/diy/ - Do It Yourself

bump help me please diy :(

bump

>>615978

/diy/ is slow. Your thread will be around for days. Don't bump.

I don't know the answer to your question though but be patient.

>>615950
your machine generates ~1.53 watts of power in mechanical motion.

If you consider that your output, and the solar energy of 900 w.m^-2 your input at 0.24m^2 surface area,
net efficiency ~= 0.7%,

However, even if your vehicle only takes 5N to pull at constant velocity, that's not to say the motors have 100% power input available- i.e. using PWM to drive them means you only have the duty cycle as your input to begin with.

>>615950
>What is the efficiency of the whole process?
>involves electrolysis of water
"Extremely low".

>>615950
>I'm a little confused with the Watt/hours
basically
work(joules) = force(N)*distance(m)
power(joules.sec^1 or watts) = work(joules)/time(seconds)
energy(watt.hour) = power(watts)*time(hours)

>>615981
but I have 3.85 Watts available for the motor and if it's only doing work for 1.53 watts and an efficiency of 66%, could it mean I use a wrong motor, wrong torque or am I way off?
>>615982
sorry
>>615983
we.. at 2 volts and 0.25 amps its 78% efficient , at 6 volt 4.5 amps only 27% so yeahh..but sun is kind of free so doesnt matter in the end
>>615984
thanks

>>615981
by the way, isnt it only 0.5% efficient?
the 216 watts are provided for a full hour, while 43 minutes are not a full hour.
so 1.095/216 ~ 0.51% efficiency?

Those 900 W m^-2 are for surfaces perpendicular to the light rays. If your solar panel is tilted, then your captured power goes down. The area you must use for your computations is the projected area in the direction of the solar rays. That might explain part of your apparent loss of efficiency.

>>616011
the panels are orientating themselves towards the sun with 2 servo motors, horiizontally and vertically

bump

>>616000
>but I have 3.85 Watts available for the motor and if it's only doing work for 1.53 watts
that's not how it works. You don't have power available, you have potential energy available for the motor to use. You might have a 12V battery capable of supplying 100A, running a 10 watt 12v motor, but it doesn't mean your system is 0.8% efficient.

>>616000
Please explain that figure of 3.85 W available power.
Also, how did you measure the motor efficiency? Because a good bet is that the efficiency of the motor is wrong. If the torque is too high, the efficiency will go down quickly. Your figure is compatible with a ~10 % motor efficiency.

bump

>>615950
Work = force*distance
Work = 5*788 =3.94 kJ

Energy_in = insolation*area*time
Energy_in = 900*0.24*3600 = 777.6 kJ

Efficiency = work_done/energy_in
Efficiency = 0.00506 = 0.506%

>>615983
Electrolysis is relatively efficient.
You can easily hit 70% efficiency with common materials, and over 80% if you use platinum.

bump

>>616315
what are you bumping for?

>>616242
thank you, what i was looking for
>>616232
>>616226
ok understood. the 3.85 come from the fuel cell to the motor. producer says its 66% efficient motor. i guess i have a wrong toruqe, how do i find the right one?

>>616315
>>616318
>>616241
thats not me.

>>616409
torque=force*distance [see pic]

Your numbers are all reasonable.

There is always a difference between theoretical and experimental.

Theoretical Efficiency = product of all the component efficiencies

Expirimental = work done / energy in

This is a very important concept to understand.

>>616414
thanks for the picture
how would i claclulate it in pic related? the axis + 5 newton?

my motor says 90Ncm
i think thats a bit too much..?

>>616439
axis is 30 cm (15 cm each from middle where motor is attached with belt)

>>616442
oh and that 9gk/cm is max torque, it says 3kg/cm should be maximum permanent

>>616446
lol sorry for the spam but here is the technical datasheet if that helps
http://www.produktinfo.conrad.com/datenblaetter/225000-249999/227552-da-01-ml-GETRIEBEMOTOR_RB35_1zu50_de_en.pdf
and pic related

bump

>>616468
well thanks for the bump

>>616232
You're assuming the difference in efficiency is entirely in the motor.

>>616409
Its not that your torque Is wrong, its that your motor may not be under ideal loading, so you're getting less than the rated efficiency. It depends on the type of motor and motor controller you're using. If you use pulse with modulation ( pwm) you can increase the efficiency vs hitting the motor with pure direct current. You'd have to ask an electronic s guy for more detail.

I suspect you're losing a bit of energy transmitting power from the motor to the drive shaft. Gear trains are typically better than 95% efficient, but belt drives are notoriously lossy. If you're using a stiff belt over small pullies you could easily be losing 50% of the motor output power.

>>616497
i am using PWM but always have to hit full power to let it drive, otherwise it just makes this high pitched noise in the motor.
>Gear trains are typically better than 95% efficient, but belt drives are notoriously lossy.

but belt drives are 98% efficient according to wiki
>Timing belts with a helical offset tooth design are available. The helical offset tooth design forms a chevron pattern and causes the teeth to engage progressively. The chevron pattern design is self-aligning. The chevron pattern design does not make the noise that some timing belts make at certain speeds, and is more efficient at transferring power (up to 98%).
my tires are quite soft.. that may play a big factor i suppose but not sure.

anyone know how to calculate perfect torque?

>>616627
>i am using PWM but always have to hit full power to let it drive, otherwise it just makes this high pitched noise in the motor.
In that case you are probably getting close to the rated efficiency.

>but belt drives are 98% efficient according to wiki
No. That one style of highly engineered belt drive is 98% efficient. Unless your belt drive is appropriately sized and very well constructed I doubt you're a getting better than 90%.

What do you mean by 'perfect torque'?

>>616631
>No. That one style of highly engineered belt drive is 98% efficient. Unless your belt drive is appropriately sized and very well constructed I doubt you're a getting better than 90%.
well, i dont know how much power i am losing but I don't think i have a 50% power loss here, not sure though.
pic related, this is how it's installed. its quite tighted and mot much room to slide around.
>perfect torque
I mean what is the best motor for my vehicle, i think i have a wrong one, it's rated for 12 volt but im only running it on 8.75 and i just dont know how i chose the perfect motor, i picked this one almost randomly and it worked but there was not much maths/physics behind the decision.

Just from eyeballing I would guess that belt drive is about 90% efficient for your application.

Choosing the optimal motor is outside my knowledge base.

Are you trying to hit some efficiency mark? Because you're going to have an extremely difficult time getting a total system efficiency better than 2-3% with all the steps you're using. This isn't a bad thing. Just getting everything assembled and working is pretty impressive. Cool project.

What exactly are you trying to achieve at this point?

>>616643
90% is quite fine to me
hmm ok, gotta look into that or maybe somebody on here knows?
nope im not trying to hit a specific mark, just trying to reach the best efficiency possible.

>What exactly are you trying to achieve at this point?
actually I did it for my graduating project for school and grade was already made.
It interests me a lot though so here and there I'm reading up on some related stuff, watching vids, maybe planning similar projects soon but more real-life orientated.
Next thing would be to build a very efficient, high-producing electrolysis cell that can reach an inside pressure of 435 psi so I could fully fill the metal hydrides.

also trying out other electrode concepts and different materials, pic related, those are stainless steel electrodes, during the time the anode (right one) turned yellow-blue from the oxygen and the cathode (left one) turned blue from the hydrogen...no idea how.
platinating the electrodes would further increase the process but might be quite costy.

>>616449
The technical datasheet helps a lot.

>>616637
Look at the charts of the technical datasheet. You have straight lines relating the torque, the regime and the current. Use the left chart but scale the torque axis to the 0-8 kgf cm (0-0.784 N m) to reflect the fact that you are using the 1/50 gearbox. What you obtain is for 12 V. If I = current, n = regime, T = torque: I = I0 + I1 T; n = n0 - n1 T. I0 is the current under no load (T = 0), I1 is the slope of the current vs the torque, n0 is the regime under no load (T = 0) and n1 is the slope of the regime vs the torque.

Now assume you feed your motor at a voltage U. I0, I1 and n1 are fixed constants in this model, but n0, the regime under no load, depends on the voltage: n0 = n0(U). What's the dependency? This one: U = n / (dI/dT) + I R. You get R from the stall current (840 mA from the left chart): R = 12 V / 840 mA. Now you get n0(U) = dI/dT * (U - I0 R).

Finally: I = I0 + dI/dT * T; n = dI/dT * (U - I0 R) - dn/dT * T.

The efficiency is output power (T * n with n in rad/s) divided by input power (U * I): eta = T * [dI/dT * (U - I0 R) - dn/dT * T] / [U * (I0 + dI/dT * T)]. Whatever it gives. I will plot it later for U = 8.75 V.

>>616682
Replace dI/dT with I1 and dn/dT with n1. It took me a few minutes to write this and I mixed up the notation.

>>616682
>>616683
Pic related shows the efficiency curve of your gearmotor.

>>616439
As you know the force (5 N) and it seems there is no additional gearing (one revolution of the output shaft of the motor makes the wheels turn one revolution) the torque given by the motor is that force times the wheel radius (assuming no friction losses at that belt transmission and the axle).

>>616682
>>616683
>>616689
>>616691
first of all thanks for your effort, very appreciated.

>Use the left chart but scale the torque axis to the 0-8 kgf cm (0-0.784 N m)
I think you shifted 1 column, shouldn't it be 0-1.2 kgf cm looking at the 1/50 motor?
>What you obtain is for 12 V. If I = current, n = regime, T = torque: I = I0 + I1 T; n = n0 - n1 T. I0 is the current under no load (T = 0), I1 is the slope of the current vs the torque, n0 is the regime under no load (T = 0) and n1 is the slope of the regime vs the torque.
regime = revolutions per minute? sorry dont know that word in english

>This one: U = n / (dI/dT) + I R. You get R from the stall current (840 mA from the left chart): R = 12 V / 840 mA. Now you get n0(U) = dI/dT * (U - I0 R). [...]
lol that seems too complicated to me, nice work anon, shouldve paid more attention in physics but will try to understand and read it dozens of times again

>>616689
that chart is awesome, thanks anon. but should scale from 0-1.2 not 0-0.8, right?
>>616691
there is an addidional gearing, the motor has a 15 tooth gear and there's a 10 tooth gear attached to the axis.
so assuming there was no gearing and no friction losses it would be 0.3-0.35 Nm?

so this is not the perfect motor. right? (with the 0-0-8 scaling, wonder how it is with the 0-1.2 scaling))

thanks anon

>>616726
tire is 6-7cm in diameter (dont have it here right now) and very soft, so i assume there's friction loss there.

>>616728
and considering the 15 to 10 ratio this would mean the motor is then 1/33 ?

>>616726
>I think you shifted 1 column, shouldn't it be 0-1.2 kgf cm looking at the 1/50 motor?
I see. That datasheet is driving me nuts. The "geared motor torque/speed" states that the rated torque of the 1/50 gearmotor is 1.2 kgf cm, but the "geared motor characteristics" chart shows about 8 kgf cm (the stall torque at N = 0). Those charts express the torque in kgf / cm, but torques are force times length (kgf cm, N m...).

>regime = revolutions per minute?
Yes, the speed. I took the word from my native language. For the efficiency calculations, it should be expressed in rad/s: n(rad/s) = n(rpm) * (2 pi rad / 1 revolution) * (1 minute / 60 s).

>>616728
So 5 N times 6 cm is 0.3 N m. That's the torque at the wheels. The gearing ratio is 10/15: the torque at the motor is 15/10 times 0.3 N m = 0.45 N m. Assuming that the 0-0.8 N m rating is correct and not the 0-1.2 N m (mostly because it's kind of late and I don't have time to plot another efficiency curve), that leaves you with 20 % motor efficiency. The global efficiency without more losses would be 1 %. If we account for 90 % efficiency of the belt as some anon said and perhaps another 90 % because the wheels are very soft, we go down to 0.8 %. We are almost there...

>>616726
>>616740
It's me again, DC motor modeling guy. 0-0.8 N m is the correct range. That 1.2 kgf cm (~0.12 N m) rated torque is for maximum efficiency at the nominal 12 V, which coincides with my graph.

Last message for today.

So you would need a gearmotor giving around 0.45 N m (~4.6 kgf cm) for peak motor efficiency because you want to output 5 N with a 6 cm radius (~3.1 kgf cm) and a 15:10 reduction ratio at the pulleys. The slope of the efficiency curves is flatter for high torques than for low torques, so we should go with the gearmotor with maximum efficiency torque nearest to 4.6 kgf cm, but not bigger than that. That's the 1/200, which has its peak efficiency at 4.1 kgf cm.

>>616740
>1/50 gearmotor is 1.2 kgf cm, but the "geared motor characteristics" chart shows about 8 kgf cm (the stall torque at N = 0).
according to pic in >>616449 stall torque should be 9kgf cm but chart seems like it's 8 kgf cm
>Those charts express the torque in kgf / cm, but torques are force times length (kgf cm, N m...)
i think they made some mistake in the lower charts writing kgf / cm because in "geared motor torque/speed" they write "kg - cm"

>Yes, the speed. I took the word from my native language. For the efficiency calculations, it should be expressed in rad/s: n(rad/s) = n(rpm) * (2 pi rad / 1 revolution) * (1 minute / 60 s).
ok thanks

>The gearing ratio is 10/15: the torque at the motor is 15/10 times 0.3 N m = 0.45 N m
im confused with this one, the motor has a 15 tooth and axis a 10 tooth gear, that means instead of 120 rpm it will make 180 rpm and 1/33 rated torque?
how does it change the torque of 0.3 Nm since force and diameter stay the same?
shouldnt the curve change to the right or left?
either that or im really confused right now lol
its late here too...soon 2 am

>>616753
ok thank you very much. very helpful.
maybe you could explain me my last post tomorrow? good thing threads dont 404 here in a long time

>>616755
The motor has 15 teeth and the axle has 10 teeth, so it takes 2 revolutions of the motor to rotate the axle three times. Torques go the other way around: it takes 3 units of torque at the motor to output 2 units of torque at the axle. That's assuming no friction, of course: if there is friction, you need even more torque at the motor to give the same torque at the wheels.

>>616757
Also, if you switch the gears, so the 10 teeth gear is at the motor and the 15 teeth gear is at the axle, then the required torque at the motor would be 0.2 N m (2/3 times 0.3 N m instead of 3/2 times 0.3 N m) and the motor efficiency would be about 55 % if I didn't make any error in my plot.

>>616757
ok thanks very much. understood.
ill review all this tomorrow again, i really need to learn to make such awesome plots and find out point of max efficiency. thanks again. night.

>>616758
i should totally do this. thanks

bump to not let it die. still need to figure out how to make those plots.

>>617103
First of all, the stall current of on the second page of the datasheet is wrong. It cannot be higher than the stall current without gears unless there is some magnetic gear magic going on.

Second, this is what you need to model micromotors. If N is the speed, I is the current, T is the torque, and U is the applied voltage, you have these equations:

U = c1 N + I R;
I = I0 + c2 T.
N = N0 - c3 T.

R, I0, N0, c1, c2 and c3 are constants you can deduce from taking points from the datasheet.

The efficiency of the motor is the output power divided by the input power: (T N) / (U I).

>>617209
My bad. The speed under no load, N0, is not a constant; it depends on the voltage U.

U = c1 N + I R.
Under zero load (T = 0):
U = c1 N0(U) + I0 R => N0(U) = (U - I0 R) / c1.

>>617209
>>617213
And now to the datasheet. I eyeball-measured the following points from the charts:

Nominal voltage: Unominal = 12 V.
Current under no load (T = 0): I0 = 40 mA.
Speed under no load (T = 0) at the nominal voltage (U = Unominal): N0(Unominal) = 12.6 rad / s.
Stall current at the nominal voltage (U = Unominal): Istall(Unominal) = 840 mA.
Stall torque at the nominal voltage (U = Unominal): Tstall(Unominal) = 0.785 N m.

We go to the voltage equation when the motor is stalled to get the winding resistance R: Unominal = Istall R => R = 14.3 ohm.

Now we use the same equation under no load to get c1: Unominal = c1 N0 + I0 R => c1 = 0.907 V s / rad.

Now we go to the current equation at the stall condition to get c2: Istall = I0 + c2 Tstall => c2 = 1.02 A / (N m).

Now we go to the speed equation under the stall condition to get c3: 0 = N0 - c3 Tstall => c3 = 16.1 rad / (s N m).

Now remember that N0 = (U - I0 R) / c1. The efficiency is eta = (N T) / (I U) = [(N0 - c3 T) T] / [(I0 + c2 T) U] = {[(U - I0 R) / c1 - c3 T] T} / {[I0 + c2 T] U}. You can plot it versus the torque T for a given voltage U.

>>617209
>>617213
And finally, the theory behind this.

Permanent magnet, brushed, DC motors behave electrically like a resistance (the winding resistance R) in series with a voltage drop that grows linearly with the speed (the c1 N term): U = c1 N + I R.

The torque and the current are related by a straight line. There is a small current, I0, that is needed to keep the motor running even with no output torque. I = I0 + c2 T.

The speed and the torque have a straight line relationship, too. You could deduce it from the U = c1 N + I R and I = I0 + c2 T equations or get the N0 and c3 parameters directly from the datasheet. If I compute those parameters from the previous ones instead of going back to the datasheet, I obtain c3 = R c2 / c1 = 16.1 rad / (s N m), which is the same value I obtained from the datasheet. This confirms that the model is consistent and fits well what the datasheet says.

>>617209
>>617213
>>617215
>>617223
thank you very much. will look closely into this later, have school now.

That is what makes me come to /diy/.

Thanks.

An efficiency map of the 1/50 gearmotor because /diy/ rocks.

op bump here, hadnt time to look at it closeley yet, was very busy last few days but have time tomorrow.
so bump that it doesnt die.
>Now remember that N0 = (U - I0 R) / c1. The efficiency is eta = (N T) / (I U) = [(N0 - c3 T) T] / [(I0 + c2 T) U] = {[(U - I0 R) / c1 - c3 T] T} / {[I0 + c2 T] U}. You can plot it versus the torque T for a given voltage U.
>[(N0 - c3 T) T] / [(I0 + c2 T) U] = {[(U - I0 R) / c1 - c3 T] T}
mind=blown, lol ill need some time to process this information, shouldve paid more attention in physics

>>617591
wow man, thanks for all your effort. that's really incredible.
what program are you using to make these plots?

>>618035
I use gnuplot. I started using it many years ago. I do a lot of plots at work, so it's paid off learning to use it.

>>617209
>First of all, the stall current of on the second page of the datasheet is wrong. It cannot be higher than the stall current without gears unless there is some magnetic gear magic going on
I tested the motor today: no load current: 0.04A
8.75 Volt: Istall = 1.15A
12 Volt: Istall = 1.55A
also I measured my wheel, it's 8.2 cm in diameter, so 4.1 cm radius, so we are getting closer to like 0.2 Nm, thus the efficiency must be higher than assumed. if we consider the 10/15 gearing that's 0.35 Nm
so>>617215
R = 7.75
c1 = 0.9277
c2 = 1.923
c3 = the same

>>617209
>>617213
>>617215
>>617223
still do not fully understand the physics/maths but will try.
do you have any recommendation on literature or websites with easy examples?

>>617591
>>616689
>>618050
could you copy&paste the gnuplot code/input of both plots, so I can experiment with it? I installed the programm now but it seems a bit complicated to me lol but I feel that it's gonna be useful if I understand it well

>>618050
what did you study/are you studying by the way?

>>618376
Aerospace engineering. We had a subject on electrical machines.

>>618375
Perhaps we need to take dynamic measurements here in order to get more information on the motor because with these parameters, the maximum efficiency is kind of low and mismatches other figures from the datasheet.

gnuplot code for plotting the efficiency-vs-torque graph:

Unominal = 12.
I0 = 40e-3
N0 = 12.6
Istall = 1.55
Tstall = 0.785

R = Unominal / Istall
c1 = (Unominal - I0 * R) / N0
c2 = (Istall - I0) / Tstall
c3 = N0 / Tstall

eta (T, U) = (((U - I0 * R) / c1 - c3 * T) * T) / ((I0 + c2 * T) * U)

set xlabel 'Torque [N m]'
set ylabel 'Efficiency'
set grid

plot [0 : Tstall] [0 : 1] eta (x, 12.) linewidth 5 title '12 V', eta (x, 8.75) linewidth 5 title '8.75 V'

>>618375
gnuplot code for the map:

Unominal = 12.
I0 = 40e-3
N0 = 12.6
Istall = 1.55
Tstall = 0.785

R = Unominal / Istall
c1 = (Unominal - I0 * R) / N0
c2 = (Istall - I0) / Tstall
c3 = N0 / Tstall

eta (T, U) = (((U - I0 * R) / c1 - c3 * T) * T) / ((I0 + c2 * T) * U)

set xlabel 'Torque [N m]'
set ylabel 'Voltage [V]'
set view map
unset surface
set cntrparam levels incremental 0, 0.05
set contour
set grid
set key bottom right
set samples 100
set isosamples 100

splot [0 : Tstall] [0 : Unominal] eta (x, y) linewidth 5 title 'Efficiency'

>>615950
>If the sun is shining on my panels for 1 hour it produces 216 Watt/hours, right?

no

I'll focus on the solar part of efficiency.
The sun emits an nice near black body radiation spectrum and the atmosphere absorbs some parts of it. The total watts that actually hit the ground depend on what time of year (6 % variation due to the suns angle at noon and the distance away from the sun), the part of the globe you are in (the atmosphere is thicker is due to sun's angle), the time of day (obviously), and the angle of the solar panel (should be directly to the sun).

The most accepted value for the US is AM 1.5 G which is 1000 watts. So that is 1000 joules per second. Multiply that by 1 hour 3.6mega joules or simply 1000 watt*hours. Times that by 15% efficiency you 150 Watthours or 540kilojoules.

watt/hours is a power rate or (1/3600 joules/seconds^2) it can also be called energy acceleration(notice the seconds squared in the denominator). watt*hours is energy.

usually 15% is a rounded number. What type of solar cell is it?

>>618422
>>618427
thanks works great.
now I wonder how the stall current can be 1.55 Ampere as now the max efficiency under ideal load is 40%?
>dynamic measurements
what do you mean by that?

>>618631
hey there
hmm but so 216 Wh is still correct because max sunshine is 900 W here (looked it up) and the surface area is 0.24m^2.
(32.4 Wh considering the 15%)

cells are polycristalline according to manufacturer but in my eyes some seem like monocristalline.

>>618422
>Aerospace engineering.
Very nice. Kudos!
any private project you're working on?

>>618960
>>618631
also the panels are self-orientating towards the sun. so always 90°angle

>>618960
>>dynamic measurements
>what do you mean by that?
I mean that perhaps you should measure the current when the motor is running and also measure the speed of the motor. That way, you will have a data point to work with it. The datasheet contains inconsistent data or perhaps the motor is sort of nonlinear in its behaviour and the straight-line model I applied is invalid (but the curves of page 1 are the usual straight lines). If I compute the c3 parameter with the stall condition, I obtain one value which is ~10 % lower than the value I would obtain from the 100 RPM, 250 mA, 1.2 kgf cm condition from page 2.

>>618962
No fancy flying machine yet. These days I'm mostly doing cheap, ground-based and hand-held, electronic and electromechanical gizmos because busy saving ant mode.

>>619035
current is between 0.35 and 0.44 Ampere when running in my vehicle. diameter of wheel is 8.2 cm and it's running at 1.091 km / hour. Force is 5 Newton and weight of vehicle 4.6 kg if that helps anything.
so thats like 71 revolutions per minute - the 10/15 gearing?
hmm so with which data would you go with if you wanted the most precise calculations?

>>619411
also because the wheel is so soft it'll be more like 7.7-7.8 in diameter.

>>618960
does your surface area take into account the empty spaces between the active device?

this matters.

>>619489
Yeah it's taken into account, otherwise it'd be 0.35 m^2

>>619411
5 N at 1.091 km / hour means the output power is 1,52 W. 0.35 A at 8.75 V would be 3.06 W of input power at the motor, so the efficiency of the motor + transmission + soft wheels would be about 50 %. 0.44 A at 8.75 V would be 3.85 W, so the efficiency would rop to 39 %.

Now we can model the gearmotor. 1.091 km / hour with a wheel radius of 4.1 cm requires an axle speed of 70.6 RPM or 7.39 rad / s. The speed at the shaft of the gearmotor would be 47.1 RPM or 4.93 rad / s. The torque at the axle is 5 N times 4.1 cm = 0.205 N m; the torque at the shaft of the gearmotor is 0.308 N m. Go with these parameters to the model and the maximum efficiency will be so low and the model fails so hard that it is not even funny. I'm puzzled by this. I will take a closer look at this tomorrow.

>>619583
I can't into motors at this time in the night. I introduced wrong numbers in the model. Now I fit your data.

The model I am using is somewhat invalid, though, because I am not taking into account the losses at the soft wheels. Part of the output torque of the motor goes into straining those wheels and the driving belt, so the torque at the motor is actually higher than what gives the 5 N output force. This means that the input data of the model is plain wrong and the curves we can plot, although legit-looking, can be rubbish.

>>619583
>>619609
hmm seems really complicated then
but I think 5 Newton must be right, because when I measured it, i also fastened the toothed belt really tight.
So when pulling my vehicle at constant speed the motor acted as a generator and therefore as a resistance like it would be running the other way round.

Or did I actually measure something wrong / wrong method?

Here's a gif for you, it's how I measured the sum of all counter-forces (axle friction, belt loss, motor/generator friction, gravity pull, air resistance, wheel friction/softness...) for pulling the vehicle at constant speed.
http://i.minus.com/ioOFtjTxFUnz8.gif

you can see it fluctuating around 5 Newton

>>619871
bump

Bump

Bump

dc motor guy are you still alive?

Hi. I'm back.
>>619871
How was the motor wired when you towed the vehicle to measure the resistance? Did you short the leads with an ammeter to measure the current?

>>622846
hey, wb
it was wired as if it was wired when I'd run it normally with my hydrogen fuel cell just that it was not connected to the hydrogen fuel cell. the current had nowhere to go.

i could have measured the voltage and amperage of the motor when towing it but then the speed of the vehicle towing my vehicle would determine those values? should I have towed it with the exact speed my vehicle reaches at max speed?

>>623031
An ideal DC motor has an output torque that is directly proportional to the current. As a real motor has some friction, the actual relationship between the output torque and the current is not a direct proportionality, but a straight line with a small offset (the current under no load). What happens is this: the output torque is the electromagnetic torque of the ideal motor minus the frictional loses.

You measured the force with the motor terminals in open circuit: no current. That means that you measured the resultant of the rolling resistance, the loses at the belt, and the friction at the gearmotor. We cannot distinguish the contribution of the friction at the motor from the other contributions, so we end up counting it twice if we use the current under no load (actually, the current needed to offset the friction at the motor) as a data point.

The uncertainties we are handling here (what is the friction at the motor, the range of values of the current and of the force...) are large enough to make our computations vary noticeably. What we have seen so far is that we can roughly match the order of magnitude of the measured efficiency of the vehicle. If we had more accurate measurements of the individual components of the mechanism, we could get to the exact global efficiency by taking the product of the individual efficiencies. Inaccuracies in the knowledge of the individual efficiencies build up.

So what should I do concretely to get the most exact measurements?
1. should I measure the force without attaching the motor
Or
2. should I put an electric resistance equivalent to the motor usually has at 8.75 volt and 0.44 A when it is running and theb re-measure the force?

Or how would you do it/what would you recommend to do?

>2. should I put an electric resistance equivalent to the motor usually has at 8.75 volt and 0.44 A between the terminals when it is running and then re-measure the force?

Sry on my mobile atm

>>623362
I don't know about anyone else, but I am profoundly impressed and thank God for anons like you with expertise and knowledge to contribute to any OP's inquiries for solutions. That is what this site is all about,

>>624487
>>624488
You can measure the force without the motor. That way, you will almost isolate the component that is independent from the motor. I say almost because it will be tricky to obtain the same conditions for the belt when the motor is not connected to the pulley.

You can put several resistances between the terminals and measure the voltage and the force at different speeds. That way, you will have several data points to fit your curves. If you do this, you will have the curves of the system motor + belt + wheels on terrain. If you can fit well your points with straight lines, then it's cool. If you need more complicated curves, that's cool too because in any case you will be fitting your model to real data and you can't be more realistic than that.

There's the possibility that your data will be so noisy that the observed variations will give you wildly varying curves. Sometimes, you can improve your knowledge with elementary statistical tricks like taking the average of many measurements.

>>624938
hmm it is indeed tricky to obtain same conditions for the belt.
maybe if i detached the belt completely and assumed that it is >90% efficient?
>You can put several resistances between the terminals and measure the voltage and the force at different speeds
how would I go about that with the different forces?
would it be a solution to attach a string to the pulley and different wights at the end of it and measure distance, time and the given force (gravity)?

and if i didnt do any of this, which curve would you go with?
do you think >>617591
is accurate enough?

>>625882
or maybe if I turned my whole vehicle backwards so that it is facing the ground with its rear end and somehow fix it in the air, then attach a string to both wheels and lift different objects that i know the mass of?

almost 404d
keeping it alive
didnt save the posts yet

>>625882
>how would I go about that with the different forces?
The output torque is the sum of two things: one that does not depend on the current and one that is a constant times the current. The first summand depends on friction and other nasty things that makes us depart from an ideal motor. The second summand is due to the electromagnetic interactions between the rotor and the stator of the motor. You have one data point taken with the dynamometer: when there is no current (effectively infinite resistance between the terminals because there is only air). If you put known resistors, these resistors will let more or less current pass and the force you measure with the dynamometer while towing the vehicle should vary in consequence.

>would it be a solution to attach a string to the pulley and different wights at the end of it and measure distance, time and the given force (gravity)?
Tricky stuff if times are short because human reaction time is what it is, but doable. You will not have the rolling resistance on that bumpy terrain, but that's something that depends on the terrain and that you can measure at any time independently by just towing the vehicle without the motor.

>and if i didnt do any of this, which curve would you go with?
>do you think >>617591
>is accurate enough?
The only point that we know that matches the datasheet is the optimum efficiency point at 12 V. The graph should be taken with a grain of salt because, for starters, I made it taking the stall current at 12 V as 840 mA and not 1.55 A.