Using gravity to generate power

I believe that a 17.18 lb weight pushed off a 32ft ledge is one horsepower being that 550 foot pounds/ second is equal to 1hp and G is 32ft/s squared. Thus dropping 17.18 lbs for 32 feet equals 746 watts (one hp). Then if I was to use a pendulum escapement mechanism to control the duration of the fall to last 6 hours (this duration is only important in maintaining low temperature/ high efficiency and endurance of materials) to generate 2200 watts output for 6 hours, I would need to drop 310lbs from 32 ft. Or more realistically, 775lbs from 16feet. Since gravity is acceleration at T squared, I think that if I cut the distance in half the weight must increase by 2 plus half. Anticipating 10% loss in efficiency, I could run 2 gravity driven generators based on this configuration and produce 4Kw for 6 hours of use. Am I right on this? Should I put this in practice?

• have you looked at hydroelectric power? en.wikipedia.org/wiki/Hydroelectricity The water was at a high elevation from sun and wind and rainfall. How will you get your continuous falling weights? To get them to an elevation you would have to spend energy, to roll them to a drop too. Commented Apr 17, 2017 at 4:21
• There is a good deal more to designing power generation stations than knowing how far you would have to drop a weight to get the required power. How would you connect the weights to a generator? What makes you think the losses would be 10%? Commented Apr 17, 2017 at 4:52

The way to use gravity as a source of energy would first need to know how much weight do you need and what gear ratio would you use to have the weight drop at, say, a speed of a $$\text{ft/h}$$, and create enough torque to turn a pulley at $$10\,\text{hp}$$ to turn a $$7\,\text{kW}$$ generator at $$3500\,\text{rpm}$$, etc. The generator could charge batteries to use a winch to reset the weight at $$24\,\text{ft}$$ of course. Even at $$2\,\text{ft/h}$$, you would only have to reset it twice.