Tricky question

Tygra

New member
Joined
Mar 29, 2021
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1
Hi there,

I was wondering if somebody could help?

I am working on my coursework for University, and on part of it, I have to calculate the production rate of an excavator. The excavator moves soil at a rate of 1.54m3 per 17 seconds. But this is not the actual real-life working rate as it assumes perfect conditions constantly. For this reason, the quantity needs to be multiplied by 0.7 for a job factor, and also a human factor of 50 minutes per hour.

Thus we have 1.54m3 * 0.7 * 5/6 = 0.898 m3 per 17 seconds. This is the real-life actual output of the excavator.

My question is, rather than calculating the bucket load value, how do I calculate the new time? The time is inversely proportional to the bucket load - the bucket load goes down, so the time would therefore increase. So do I instead divide the time of 17 seconds by 0.7 and 5/6?

Could someone help please?

Thanks
 

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
11,540
I am working on my coursework for University, and on part of it, I have to calculate the production rate of an excavator. The excavator moves soil at a rate of 1.54m3 per 17 seconds. But this is not the actual real-life working rate as it assumes perfect conditions constantly. For this reason, the quantity needs to be multiplied by 0.7 for a job factor, and also a human factor of 50 minutes per hour.

Thus we have 1.54m3 * 0.7 * 5/6 = 0.898 m3 per 17 seconds. This is the real-life actual output of the excavator.

My question is, rather than calculating the bucket load value, how do I calculate the new time? The time is inversely proportional to the bucket load - the bucket load goes down, so the time would therefore increase. So do I instead divide the time of 17 seconds by 0.7 and 5/6?
What do you mean by "bucket load value"? And what time do you want to calculate? The time for moving one cubic meter, or what?

It would be helpful if you showed us the actual exact problem as given to you.
 
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