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power required to lift a mass

how do i calculate the motor required to lift a mass eg mass = 75kg what hp is required


How far do you want to lift it, how quickly, and how efficient is the mechanism you intend to employ? Time to dig out those high school physics text books!  Sometimes I only open my mouth to swap feet... 

thanks for your response i have dug into my physics books and found the this formula power= kg lifted xlifting speed in m/s divided by 9.81x efficency . is this the corect formula to use if i was not going through a gearbox and direct off a motor shaft . and is ans in watts or kilowatts i want a lifting speed of 1m/s to a hight of 15 metres eff=.7 mass=75kg just trying to work out what size motor it would take to lift my weight


CJCPE (Electrical) 
24 Jun 07 20:09 
That is correct. Power in Joules per second or Watts = mass in grams X speed in meters per second divided by the acceleration due to gravity (9.80665 m/s^2). 

Acceleration of gravity goes in the numerator, not the denominator. F = m * g P = v * F = v * M * g If v is m/sec, M in kg and g in m/sec^2 then P is in kg*m/sec^2 / sec = Joule/sec = watt this is for constant speed... acceration requires additional force. Also as was mentioned, account for the efficiency of the power transmission. ===================================== Engtips forums: The best place on the web for engineering discussions. 

"That is correct. Power in Joules per second or Watts = mass in grams X speed in meters per second divided by the acceleration due to gravity (9.80665 m/s^2)." There are at least two mistakes in that equation, and another in the preceeding sentence. Cheers
Greg Locock
Please see FAQ731376: EngTips.com Forum Policies for tips on how to make the best use of EngTips. 

thanks for takeing the time to respond to my question it is nice to know there are people that are prepared to put themselves out and give their knowlege
thanks turbine f14 

IRStuff  your equation is not correct. Kinetic energy should be included within the energy. This would give rise to an acceleration term dependent upon rate of change of velocity ===================================== Engtips forums: The best place on the web for engineering discussions. 

The corrected version would be as follows: Power = d Energy(t)/dt = m*g*dh(t)/dt + m v(t) dv/dt = m*g*v(t) + m v(t) a(t) = v(t) * [g + ma(t)] = v(t) * F(t) where F(t) = [g + ma(t)] as expected ===================================== Engtips forums: The best place on the web for engineering discussions. 

Add at the beginning of my previous post: Energy(t) = m*g*h(t) + 0.5 * m*v(t)^2 ===================================== Engtips forums: The best place on the web for engineering discussions. 

Consolidating the two previous posts, the corrected version would be as follows: Energy(t) = m*g*h(t) + 0.5 * m*v(t)^2 Power = d Energy(t)/dt = m*g*dh(t)/dt + m v(t) dv/dt = m*g*v(t) + m v(t) a(t) = v(t) * [g + m*a(t)] = v(t) * F(t) where F(t) = [g + m*a(t)], as expected ===================================== Engtips forums: The best place on the web for engineering discussions. 

A minor correction (final answer): Energy(t) = m*g*h(t) + 0.5 * m*v(t)^2 Power = d Energy(t)/dt = m*g*dh(t)/dt + m v(t) dv/dt = m*g*v(t) + m v(t) a(t) = v(t) * [m*g + m*a(t)] = v(t) * F(t) where F(t) = [m*g + m*a(t)], as expected ===================================== Engtips forums: The best place on the web for engineering discussions. 

DickDV (Electrical) 
25 Jun 07 17:58 
turbinef14, if you are lifting your load with a cable winding around a drum, as in a common hoist, I think I can make it a bit simpler for you.
First, convert kg to lbs by multiplying kg by 2.2.
Find the circumference of the winding drum in feet and the radius of the drum in feet.
The torque in ftlbs on the drum drive shaft is your load in lbs times the drum radius.
The rpm of the drum drive shaft is the lifting speed in feet divided by the drum circumference.
Once you know the drum shaft torque and speed, multiply them together and divide by 5250 to get hp.
This is constant speed lifting hp. You will need to add some extra torque for acceleration upward, however, the motor will have at least 200% shortterm overload capacity if its a NEMA A,B, or C motor across the line and, if on a VFD, as much % shortterm overload capacity as the drive shortterm current divided by the motor FLA. You would want at least 150% shortterm overload in the drive for most hoisting jobs.
Likely, you are fully aware of this but, the braking hp must be calculated the same way with double care. Either the drive must have full capacity braking or the mechanical brake will need to do all the braking. Much better to have the drive do all the dynamic braking and the mechanical brake do the holding braking and serve as backup for the drive braking. 

DickDV (Electrical) 
25 Jun 07 18:04 
One other thing, the drum shaft hp will only match the motor nameplate hp if the shaft speed matches the base speed of the motor. Otherwise, you will need a gearbox or equivalent or, if on a VFD and direct drive, the motor hp will be the constant speed torque times the motor base speed. You would need to pick a motor with enough poles so the motor base speed is close to the max drum speed but not less than the max drum speed. 

don01 (Electrical) 
25 Jun 07 23:08 
hi all Pete ,as usual, is on the money again. But here it is straight from my physics text book (which is soooo old it is a national relic) PROBLEM: a plane weighing 5000kg can climb vertically at a rate of 200km/hr. What power is expended. (9.8m/s)
SOLUTION: power = work/time = (force x distance)/time = force x velocity
F=mg = 5000x9.8 newtons v=200km/h = 55.6m/s
power = 5000 x 9.8 x 55.6 = 2724,000watts = 2.724kW
now in the way of all good plagarists cut out their values and then paste in yours = solution (Ihope)
anyway enjoy the journey & grab the memories Don
(Physics for Engineers  Schofield 1970) 

Wow, 4 hp to lift a 5 ton plane at 120 mph vertically. I'll have some of that. Cheers
Greg Locock
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//2724,000watts = 2.724kW// You might want to edit that. 

ozmosis (Electrical) 
26 Jun 07 4:36 
turbinef14 Another point. If your 1 M/s was purely an arbitrary figure, then do your calculations based on longer velocity rates. Your final power calculation will alter quite a lot and therefore cost of equipment. And do not forget the point that DickDV makes about bringing the load back down, if it is. This time you have gravity working with you and this is not always favourable.


sreid (Electrical) 
26 Jun 07 16:07 
75 kg = 165 lb
1 m/s = 3.3 ft/sec
Power = 165 x 3.3 = 541 lbft/sec
550 lbft/sec = 1 HP
541/550 = approx 1 HP


Since everyone seems to have had a go on this... "Our" (European) definition of one horsepower is "The power needed to lift a mass of 75 kg vertically 1 m in 1 second" This is suspiciously similar to the OP that says that he needs to lift 75 kg at a speed equal to 1 m/s. Using SI, we get: P = 75*9.8065*1 = 736 W (the European HP). The efficiency stated is 0.7, so you would need something like 1050 watts. Then, there are lots of other things to consider. But your original question has this answer. Gunnar Englund www.gke.org  100 % recycled posting: Electrons, ideas, fingertips have been used over and over again... 

This is reminding me of a junior high physics lesson I got, way back when. The teacher was trying to make the concept of power real to us. In the lesson, each of us had to be weighed, then be timed in the 50yard dash. From that we were supposed to calculate the power we could each generate.
In approximate numbers, I weighed 100 pounds then, and I ran 50 yards in 8 seconds. So I was supposed to calculate:
100lbs * 150ft / 8sec = 1875lbft/sec
1875lbft/sec * 1HP/(550lbft/sec) =3.41HP
Now this bothered me  I knew that a horsepower was in some sense the power that a horse could generate. As a scrawny little 12year old, it didn't seem to me that I could generate more power than 3 horses.
But it turned out to be a great "life lesson". It got me started on really understanding the difference between mass and weight. (This has saved me a lot of grief in momentofinertia calculations.) More importantly, it got me verifying and doublechecking the information supposedly knowledgeable people gave me. I realized that if I wanted really to understand things, I was going to have to build up my own knowledge from first principles. Even on this forum, I've seen people [mis]use "plugandchug" formulas without understanding what goes into them.
Curt Wilson Delta Tau Data Systems 

Well no wonder you were confused. He was assuming that the resistance to motion, horizontally, equalled your weight. Cheers
Greg Locock
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You're so picky Greg, worrying about minor details like physics all the time. I mean really...how much harder could it possibly be to run vertically up a ladder at 6 yards/sec than it is to run accross a sidewalk at 6 yards/second! ===================================== Engtips forums: The best place on the web for engineering discussions. 

I hope no one was offended by my moment of giddyness. ===================================== Engtips forums: The best place on the web for engineering discussions. 

We did a similar experiment at school, running up a flight of stairs. Fit fat rugby players did manage around 1 hp of vertical PE gain. Incidentally (and obviously) a horse can produce far more than 1 hp, but on average over a day that is about what you'll get. wiki says 15 hp, briefly Cheers
Greg Locock
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aolalde (Electrical) 
28 Jun 07 0:12 
By definition: 1 HP = 33000 LB x FT/1 min
75 Kg = 165.35 Lb 1 m/s = 196.85 Ft/min Eff = 0.75 Your power requirement:
P = 165.4*190.85/(0.75*33000) = 1.315 HP = 0.981 kW


...or, using the OP's efficiency: 1051 watts. Which is something you arrive at with no other constants and factors than g, if you use SI. Gunnar Englund www.gke.org  100 % recycled posting: Electrons, ideas, fingertips have been used over and over again... 



