Tractor mounted fork lift

[chandra007]

I am trying to build a forklift attached to tractor.

My specifications are,4000 lbs of weight to be lifted to 8ft. with in 5 seconds.

what should be my engine horse power and pump specifications for the same.

What should be the cylinder specifications.

Learned members may give some calculation relevant to it.

Thanks in advance.

chandra007

 

[Leejohn]

I would look for a old forklift boom and fit to your tractor. 4000lbs to 8 feet is a task. Or do you have one and just wanting to know what cylinder to get?

 

[swick1]

This should get you started...

Power is work/time. Work is force * distance. Since you want to raise 4000lbs in 5 seconds to a height of 8ft, we can calculate power after calculating work.

The work to be done is force * distance = 4000 lbs * 8 ft = 32000 lb-ft

The power necessary to do the work in 5 seconds is work / time = 32000 lb-ft / 5 sec = 6400 lb-ft/sec

To convert this to horsepower, we divide by 550 lb-ft/sec = 1hp

This gives us a needed horsepower of 6400/550 = 11.63 hp

An external gear pump has an efficiency of 85%. Divide the horsepower by efficiency to get the actual driven horsepower needed = 11.63/0.85 = 13.68 hp. This is your hp requirement at the pump. If your drive is 95% efficient then your hp requirement for the engine is 13.68/0.95 = 14.4 hp

J_J, It's your turn!

 

[J_J]

I have already replied to him via PM.

He is wanting someone to do all the leg work and designed a forklift for him.

How fast something can lift is based on the cyl, GPM and the mechanical, chains and pulley's, etc

You can operate a cyl with just about any GPM the pump can produce with the pressure required.

The more GPM you feed the cyl, the faster it will move.

The other thing is how safe can he build it . 4000 lbs, 8 ft in the air.

Once he has the GPM requirement for the GPM, he can figure out the pump size, and pressure, and the HP to run it.

He should go and look at some forklifts and get a working knowledge of the setup.

They make forklifts that will work on the back of a tractor.

http://www.edwards-equip.com/SiteDocuments/tt36-416.101ab.pdf

http://www.edwards-equip.com/SiteDocuments/tf79-426.1011ab.pdf

 

[swick1]

Also I saw one for a mini skid steer that mounted to the loader arms. It had its own wheels too.

You are right about the safety aspect. Hydraulically speaking he should look into pilot operated check valves to keep it from falling if someone grabs a lever while the load is raised.

Mechanically he should rig some failsafes and overbuild

 

[chandra007]

Thank you swick and JJ,

As per your calculation , total power required to lift 4000 lb is 14.4 hp(if 95% efficient)

power = pressure*flow/1714.

what should be the flow? and cylinder specification?

chandra007

 

[masseyR1327]

There is a forklift on the front of my tractor. It is designed to lift 1 metric tonne. (1000 kilograms). It will lift almost 3 metres. It has three rams. Two for tilt and 1 for lift.

Not a very clear photo but I can take more if you are interested.

 

[chandra007]

hi,

what is the specification of your hydraulic pump.if possible please write and what is the stroke length of the lifting cylinder and bore size?do you have chain arrangement in your forklift?


chandra007

 

[swick1]

Thank you swick and JJ,

As per your calculation , total power required to lift 4000 lb is 14.4 hp(if 95% efficient)

power = pressure*flow/1714.

what should be the flow? and cylinder specification?

chandra007

Use algebra:
Flowrate=(power*1714)/pressure

You will want to build in a factor of safety for each part. The numbers we have given you are theoretical minimums. In practice you will want more power, flowrate, etc.

 

[chandra007]

swick,the formula is flowrate =power*1714/pressure

pressure=Force/area

4000lb/3.14*d*d/4,if i select 2 inch bore dia.then pressure=4000/3.14=1273psi.

14HP=(1273*Q)/1714 >>Q=(14*1714)/1273=18.84 GPM

flowrate=RPM*displacement/231

If RPM is 1500,

18.8 GPM=1500*displacement/231

Displacement=18.8*231/1500=2.88 cu.in
So a pump with 2.88 Cu.in of displacement and flowrate of 18.8 GPM running at 1500 rpm and rated pressure of 1273 psi,will be suitable for lifting 4000 lbs.

For timing 8ft=8*12=96 inches

Q=AV>v=Q/A=18.8/3.14*2*2/4=18.8/3.14=5.98 nearly 6 inches/second

for 1 second it can move 6 inches for moving 96 inches it takes 12 seconds,but i wish to lift the weight in 5 seconds,whether i have to change flow requirements?

For lifting power required is 14.HP,for drive how much power is required?

when 4000lb 's of weight lifted to 8 ft ,counterbalance valve is required to make it stay at top even power is off?