The discussion centers on calculating the torque required to throw a mass using a snowblower equipped with two pulleys and a belt. Participants clarify that 500 kg/m³ refers to the density of snow, not its mass. They emphasize the importance of defining the parameters of the throw, such as height and speed, and suggest using the formula power = torque × RPM to determine the necessary torque. Additionally, they recommend calculating the power requirements of the snowblower, which can range from 2 hp to 15 hp.
PREREQUISITESEngineers, DIY enthusiasts, and anyone involved in snowblower design or modification, particularly those looking to optimize torque and performance in snow removal applications.
anorlunda said:500kg/m(3) is a density, not a mass, not a force.
There is not a single formula, but a collection of formulas. See https://en.wikipedia.org/wiki/Equations_of_motion
There are also a number of online calculators you can use. See https://duckduckgo.com/?q=suvat+calculator&ia=web
jayjay said:ok so i need to find the mass like in kg? like for an exemple a 5 gallon bucket full of snow
anorlunda said:Yes. More important, you need to define "throw" numerically. How high? How fast? Is it fast enough that we need to consider air drag on the thrown object?
That puts in context. It is far easier to work with power than speed, acceleration and those equations of motion.jayjay said:well just trying to figure out how much force i need to the pulley to throw snow with a snowblower.
anorlunda said:That puts in context. It is far easier to work with power than speed, acceleration and those equations of motion.
First, figure out how much power your snow blower needs? Then, power=torque*RPM is the remaining formula.
You can buy snowblowers in the store with power ratings from 2 hp to 15 hp or more. Which of those fits your need?