Drive Train

Drive train component spreadsheet

Updated 21 Feb 2008

Gear ratio calculations

Updated 21 Feb 2008

Fixie considerations

How to do a multi-speed fixed gear bike

Jackshaft design

jackshaft-ray.jpg
Sketch made by R. Canzanese. Idea is that pedals, and motor will all drive the same jackshaft, which will drive the rear wheel. If the rear wheel contains a freehub, and the pedals' cogs at the jackshaft contain freewheels, the wheels can spin independently of the motor. This design assumes NO regenerative braking.

Transmission suggestions

trans.ppt (Layton)
trikedrive.ppt (Sadler)

Gear Ratio Calculations

Gear_Ratio_Spreadsheet.xls

Types of clutches

1. Tradtional cable/rod activated
2. Electromagnetic
3. Centrifugal
4. Free wheel arrangement

clutches.xls

Discussions

Brad, 2008.01.27

These are great questions with no simple answers. This is indeed where
the hard engineering comes in and probably explains why the Prius and
the Segway cost way more than they should. The first thing to do is to
determine what the mass penalty is for an overall optimization function.
There is probably a way to directly compare the energy storage of all
components of the vehicle as a function of time and to evaluate how much
each contributes. For example, if the yield stress of the steel in the
frame is 30ksi (200 MPa), then this represents an energy density of 200
MJ per cubic meter. As a comparison, the energy density of gasoline is
40 MJ per gallon. The power density of a human is 1 W/kg. Obviously
these units must be normalized by mass so that "dead weight" may be
trimmed from the vehicle. If the vehicle is to be traveling on flat
terrain with no stops and starts, then a regen system is not likely to
contribute since its energy will not be called upon. However in hilly
terrain, where a lot of braking occurs or in stop-and-go traffic where a
lot of braking occurs, the mass penalty may be worth the added energy
density.

It sounds like you already have a decent grasp of the efficiencies
involved. If you want to pursue this mathematically, I do have access to
a few good design books that could help frame the discussion.

I just rode my wife's Trek Soho to the grocery store with 80lbs of food
and was thinking several times how having an efficient regen system
would make the ride easier.

Rich Sadler, 2008.01.27

Ray,
If the motor can connect with the pedals during regen, couldn't you end
up in a situation where the wheel wants to drive the right & left
chains at different speeds? Let's say the wheel during regen is
rotating at 240 rpm & drives the motor with a chain on the left side of
the wheel thru a 10:1 ratio at 2000 rpm. There is either no freewheel
on the left side or there's one that's set up so the wheel drives the
motor when the motor's rotating slower than the wheel. There is a
freewheel between the motor & pedal drive train, but it will engage when
motor speed exceeds pedal speed, so it will start to turn the pedals.
Let's say the motor speed is geared 30:1 to the pedals, so the pedals
will be rotating at wheel speed times motor-wheel ratio divided by
motor-pedal ratio, or 240*10/30=80rpm. Now the pedal/wheel ratio varies
with what gear I'm in. Lets say I'm in a gear that gives me a 3.375:1
ratio- the wheel turns 3.375 revolutions every time the pedals turn one,
so with the pedals turning at 80 rpm, the wheel will want to turn 270
rpm. It can't, though- I said at the beginning that the wheel is turning
at 240 rpm, & it's the wheel that's driving the pedals around. I can't
have the left side going slower than the right side, so like the song
says, something's got to give. Note,this would cause no problem if the
pedals were in a ratio that makes the right side go slower than the left
side, because the wheel freewheel won't allow the wheel to drive the
pedals. The wheel freewheel will catch if the pedals want to drive the
wheel, though, as in the first scenario, & that's when the chain will
break. That's why I think you need a separate motor drive- you really
can't guarantee which gear you're pedaling in when you regen.
Rich

Ray, 2008.01.27

Brad -

Joe and I were discussing the implementation of regenerative braking
yesterday afternoon, trying to determine the simplest way of
implementing it. One question we didn't have an answer to was how easy
it would be to disable the freewheel on a hub.

The other topic we were discussing was the effectiveness of regenerative
braking in our vehicle. A market survey of other vehicles with
regenerative braking shows that these vehicles have one thing in common
— no transmission on the electric motor. This inevitably simplifies
the the operating of regenerative braking, but also, perhaps more
importantly, limits losses due to the transmission, and added mass of a
transmission.

Rich, regarding the seperate transmission for the pedaling and motor, we
wanted to avoid that to avoid overloading the driver. While having the
stoker shift will do that, it also requires a stoker in order to change
gears. We wanted the vehicle to also be pilotable by a singe rider.

Joe and I were tring to do some simple and dirty math to estimate the
efficiency of the regenerative braking. We estimated 10% losses of
energy during braking from external forces, 10% losses from the
transmission, and 70% motor efficiency. This yeilds a 55% efficiency of
regenerative braking, but still offers a best case scenario. There will
also be losses in the circuitry, and the electric motor, as it spins at
lower RPMs as the vehicle slows will inevitably fall below this 70%
efficiency. I have read in some sources of questionable validity that
the efficiency of regenerative braking on an electric bicycle is in the
area of 30%. We debated whether the added mass from additional gear
hubs, etc. and added financial cost of adding these components to the
vehicle is worthwhile.

I crunched some numbers for a vehicle driving river-to-river on Chestnut street and stopping at 1/4 of the traffic lights, but didn't come up with results that made sense, so I will rerun the numbers this afternoon to try to better quantify the effects of regenerative braking. I would like to see numbers that legitimize the additional cost, weight, and complexity of regenerative braking, as I continue to remain skeptical.

Ray

Rich Sadler, 2008.01.27

Brad,
That could work if you could somehow disable its freewheel so it could
drive your motor for regen as well as be driven by it. I had thought of
that but was hoping instead I could convince you to give up on regen,
for simplicity. Since your design seems to always have the motor doing
something- driving or regenerating but never just off, I think the
best method to use it would be to have the 3 speed hub frame
mounted, & chains running from motor to hub sprocket & hub shell to
right side of wheel. The two stage gearing this gives you would
facilitate large reduction ratios & the 3 speed could be shifted to
keep the motor somewhat near its power or efficiency peak. I'd then keep
the pedaling drive train entirely separate. The motor would drag if you
ever decided to pedal only or coast, but from your description of the
multi-freewheel system, you never operate in those modes. Maybe you
could have the stoker be in charge of shifting the motor so as not to
overload the captain.
Rich

Bradley E. Layton wrote:

Rich,
The roller idea is novel. What about a 3-speed internal hub instead?
-Brad

Rich Sadler, 2008.01.25

Joe,
Thanks for the explanation. I figured you could drop the generator
voltage if it was too high, but I didn't know you could also raise it if
it were too low. I had a kind of wacky idea for a motor drive that would
let you disengage the motor & also change its gear ratio. If you have a
spring loaded roller, driving the motor by belt, that can be pressed
against the wheel when you want regen, that could serve as your clutch.
If the roller were conical instead of cylindrical, and you can move the
roller sideways across the face of the wheel, you could vary the ratio
of wheel/motor speed (see attached powerpoint). If you wanted to pedal
without the motor, you can lift the roller off the wheel. It's probably
not as efficient a drive as a chain, but if you're just using it to get
rid of excess kinetic energy, who cares.
Rich Sadler

rollerdrive.ppt

Joe P. 2008.01.24

Kind of.
If motor-as-generator is used to charge a battery, a voltage regulator would be required. To hook up the generator directly to the battery without a regulator would be a potentially hazardous situation (battery overload leads to explosion/leak, fire, fumes, damage to motor or other electrical components, etc). The voltage regulator acts as a voltage-controlled current source to force a fixed voltage to appear across the regulator's output. The voltage controlled current source uses a feedback loop to monitor the output voltage and adjusts the voltage-controlled current source accordingly. Also, note that capacitors across the input (Vin to ground) can improve the voltage regulator response. See the following diagram.

Modern voltage regulators are capable of operating with input voltages below, at and above the desired output voltage, depending on current needs. This means that the lower the input voltage the lower the output current the converse is also true, the high the input voltage the higher the output current.

Most batteries (all but a few specialty batteries) are charged at a constant voltage where the current determines the speed at which the battery charges. (Take the chargers for dragonwagon 1.0, these chargers produce a constant voltage with a current that diminishes as the battery charges). Please see my note on batteries at the end for a more detailed explanation of batteries. So that is essentially what we would be doing with the voltage regulator. The voltage regulator would be able to operate within a wide range of input voltages so the voltage generated by the motor equaling the battery voltage is not an issue.

One issue that will come up is if the battery is "full" at this point the battery is unable handle any more current. At this point a short-circuit circuit will be implemented. This will discharge any further generated potential across a mostly resistive load. This will allow the regen breaking to still function without overloading the battery, yes energy is lost here but if you no longer have storage capacity for it what are you going to do.

Please note that a voltage-controlled current source is a power device which means that it requires energy to operate in most cases voltage regulators are only about 75% efficient. This leads to charging a battery is not an efficient option.

The alternative, using capacitors. The most common device used to store energy generated by regen-breaking is capacitors (generally of the super or ultra variety). Capacitor store charge because of this they work nicely for storing electrical power. In addition they are passive devices which means that their only losses are due to internal resistances which leads to a 95%+ efficiency. In the case of capacitors the voltage is kept low, around 2 volts which allows them to store large currents and power since Power = I^2*R (Power = V x I, V=I X R, ohms law, => Power = I^2*R). In this case a capacitor would need to be selected that is capable of storing much more energy then we could hope to produce via regen-breaking and in addition the same fail-safe for short-circuit circuit from the battery option will be implemented.

Note on batteries:
Batteries store energy in the form of chemical potential. This potential equates to an electrical potential (voltage) across the terminals of the battery. The chemical potential is set by the designer/manufacturer of the battery and, neglecting decay, this potential will remain constant over the life of the battery. The capacity of the battery then becomes the current*time in the battery (this is why car batteries are rated in amp-hours). The reason that the batteries tend to drop in voltage as they are used is because the current from the battery runs down and no-longer has the ability to sustain the Power and by extension the voltage (P=I*V remains constant). The conversion between chemical and electrical potential an inherently inefficient, but well understood process. On the other hand capacitors store power in an electrical field which means their is no conversion between energy types (ie no chemical to electrical or visa-versa) which leads to a much more efficient process. The traditional problem with capacitors is two fold, first, until recently the die-electric materials used to make them severely limited the ability to store power and the second is that they are typically non-linear devices which makes them hard to model.

Rich Sadler, 2008.01.24

So am I all wet with my contention that because motor-as-generator
output voltage is proportional to motor speed, below some speed the
voltage the motor generates will equal battery voltage & no more battery
charging will occur, or do you have something going on in the motor
controller to boost the voltage as the motor speed drops? Inquiring
minds need to know….

Brad, 2008.01.24

Thanks. I appreciate the advice to “ditch” the regen, but this is central to our design. I know it is complicated as we must figure out a way to allow the motor to either 1) drive, 2) be driven, or 3) be disengaged.

Yes, the rear wheel must be built symmetrically without dish.

Rich Sadler, 2008.01.24

I'm Rich, the other grumpy old man, not John. I'm the one who sent the sketch of the jackshft drivetrain. I would also run the motor to the jackshaft & ditch the regenerative braking- in fact, that's what I did on my own bike with electric assist. I think I put a picture of that in John's & my powerpoint presentation. My jackshaft was a rear hub with freewheel. The motor drove the hub shell thru a 6:1 reduction timing belt, then the hub drove the left crank thru a chain & sprocket attached to its freewheel. When I pedal with the motor off, the freewheel turns but the hub doesn't, so I don't have to overcome the frictional torque of the motor. On my system I have no controller- just an on/off motor switch. I only switch the motor on while the bike is moving, so the motor drivetrain never feels the stall torque of the motor.

Brad, I didn't quite understand your freewheel system for driving the motor in regen when I wrote my last post, but now I think I do. One unexpected mode of this is that if the motor is being driven by the wheel thru its one way clutch during regen, & the motor can drive the pedals thru another one-way clutch, then the pedals will be driven during regen. This might be disconcerting to the riders when descending or braking. If you must have regen, the cheapest clutch would be a roller that could be pressed against the rear wheel by a spring for regen, lifted off & latched up during other times. John Tetz made a system like this for a previous power assist using a weedwacker gasoline motor on a Lightning recumbent.

I see a lot of e-mail about wheels. One element of wheel construction that makes a rear wheel weak against side load is the non-symetry of the spoke angles. To keep the drive train close to the bike centerline on an upright bike (so the rider's legs & feet don't hit the chain), the hub flange that the spokes attach to are offset toward the wheel centerline on the chain side. This means those spokes run close to perendicular to the wheel axle, and are less able to handle side loads. In bicycle parlance, this offset is known as "dish". Since hubs don't get wider when the wheel diameter increases, the larger the wheel, the closer to perpendicular those chain side spokes are & the weaker they are against side loads. This is one reason why you are more likely to break a spoke on the chain side of the rear wheel. Recumbents don't need dish since your legs & feet aren't near the rear wheel, so can be built without dish & be stronger- y ou jus t have to allow for it in the design of the rear fork

Ray, 2008.01.24

Thanks for the sketch of the drive train. When I was doing the gear
calculations the other day, I made a sketch of the drive train that
looked very similar to what you have just sent. The difference in my
drawing was that the motor was also attached to the jackshaft. I
thought that this way, we could better utilize the motor by stepping it
through the gears on the rear wheel's casette during acceleration. A
gearing system for the electric motor is a necessity for us, since our
goal is to have the motor make a large contribution during acceleration.
I have yet to get together with Joe and compare models, but it appears
to me that the motor will be doing 75% of the work during acceleration.
I imagine a motor attached to the jackshaft such that it is rotating the
jackshaft at ~90 rpms at its ideal RPMs, whatever they may be (we have
yet to select a motor). The only caveats i see here are the startup
torque of the motor causing a lot of stress on the jackshaft, chain,
derailleur, casette, etc. I foresee a controller designed to mitigate
these effects.

As far as disengaging the motor, I feel that it is becoming more and
more evident that regenerative braking is going to be highly
complicated. My understanding regarding regenerative braking is that:

- We can't use derailleurs in the transmission design for the motor,
since they would undergo too much stress
- The transmission introduces a unique obstacle — how do you decide
which gear to "brake" in?

Taking these two issues into account, I foresee a big, expensive
transmission for performing regenerative braking. I have not run any
math to see how much energy we can even hope to recapture. I'll look
into the math this afternoon. But, if we decide not to have
regenearative braking, can't we just have a freewheel of the jackshaft
for connecting the electric motor. This seems to me to be the simplest,
lightest, most cost effective system.

Rich Sadler, 2008.01.23

Hi Brad,
I didn't realize you were going to have sprocket clusters on both sides
of the rear wheel & two derailers. Exercising my prerogative as a grumpy
old man, I'm saying I don't like it- too complicated, too much custom
stuff (left & right freewheels, upside down derailers, etc). I've
sketched out your attached powerpoint a drivetrain where both riders
drive a jackshaft which then drives the wheel as God & Shimano intended,
on the right side. The jackshaft also steps up the gear ratio so you
don't have to go looking for two72 tooth chainrings. This frees up the
left side of the wheel for a chain to the motor to drive & be driven by
the rear wheel. You lose the independent control of gear ratios by each
rider, but all other tandems have the captain decide when to shift-
especially with this bike he's the only one who can see what's coming up
the road. I left the idlers to maneuver the chains around obstacles as
an exercise for the student. See if this floats your boat.

BTW, I saw you're having trouble finding front wheels. Dan's Comp sells
an Odyssey M7 front wheel with a 20mm aluminum axle- maybe they or you
could swap the axle for a 20mm steel one. Here's the link.
http://www.danscomp.com/401120.php?cat=PARTS

Rich Sadler, 2008.01.21

No, my motor doesn't spin when I'm pedaling forward with the motor
switched off. Because the freewheel that isolates the motor from the
pedals catches when I pedal backwards, though, it does spin if I
back-pedal. I measured the torque the drive train & the motor absorb by
pulling on the pedal at a slow, steady speed with a hand-held spring
scale. The results are:
Pedaling forward in the lowest gear: 1.47 in-lb
Pedaling forward in the highest gear: 5.16 in-lb
pedaling backward: 26.8 in-lb
If I pedal at 70 rpm, the frictional drag of the motor will consume;
P=T*N/63000= 26.8*70/63000= .030hp=22.2watts.
This seems like a lot to throw away- what's so hard about including a
bicycle freewheel in your drivetrain? They're cheap & designed to handle
high torque.
I wonder how much energy you'll get back with regenerative braking. You
can't use the pedal drive train for this- if you try to pull on the
lower part of the chain, the derailer won't take up the chain slack
anymore & the chain will wrap itself in knots. You'd have to run a fixed
ratio motor drivetrain, and as the bike decelerated, the
motor-turned-generator would put out less & less voltage until its
output voltage was less than battery voltage & regeneration would cease.
Meanwhile when pedaling, you're stuck with a constant 22 watt loss &
when the motor is running , it will only be at peak efficiency only at
one vehicle speed, so you'll chew through the battery faster. Both
John's & my bike put the motor through the bike's drivetrain so we can
use the gears that impedance match our legs to also impedance match the
motor.

Eric, 2008.01.21

Doesn't this bring our "torque" sensing system into effect - the motor is almost always active - if we are pedaling the motor is working as well. Also, a freewheel will not allow us to use regenerative braking. How much loss does really translate to when compared with the aero-drag? I understand the need for gearing, how high rpms will we see with the "magmotors'?

John Tetz, 2008.01.21

…..Absolutely not. The drag is way too high.

I saw a statement something about a 1500 rpm motor having no appreciable drag.

That 1500 rpm motor has to go thru a gear reduction. If you driving the chain rings they turn over around 70 to 100 rpm. So its not just the motor but the gearing you have to spin up to speed.

Another statement

….Yikes !!!! These are huge numbers.

Look at my Human Power Capability plot. The rider sustainable is do-able. 150 watts per rider means the time to fatigue is 3 hours for the Healthy Humans.
The 600 to 800 is high. At 300 watts per rider that brings the time to fatigue down to maybe a few minutes.
400 is seconds. Even for trained riders - not first class athletes. .

Also remember these numbers come from very active people. Put them on a recumbent that they haven't had time to build the muscles those numbers need to be cut way down. Even if you are riding an up-right a lot the recumbent muscles are quite a bit different.

Ok tying the above numbers into a motor needing to be in the 400 watt range. A 1500 rpm motor will be HUGE and heavy. So that’s out. You could find a smaller one but the rpm goes up. That means the gearing ratio has to be higher. Spinning thru the gearing and turning over a even medium size motor will do you in.

Over the years I have found that a rider can easily feel an increase in power of 8 watts, so much so they will often drop down one gear to reduce speed and reduce that power load. Lets say they are pushing at 140 watts. Load them down to 148 watts and they will for sure want to drop speed to reduce that load.
Have that power drag on all the time simply will wear out the rider. The cost of a freewheel is extremely low.

I have made a power measurement on my chain drive system with the rear wheel of the ground. It takes about 30 watts to spin the motor/gear box (19:1) and the entire chain line. Unloaded the motor was running at its max at 7,400 rpm. So of course the 30 watts is abnormally high. I don’t know what the motor drag by itself is yet. I'll make that measurement in the summer when I take the trike out of the shell.
Part of the reason is my motor has a lot more drag because it has permanent magnets in it. Its acting as a generator with various losses. A simple brush motor may have a bit less drag but not much less (brush drag etc).

Don’t forget on a bigger motor the mass of the armature needs to be accelerated up to speed. Accelerating power is short time thing but its huge. And your accelerating over and over and over again during a ride. Yes the motor system is much less mass but its spinning quite fast. I bet Rich could do all the comparative calculations.

I sent Brad my Power calculating spreadsheet which I didn't have time to go over with you. Put in some of the user values (underlined cells only - make a copy of the program first). Then see what kind of power is required to sustain a particular speed. Then type in an acceleration value such 0.3 mph per sec and see what the power has jumped to.

For an unfaired trike the CdA will be 3 to 3.4 sq ft. The Crr (rolling resistance) you can use .006
A decently faired one will be 1.5 sqft.

Joe P, 2008.01.20

To give you an idea of the losses associated with free-spinning the
motor the frictional torque acting on a free-spinning typical motor
shaft is around 1 ox-in (0.0075 Newton-meters). The power loss would
depend on the particular angular-speed of the shaft. If we assume the
motor is running at 1500rpm -> 25Hz then the power loss would be 0.0075
Newton-meters * 25 Hz = 0.19 Watts. How does the clutch option compare
to this?

Eric, 2008.01.20

Attached is a sheet of clutch options. I believe we may have wavered from our initial plan for electrical disconnect. Initially we had talked about electrically separating the motor and figuring on free wheels for both passenger and driver. This is the best route as it keeps our weight (rotational and stationary), complexity and excess power consumption to a minimum. In addition adding a clutch will only add cost to our already limited budget. I am not avoiding the clutch, but I am trying to rationalize our spending.

The current dragon wagon does not posses a clutch on the motor (or anywhere), and the motor shaft spins freely during non operation. While I do agree the motor *may* impose a slight resistance when disconnected, when I did my calculations which were presented at the end of fall term, the loses from that setup were marginal compared to wind drag on the vehicle. Motor design dictates this matter greatly.

Perhaps John Tetz can provide some insight here. I also believe Joe feels the same way about motor freewheel loses, and I included his email below.

The best option for simplicity sake is a electromagnetic clutch, but then you lose power to the electrical coil. As for a centrifugal setup, I am not sure if we will have rpms high enough to generate enough force for it to function properly. Mechanical is probably out of the question the system would have to fixtured to the vehicle, and controls added that would have to be operated by the rider, etc.

Something of interest is that combined clutch brakes are available, which would make the system more feasible, but I have not inquired into the pricing on that. I will try on monday.

If we must have a disconnect for the passenger, which was in the original email, then I would recommend including freewheels for both sets of pedals and a quick release on the passenger pedals. We are only wasting funds by including a clutch for them.

Brad, 2008.01.15

Good questions. Joe, good answers.

1) These are the penalties I see with a clutch:
Mass, driver distraction, probability of malfunction.
These are the primary advantages:
Allowing the motor to be neither driving nor driven. I personally cannot
think of a mechanism other than a clutch that will allow for both
options. A freewheel, which is a ratchet-and-pawl mechanism is
essentially a mechanical diode, allowing for transmission of torque in
one direction, but not the other.

2) Give me a spreadsheet with costs of the clutches you are considering.
The grant writing team and I will consider costs.

Joe P, 2008.01.15

My understanding as to the purpose of the clutch on the motor is that it
will prevent the motor shaft from rotating when the motor is not being
used to propel the vehicle nor is it being used in the regenerative
breaking. Such a configuration would prevent power loss due to
free-spinning the motor shaft. If this is indeed the case (someone
please correct me here if I am not understanding this correctly) then
the clutch would need to be engaged when the motor is on, or the
regenerative breaking is on, both of these events are triggered by an
electrical event (switch for regenerative breaking and throttle for
motoring). This makes me think that an electromagnetic clutch would be
the easiest to implement.

With that said, what are the power losses associated with each clutch
design? As long as we keep the motor under 4 HP and use a brushless
design the losses associated with free-spinning of the motor shaft are
near negligible (I would imagine that if you have any part of the clutch
that has to continuously rotate then the losses associated with this
would be greater then or equal to the losses associated with
free-spinning the motor). So I think it might be best to leave the
clutch out of the motor side of things.

As for the passenger drive clutch I believe the question is four fold,
first what is the efficiency(power) of each clutch option, second what
is the mass penalty of each clutch design, third what is price and forth
will it do what we need?

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