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hydraulic traction boat trolley

  • Thread starter Thread starter maurogir
  • Start date Start date

maurogir

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hello to all, I want to build an oleodynamic traction boat cart, that is a cart consisting of a axle with differential with coupled an oleodynamic motor. I would like to know now who can give me some advice on how to dimensional the oleodynamic system, consisting of an oleodynamic motor coupled to a pump in turn coupled to a diesel engine of about 6 cv.
the full load vehicle weighs about 1000 kg and must circulate also on sand.
question: how to dimension everything? what components to use with what values? :confused:
 
maurogir said:
hello to all, I want to build an oleodynamic traction boat cart, that is a cart consisting of a axle with differential with coupled an oleodynamic motor. I would like to know now who can give me some advice on how to dimensional the oleodynamic system, consisting of an oleodynamic motor coupled to a pump in turn coupled to a diesel engine of about 6 cv.
the full load vehicle weighs about 1000 kg and must circulate also on sand.
question: how to dimension everything? what components to use with what values? :confused:
Click to expand...
a nice job from a few hundred hours of work.... :tongue:
 
exatem said:
You anticipated me.

however determined the sum of forces that oppose motion f, defined speed v and performance n, the necessary power is:
n= fv/75n (in cv).
and it's just the beginning.
Click to expand...
How... well... f=ma, e=mc_2 of x=x0 + v0t + 1/2at502...
MBT said:
a nice job from a few hundred hours of work.... :tongue:
Click to expand...
Well, a hundred if you know how to do the job, but just executive accounts, if you already have all the technical solutions in the drawer. . .
 
Well! I want to point out that the job is already over, the problem I found when I saw that the vehicle was moving with hemorphic effort, such as to make the engine go under stress, the problem I think is a bad dimensional pump-oleodynamic motor. I got myself a gearbox (1/3) to apply to differential. I think he can fix this. My doubt was to know if the coupling had been directed, what values (oleodynamic motor) should I have considered? :smile:
 
maurogir said:
Well! I want to point out that the job is already over, the problem I found when I saw that the vehicle was moving with hemorphic effort, such as to make the engine go under stress, the problem I think is a bad dimensional pump-oleodynamic motor. I got myself a gearbox (1/3) to apply to differential. I think he can fix this. My doubt was to know if the coupling had been directed, what values (oleodynamic motor) should I have considered? :smile:
Click to expand...
Then say it, don't you? ? :mad:
pig world, look at engine absorption, read out pressure to the pump and you have the answer to your problem!! !
 
I still want to point out that I am a fan of bricolage-far from myself, but not an expert, unfortunately I also show some gaps. the absorption of the motor as I quantify it and then to measure the output pressure of the pump, do I have to find it from the plate data or applying a pressure gauge? I wouldn't mind getting it. If someone can send me some computer program, I would be infinitely grateful.:smile:
 
You were talking about a diesel engine if I wasn't wrong. . .
you can easily calculate the power transmitted to the fluid of the hydraulic circuit by measuring the pressure. by multiplying pressure x pump flow you get the power transmitted to the fluid. the units of measurement must be consistent: pa and m^3/s . excuse if they are not exactly technical use. . .
the problem is to estimate the rotation speed of the electric motor. you could do so considering the rotation speed of the wheel and the ratio between the pump displacement and the hydraulic motor... if the maximum valve is not rolling.
for the rest, if the engine goes under stress, I would have used a smaller pump rather than a reducer, both for a matter of cost and simplicity.
 
Well, actually, yes, I would like to point out that the reducer, of course that I inserted it between the engine and the differential, not to the pump, as to the smaller pump, should know the value of the current one I have.
I'll tell you as soon as possible.
 
I appeal to this debate, so I do not open another one. the question is always that, dimensional pump and motor/motor for an alaggio trolley.
I would like to know if my calculations are correct or not and if there is something missing. I take the general case of the sloped plane.
I break the weight force p [N] in the two components, parallel and perpendicular to the fpa and fpe plane.
♪ [N] = p * sen(angle)
♪ [N] = p * t (angle) * coeff. friction
♪ [N] = = sync, corrected by elderman = = @ elder _ man
m [Nm] = ftot * radius_ruota

therefore mn is the twisting moment in neutral position. I then add a torque moment to add according to the desired acceleration then macc [Nm] = (p / 9.81) * acc * radius_ruota.
mtot [Nm] = mn + macc

correct?
Thank you.
 
how will you ever add a parallel force with a perpendicular you know only you.

then to tell the truth if p is in kg you have to multiply it for 9,81 to get n and not to divide it.
try again.
 
meccanicamg said:
how will you ever add a parallel force with a perpendicular you know only you.

then to tell the truth if p is in kg you have to multiply it for 9,81 to get n and not to divide it.
try again.
Click to expand...
regarding the unit of measurement, read my post well, I wrote expressly p [N] ... at my house n = newton...
about the perpendicular component... then tell me how to calculate the friction force... Or do you get the friction of the sand? I look forward to your formula for the calculation of friction.

Bye!
 
netstrike said:
regarding the unit of measurement, read my post well, I wrote expressly p [N] ... at my house n = newton...


about the perpendicular component... then tell me how to calculate the friction force... Or do you get the friction of the sand? I look forward to your formula for the calculation of friction.





Bye!
Click to expand...
I've escaped because normally p is made in kg and f in n by tradition. . .





for the total force you cannot subtract in module, and in absolute fpa value that is parallel to the plane to the perpendicular fpe as you write. that then fpe you wrote it as a perpendicular force for friction coefficient is right... a little different between saying and done... and it becomes parallel to her of opposite direction.



then the mn pair you call neutra should be the total force above multiplied the ray and so far ok but you forgot that the wheels, especially of rubber on sand have the turning friction, that is the shift of a defined amount of the vertical reaction of the wheel (synthetic shift) compared to the road running in opposition to the motion. So you also have this component that sucks power to the engine....which you beautifully overlook because you have diamond wheels, don't you?



wrong

♪ [N] = p * sen(angle)

♪ [N] = p * t (angle) * coeff. friction

♪ [N] = | fpa - fpe |



Right.

♪ [N] = p * sen(angle).......parallelelo

♪ [N] = p * cos(angle) ..... perpendicular

invoiced = fpe * coeff.attrito.......parallelelo

♪ [N] = | fpa - fattrito |



How ironic are you? .
to cure anxiety there are anxiolytics...to cure ignorance you must learn to be humble and listen without polemizing.
 
You wrote exactly what I wrote with an extra passage.... So what did I do wrong?? :confused:
Of course I'm ironic... you answered me by treating me stupid... besides wrong you...
 
netstrike said:
You wrote exactly what I wrote with an extra passage.... So what did I do wrong?? :confused:

Of course I'm ironic... you answered me by treating me stupid... besides wrong you...
Click to expand...
you're wrong by saying that fpe is the disposition of force in the perpendicular direction on the plane...... it's not just because it changes direction and direction as well as module. remains only the same application point. see if it is equal or not.

Anyway, calm down, I treated you like you wrote me, no more or less.
 
Okay, so thank you :biggrin: , now I know that the calculation was right, and that was what I wanted to know, now I can go further and choose hydraulic engines, pump and motor.
 
Remember that rubber on sand you have a friction coefficient of about 0.3 therefore fv = u/r where u is decentralization and r the wheel radius.

you have not calculated this and instead you have to consider it being required to your engine....that is, rolling resistance. so you can't choose anything yet since you didn't consider it.
 

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