forklift motor

[bisogna_rimanere_umili]

I am studying that motor use for handling a mass electric forklift about 1300 kg, the two front wheels are motions and I have to find the engines that serve...
simplistically I divide the mass between the two wheels therefore 650 kg...

the speed of regime on the floor is 4 km/h (1.11 m/s) ipotizzo acceleration from 0 to 1.11 in 3seconds (a=0.37m/s2)

wheels have a diameter of 250mm (0.25m) and ipotizzo a weight of 10kg (I have engines and gearboxes)
j=2*0.078 (ruota ant e post) f volvente 0.01, g=9.81 m/s2
calculation of durable force rolling force frot=63.76n
force weight against motion fh=0n
aerodynamic force the dark faero=0n
force inertia fi=240.74n
f inertia wheels firuota=3.70n
ftot=308.2n potenza=342wcalculation the torque that serves the wheels
omega=8.8 rad/s
n=84 r/minc=39nmI do the same for bike the ascent with v regime of 2km/h (0.55m/2)
30% slope (16.69°) and acceleration time of 4 seconds (0.138m/s2)

fat=61.07n
fh=1832.27n
lighthouse=0n
fi=90.27n
firuota=1.38nftot=1985n power =1101wcalculation the torque that serves the wheels
omega=4.44 rad/s
#43 r/minc=248nmthe forklift works with batteries, so current continues to 24-48 volts usually, I look for a reducer and I choose one with i=33

then the engine will have npiano=2805giri/min cpiano=1.18 nm
#1420giri/min csalita=7.51 nm

for the electric motors It seems like a high couple

for the hydraulic motorshypothesized 110 bar pressure
displacement v of 6 cm3
and a range of 9.46 l/min

the total flow rate is 19 l/min (two wheel)
assuming a speed of 2800 rpm the pump has a displacement of 8 cm3
if calculation of the required torque and power
17
4.73kw

Even worse than before...
I've definitely done something wrong, but I miss you can help me?


for those who want to help me also allego the formulas I used

 

[Er Presidente]

I'm too old for calculations, but it doesn't seem so bad.
The problem is that 30% slope, if you set a ramp so high the power needs.
The electric motor is generous, but it depends how long you think you're staying on those conditions.
make a 7% check at constant speed (decides which) to check the limit of continuous power (at least an hour), then check 12%, for a shorter time (half hours) and for 30% I would say to check only the point full load (seconds).
The electric motor, depends on the type you use, a chopper can get him to hunt almost anything you want, but if you don't watch out for the time you cook him.
given the tables with the power curves and the nominal powers of the engines with the service factors.
speaks to the engine builder for what the engine can really stand and then size everything.

Hi.

 

[gerod]

forklift you mean the so-called "mulet"?

If yes:
1. besides the weight of the cart I would also put the cargo carried.
2. f volved: here you have to define it well (what units of measurement? adimensional). for rubberized wheels (I suppose) seems too low.
for the record there are already the wheels ready for similar applications, just put the engine (eg. brevini)
Hi.
Hi.
Usually to find the coeff. of rolling friction:
m = fxr = pxn from which:
f = p x (n/r), where n is the weight force arm on the wheel that contrasts the torque given by fxr. your fv is n/r which is adimensional.
on the manuals, for rubberized wheels, n is about 10 mm (sometimes even 7 cm, depends on the wheel if it is pneumatic, rubber sh90, etc).
fv = 125/10 = 0.125
3. for the motor speech, you put a nice epicloidal gearbox with a nice report and you're riding!

 

[bisogna_rimanere_umili]

it is true I should consider motion to regime to find the nominal torque of the motor and consider motion with acceleration and relative inertia so as to see if the motor bears it for a few ten seconds everything...
I had already thought about this and I had seen that without inertia the necessary couple varies of 10-15 nm...

30% uphill is a garage ramp or the ramp for unloading and loading from the crate of a truck...

the coefficient of friction is low (I know that in the accounts I ran one zero) surely you have to use at least one 0.1...
the reducer can't take it to pleasure, the electric motor over 2500 turns I think it sends me to that country...

the machine in question is actually a pantograph air platform and in 1300kg should be everything, the precise total mass I will have when I have sized everything well...

the doubt that I have is that machines with similar size and flow have traction with hydraulic motors and the pump is operated by a motor alone 3.5-4 kw that works at 24 volts (read on a manual of use)...

 

[gerod]

If you tell me about high speed (ple) motorized platforms, I would say that before you check that speed, acceleration, etc. are compatible with those of en 280.
the above-mentioned norm says that the max v for semovent working piatatforms (not those mounted on vehicle or those mounted on carriage running on rails) is 0.7 m/s.
you are at 1.11 m/s. If you have risk of falling > 3 meters better than give an eye to the norm because, when you pass to the notified body, you do not pass it!
an observation for users: When you ask for information you are clear, do not say that you have to design a square and then it turns out that it is a rectangle! Thank you.

 

[bisogna_rimanere_umili]

If you tell me about high speed (ple) motorized platforms, I would say that before you check that speed, acceleration, etc. are compatible with those of en 280.
the above-mentioned norm says that the max v for semovent working piatatforms (not those mounted on vehicle or those mounted on carriage running on rails) is 0.7 m/s.
you are at 1.11 m/s. If you have risk of falling > 3 meters better than give an eye to the norm because, when you pass to the notified body, you do not pass it!

true, there is the limit of 0.7 m/s if lifted... but if it is closed at 1.11 it can go. . .
My doubt is not that of the norm but how come an exaggerated power comes out to me... I checked the calculations and I think I'll lower speed and slope... .

 

[gerod]

Did you do a competition check?
Maybe find the hippo

 

[Er Presidente]

true, there is the limit of 0.7 m/s if lifted... but if it is closed at 1.11 it can go. . .
My doubt is not that of the norm but how come an exaggerated power comes out to me... I checked the calculations and I think I'll lower speed and slope... .

for clarity, but to you what power goes out on the ramp?

 

[bisogna_rimanere_umili]

Did you do a competition check?
Maybe find the hippo

It's been a while I've been looking for hippo but I can't find it... :frown:
the fact is that similar machines have traction using hydraulic engines and pump is operated by a 3.5-4 kw engine that works at 24 volts...

for clarity, but to you what power goes out on the ramp?

I corrected the desired friction coefficient using 0.1

and I modified some data:

mass 1200kg (before 1300)
slope 25% (before 30)
speed 1.5 km/h (first 2) and wheel turning 32 turns/min

on the ramp serves a pair of 248nm and a power of 832w at regime
(with inertias 260nm and 872w)

then chosen a reducer with i=42 with yield 0.95
I need a 6nm to 1350 rpm torque engine and power of 0.921 kw...

with hydraulic engine at 140 bar and pump with hydraulic yields 0.9 and mecc 0.9 in the end I need a 2.5 kw engine from 11nm (at always regime). . .


the problem now moves to the lifting cylinder where double force values come out eek: ) compared to similar machinery... .