motorized sizing on cart and translation mechanism

[loscienziatopazzo]

Good morning to all, I ask an vs opinion for a sizing.
I am designing a trolley with a motorized base that can help in a short phase of translation.
cart has a total mass of 1600kg
has n°2rear wheels ø160 and n°2 front wheels ø80mm

the request of the customer was to insert a motorized at the base to help the translation of about 1m when the cart was in a certain position. must be just a little help on rare occasions.

input data:
m=1600kg
n°4 wheels ø160 (not correct but I wanted to simplify the calculation)
vel max= 0,3m/s
acc = 0.15m/s^2
no slope
res rolling = 0.02 (industrial resin)


from my sizing it is necessary:
p= 105w (Nominal engine capacity at level)
torque wheel operation in plan = ~25nm
torque wheel by plane = ~ 38nm


the motoruota must be insertable and desiccable, reason why I designed a sleigh (salts/scends) composed of a plate to l that guarantees the translation.
this plate has, on the lower part, the bracket of the motoruta while, on the upper part a compact/guided piston that allows the translation.
the translation is guaranteed by n°2 igus+guide trolleys from 35mm. behind it has a polyzene bar to avoid a stinging moment in progress.


spec. motor/motor gearbox (depotentiated to comply with s2-60)
p=600w
== sync, corrected by elderman ==
continus torque= 40nm
peak torque=80nm
max allowable torque= 110nm
wheel = ø160 x sp.54mm


the motoruota does not discharge part of the weight as it is compressed to the ground by a cylinder from ø50 (with unp=6bar) then with ~1200n
in such conditions I find that mr= ~1,6nm

Now considering that I should supply 25nm on the floor (over 80nm at the point) I fear that the motoruota can slip and that the f of the cylinder is not enough to transfer the necessary torque to the carriage.


how can I check the required minimum load of the cylinder? ?
Thank you in advance

 

[biz]

0.02 what is it? what is the acceleration of the motoruota in the object and what coefficient of friction desired values?what other engines contribute to the motion?

 

[loscienziatopazzo]

Good morning.
> 0.02 volve friction (estimated)
> would be a 5th wheel (the only but motorized)
> the acceleration of the stigma in 0.15/0.2 m/s^2 (can be very slow)
> only 1 engine

Thank you.

 

[biz]

0.02 in my opinion is too low

 

[loscienziatopazzo]

the data was provided to me by the builder;
wheel in urethane 95shore at - on industrial level flooring / resin

 

[biz]

All right, let's do it for good.

 

[biz]

in theory should not slip. . I have time to explain how I came to this conclusion.

 

[loscienziatopazzo]

grazie biz

 

[biz]

There is a coefficient of adherence, empirically empirical by several authors (curtius‐kniffler for example) and we can, in this case, estimate at the worst of hypotheses at 0.7.
the coefficient multiplies it to the weight force that weighs on your wheel, we say about 1800n.for which the force that acts parallel to the contact between the two surfaces in relative motion should not exceed 1260n.
force we simply revenues from m max=80nm.
results 1000 n.
We're in it.
attention, if the torque should be greater or the adhesion conditions should be altered, for a wire of oil or similar we are fried.
My advice: increased those 1800n

 

[meccanicamg]

the data was provided to me by the builder;
wheel in urethane 95shore at - on industrial level flooring / resin

is low because it is the value of the steel wheel on steel track. 95sha is higher because the rubber is compressed. probably the rubber layer is little and with the load so low does not deform enough. However try to be cautious otherwise lacks power.

the calculation of adherence and the necessary couple has been treated of etse times on the forum for trolleys of various nature. First, take a look at them.

 

[biz]

You don't have to worry about angular acceleration as it is.

 

[biz]

Excuse me, but how did you calculate 105 w?

 

[loscienziatopazzo]

There is a coefficient of adherence, empirically empirical by several authors (curtius‐kniffler for example) and we can, in this case, estimate at the worst of hypotheses at 0.7.
the coefficient multiplies it to the weight force that weighs on your wheel, we say about 1800n.for which the force that acts parallel to the contact between the two surfaces in relative motion should not exceed 1260n.
force we simply revenues from m max=80nm.
results 1000 n.
We're in it.
attention, if the torque should be greater or the adhesion conditions should be altered, for a wire of oil or similar we are fried.
My advice: increased those 1800n

Why 1800n?
the fpeso, if a cylinder ø50 can be ~1060n (106kg to 6bar)
1060x0,7= 742n

Cmax=80nm
f= cmax/r= 80/0,08=1000n

If correct, I would already be out and therefore would be able to use a piston ø63 at least 6bar (I think it is appropriate to raise the pressure for greater safety)


for 105w
(1600kg*9,81) x 0.02 x 0.3m/s x 1/returner = ~104w
(ii) a system and plan, with the exception of

 

[biz]

I'm sorry, I've been running things.
0.025^2*pigree*6*100000=1178
1 bar = 100000 pa
and why for 0.7?

 

[loscienziatopazzo]

is low because it is the value of the steel wheel on steel track. 95sha is higher because the rubber is compressed. probably the rubber layer is little and with the load so low does not deform enough. However try to be cautious otherwise lacks power.

the calculation of adherence and the necessary couple has been treated of etse times on the forum for trolleys of various nature. First, take a look at them.

ok I will try to pay more attention to previous chats.
from model cad turns out that the thickness is of the rubber extract is 30mm, so I don't think exactly slender.
What value do I recommend to consider?

Thank you.

 

[biz]

to maximise the yield to 0.9

 

[biz]

then becomes 1178*.09=1060n

 

[loscienziatopazzo]

the cylinder fpeso pushing the wheel I took it table= 106kg to 6bar
If I have not wrong to understand your analysis, I multiply 1060 x 0.7 and verify the force that acts in contact between the two surfaces and that you do not have to overcome.

the return of the reducer is 0.9 (communicated by the manufacturer)

 

[biz]

You're right, I'm sorry, but I'm working and I'm risking doing two things wrong, I'll be happy to give you a hand.However you just do,consider that 0.7 is not very conservative and just a little dirt oil etc. to lower the parameter suddenly.
However if the colleague @meccanicamg He said that it was already treated as an argument I recommend to give us an eye.
Bye.
p.s.:leave the tables,take note:
1bar = 100000 pascal
1atm=101325 pascal
these measuring units you will be confronted with more often
1psi=I don't remember looking for google

performance 0.9 conventional

 

[loscienziatopazzo]

thank you, ok wait for an answer as soon as you can
I will certainly see other items
tabs are so practical, even if they often block the ability to rationalize