Monday Madness

So… this weeks riddle is based upon what I was doing when I was offline a week ago… We are going to be spreading rock 2 inches deep, 10 feet wide on one and a half miles of road. The rock will be dug out of native material (rather than a stockpile) and hauled in a truck that holds 10 cubic yards. The material will be excavated to a depth of 5 feet (where we hit bedrock). The excavation will be 60 feet wide by whatever it takes to get the material needed. The rock has a swell factor of 10% (in other words it is more compacted in situ than in the truck). It takes 31 minutes to load the truck, haul and spread, and then return to the pit. How long does our project take? (just deal with working time. Don't consider lunch, sleeping, etc.) How big is our pit on completion?

Hi guys.

I’ll take a try at this.

We are going to be spreading rock 2 inches deep, 10 feet wide on one and a half miles of road.

Let’s find out how many cubic feet this gives us.
10 x 7920 x 2/12 which is 13200 cu. ft
Expressed as cubic yards 13200 / 27 which is 488.89 cu. yds

Which would be brought in 488.89 / 10 trips by truck, or 48.889 trips

How long does our project take? (just deal with working time. Don't consider lunch, sleeping, etc.)
It takes 31 minutes to load the truck, haul and spread, and then return to the pit.

So it would take 31 x 48.889 = 15155.556 minutes
This is 252.59 hours or 10 ½ days.


How big is our pit on completion?

The rock has a swell factor of 10%,

So this means 13200 cu. ft / 1.1 = 12000 cu ft in situ -

Hole size is 12000 cu ft
So lets express that in layman terms.
Length of hole = 12000 / (5 x 60) = 40 ft
So hole is 5ft deep . 60ft wide , 40ft long.



Thats my attempt.

Very good attempt otisx.

I won't critique your answers at this point… but I suggest that you not only review units, but also logic.

It took about 22.9 hours and you ended up with a pit 60' x 5' x 39.876' edit: Here's my reasoning. 2/36 x 10/3 x 2640 = 492.3 cu yrds That would be the volume of the space you wanted to fill. Since you won't be able to fill every nook and craney the 10% difference applied to the trucks seems reasonable. So you need 443.076 cu yrds. Total trucks = 44.31 Time would be trucks x 31 min = 1373.61 or 22.9 hrs The resulting pit would be 443.076 = 60/3 x 5/3 x 13.29228 (13.29228 is yards) So in feet the pit is 60' x 5' x 39.876'

to make it easier all calcs done in inches and minutes. and assuming the extracted gravel can be compacted to the same density as the natural rock, other wise you'd need to extract less rock and the pit would be smaller. 2inches by 120inches by 95040 equals 22,809,600 cubic inches of gravel need to fill the road. so we need 22,809,600 extracted that leaves a pit 60inches by 720 inches by 528 inches now the truck carries 36x36x36x10 less 10%(swell) = 419,904 per trip so 22,809,600 divided by 419,904 equals 54.3209876543 the truck can only make whole trips so that's 55 trips. 55 x 31 = 1,705 minutes.(assuming it takes the same time for the part truck, which of course it wouldn't, driving time would be the same but fill/empty and spread time would be less.) answers required then are time taken 1,705 minutes. size of pit 60inches by 720 inches by 528 inches

Let’s find out how many cubic feet this gives us. 10 x 7920 x 2/12 which is 13200 cu. ft Expressed as cubic yards 13200 / 27 which is 488.89 cu. yds Which would be brought in 488.89 / 10 trips by truck, or 48.889 trips How long does our project take? (just deal with working time. Don't consider lunch, sleeping, etc.) It takes 31 minutes to load the truck, haul and spread, and then return to the pit. So it would take 31 x 48.889 = 1515.5556 minutes This is 25.259 hours or 1 day 1 hour and 15.55 minutes How big is our pit on completion? The rock has a swell factor of 10%, So this means 13200 cu. ft / 1.1 = 12000 cu ft in situ - Hole size is 12000 cu ft So lets express that in layman terms. Length of hole = 12000 / (5 x 60) = 40 ft So hole is 5ft deep . 60ft wide , 40ft long.

@oldman960,

2/36 x 10/3 x 2640 = 492.3 cu yrds

check your math.

Since you won't be able to fill every nook and craney the 10% difference applied to the trucks seems reasonable

rock spread on the road is figured as "loose" yards. No further shrink or swell necessary.

@terry1966,

I think that the swell factor has you confused. The material in the truck (and on the road) takes up 10% more space than the material in the bank.

@otisx,
Each trip takes 31 minutes and delivers a maximum of 10 cubic yards.

ok simple enough then 10% less than 22,809,600 is 20,528,640 so pit is 60inches by 720 inches by 475.2 inches total gravel stays same at 22,809,600, truck carries same 36x36x36x10=466,560 trucks needed 22,809,600/466560=48.888888889 whole trucks only so 49. time 49x31= 1519 minutes. answers are pit 60 x 720 x 475.2 inches. time 1519 minutes.

Since you won't be able to fill every nook and craney the 10% difference applied to the trucks seems reasonable

rock spread on the road is figured as "loose" yards. No further shrink or swell necessary.

To clarify what I meant. The amount needed would be less than the calculated volume and the 10% would apply to the road as well.

It took about 22.73 hours and you ended up with a pit 60' x 5' x 39.6'

2/36 x 10/3 x 2640 = 488.89 cu yrds

That would be the volume of the space you wanted to fill. Since it's loose you need 440 cu yrds.

Total trucks = 44

Time would be trucks x 31 min = 1364 or 22.73 hrs

The resulting pit would be 440 = 60/3 x 5/3 x 13.201 (13.2301 is yards) So in feet the pit is 60' x 5' x 39.603'

@oldman960

2/36 x 10/3 x 2640 = 488.89 cu yrds - on the road.

That would be the volume of the space you wanted to fill. Since it's loose you need 440 cu yrds. - nope. Check your math (and this would be in the pit)

@terry1966,

ok simple enough then 10% less than 22,809,600 is 20,528,640

Material swells 10%. You can't shrink it 10%. Check your math.

Material swells 10%. You can't shrink it 10%. Check your math.

sorry but you lost me tomk.

20,528,640 cubic inches of solid rock swells 10% to 22,809,600 cubic inches of loose gravel.(which is the amount we need to fill the road.)

so nothing got shrunk as far as i'm aware and can't see where my calcs/logic are wrong with my current understanding of swell.

last shot…. 2/36 x 10/3 x 2640 = 488.89 cu yrds That would be the volume of the space you wanted to fill. Total trucks = 48.9 Time would be trucks x 31 min = 1515.9 or 25.27 hrs The resulting pit would be 440 = 60/3 x 5/3 x 13.201 (13.2301 is yards) So in feet the pit is 60' x 5' x 39.603' All I know for sure is the there wouldn't be anything in the pit.

@terry1966,

20,528,640 cubic inches of solid rock swells 10% to 22,809,600 cubic inches of loose gravel.(which is the amount we need to fill the road.)

If we excavate 20,528,640 cubic inches of material and it swells 10% we would have:

20,528,640 + 2,052,864 = 22,581,504 cubic inches. That doesn't appear to be what you said.

@oldman960,

2/36 x 10/3 x 2640 = 488.89 cu yrds

That would be the volume of the space you wanted to fill.

Total trucks = 48.9

That is the correct calculation for material required to be spread… and therefore the correct truck loads. But - how many trips?

The resulting pit would be 440 = 60/3 x 5/3 x 13.201 (13.2301 is yards) So in feet the pit is 60' x 5' x 39.603'

440 cubic yards swelled 10% would be:
440 + 44 = 484 cubic yards. Is this the number you are looking for?

oops. 20,736,000 swells to 22,809,600.. so pit is. pit 60 x 720 x 480 inches. answers pit 60 x 720 x 480 inches time 1519 minutes.

Well done terry1966!

That is the first totally correct answer.

The pit is 60 ft x 40 ft x 5 ft.

It took 48.9 loads of material which gives 49 trips to be made which takes 25 hours and 19 minutes.

otisx nearly had it but he had me making .889 of a trip and I had to make the whole trip no matter how many yards the truck was carrying. (truthfully, I actually put all 10 yards in the truck each of the 49 trips )

The trip vs load thing is what confused otisx. terry1966 and oldman960 went awry with the swell factor. Here is what I think happened:

If you swell 1 by 10% you get 1.1. However, if you take 90% of 1.1 you get .99. These are not equal and opposite actions.