Just for comparison – to get a feeling for those numbers: Average acceleration is maximum velocity over time. Assuming now that my peak scything speed is really only a tenth of the speed of a lawn-mower, 47m/2, which is still about 10 times my average speed calculated the beginning, this would result in one hundredth the energy.Ī bit more realistic energy per half an hour of scything is then : 239kcal Velocity enters the equation quadratically. … about five times the daily energy demands of a human being! So using that lawn-mower-style speed this would result in:Įnergy per half an hour if I were a lawn-mower: 133kJcal * 180 = 23.940kcal After half on hour I need to take longer break. So this is 1 arc swept per 10 seconds, 6 arcs per minute, and 180 per half an hour. I assume a break of 9,5s on average to make the calculation simpler. Scythe…1s…scythe…1s…scythe…1s….scythe…1s….scythe…longer break, gasping for air, sharpen the scythe. I pause a few seconds before the next – say 10s – on average. I feel exhausted after half an hour of scything. So they would pay us for the option to scythe our grass.īut before I crank down the hypothetical velocity again, I calculate the energy demand per half an hour: If scything were that ‘efficient’ I would put into practice what we always joke about: Offer outdoor management trainings to stressed out IT managers who want to connect with their true selves again through hard work and/or work-out most efficiently. That sounds way too much already: Googling typical energy consumptions for various activities I learn that easy work in the garden needs about 100-150kcal kilocalories per half an hour! I am assuming that this energy is just consumed (dissipated) to cut the grass the grass brings the scythe to halt, and it is decelerated to 0m/s again.ġ kilocalorie is 4,18kJ, so this amounts to about 133kcal (!!) I use only SI units, so the resulting energy is in Joule:Įnergy needed for acceleration: 5kg * (471m/s) 2 / 2 = 555.000J = 555kJ This would result in the following energy per arc swept. So the scythe on par with a lawn-mower would need to move at:ĢPi * (3000 rev./minute) / (60 seconds/minute) * 1,5m = 471m/s
The radius is the length of the pole that I use as a simplified model.
GRIM REAPER SCYTHE FOR LAWN FULL
Angular velocity is 2Pi – a full circle – times the frequency, that is revolutions per time. Velocity of a rotating body is angular velocity times radius. How fast would I have to move the scythe to achieve the same?
This page says: at 3600 revolutions per minute when not under load, dropping to about 3000 when under load.
How exactly the velocity has changed with time does not matter – this is just conservation of energy.įor comparison: How fast do lawn-mower blades spin? If an object with mass m is accelerated from a velocity of zero to a peak velocity v max the kinetic energy acquired is calculated from the maximum velocity: m v max 2 / 2. However, using this speed in further calculations does not make much sense: The scythe has two handles that allow for exerting a torque – the energy goes into acceleration of the scythe. Thus the average speed is: 2m / 0,5s = 4m/s One sweep with the scythe takes a fraction of second – probably 0,5s. All the kinetic energy is concentrated in this ‘point mass’.īut how fast does a blade need to move in order to cut grass? Or from experience: How fast do I move the scythe? In order to keep this simple, I assume that the weight of the scythe is a few kilos (say: 5kg) concentrated at the end of a weightless pole of 1,5m length. Yet I believe that in this case it is the acceleration required to bring the scythe to proper speed that matters so I will focus on work in terms of physics. Just holding a scythe with arms stretched out would not count as ‘work’. So I was curious if Grim Reaper the Physicist can express this level of exhaustion in numbers. It is utterly exhausting – there is no other outdoor activity in summer that leaves me with the feeling of really having achieved something! Yes, we tried the alternatives including a reel lawn-mower. I have a secondary super-villain identity.