything on maximum sustained rates?
Can someone explain battery discharging. Battery discharge rates are listed in coulombs and all seem to be ~3. How do I find out the maximum current output using this? I know C=As but I'm not sure how to use it to find what I want.
C is an ad hoc measure, not an SI unit. What it means is the capacity of the battery. So for example a 26 Ah battery discharged at 52 amps is being discharged at 2C. If it's discharged at 13 amps that's 0.5C. Confusingly this is also expressed as a divided by C measure. So 13 amps from a 26 amp hour battery would be 2/C. Even more confusingly, it's only in recent times that batteries could be discharged at 10 or 20C. So in the past when someone wanted to write a 20 hour discharge they'd write it as 20C because it was obvious that they meant 20/C so they just left the / symbol out.
Is there a benefit to exceeding 72V? Would you need less current so it would run cooler and last longer?
Yes, but increasing the voltage makes arcing more likely so brushed motors will be destroyed if you run them too far above their rated voltage. Brushless (AC) motors don't suffer from that so they tent to have (as you noticed) higher voltages and more power. However they're more expensive and need more complex controllers.
Is the calculation for range: Range = battery AH / max sustained amps x top speed?
See the first answer.
Cheers Jason =:)
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Topic: Researching a conversion, have some questions (Read 1814 times)