Effect on ice?

[lookingforluck]

Hello all,

I am looking at getting an atv or UTV. How does the effect on the ice based on weight change between a atv (600lbs and compact) vs. a UTV at (1200lbs and spread out)? Is the UTV that much "heavier" (per square foot) than an atv?

Thanks!

[Dave S]

depending on which ATV you're looking at, the 600 lb advertised weight is a dry weight. Add to that gas/oil/anti-freeze and packed snow/slush, and you're looking closer to 700 lbs.

If you figure the overall footprint, wheelbase x width for both machines and divide that into the weight of the machine, you can figure out the weight per square foot.

to make it easier, let's estimate this:

UTV

Width of 5 feet

wheelbase of 7.5 feet

5x7.5= 37.5 sq feet

1200/37.5 = 32 lbs/sq foot

ATV

width of 4 feet

wheelbase of 5 feet

4x5=20 sq feet

700/20=35 lbs/sq foot

Don't take my number as factual. It's only an example and real world number will vary based on machines, operators, gear, fluid resevoirs and most importantly ice conditions.

You can have 8 inches of good solid hard ice or 8 inches of soft weak ice.

With the thought behind my numbers above, figure this:

average person weighs 225 lbs.

with both feet together covering at best, 2 square feet.

225/2 = 112.5 lbs/sq foot.

So, in a totally unscientific comparison:

UTV = 32 lbs/sq foot

ATV = 35 lbs/sq foot

MAN = 112.5 lbs/sq foot

Again, the numbers above ARE ONLY AN EXAMPLE.

The best thing you can do is pick out an ATV/specs and an UTV/specs and compare.

To be safe, factor in 400 lbs extra weight of each machine for yourself/gear/liquids/snow/ice/slush

[lookingforluck]

Thank you Lep7MM!!!

[Roofer]

Atv's are actually much heavier than advertised. Here are some weights.

Copied from another site......................

#1 - Can-am Outlander 800R XT = 776 Lbs

26" Bighorns / Less the XT bumpers

#2 - Kawasaki Brute force = 777 Lbs

winch/skids/radio/cast wheels / 26" terra cross tires

#3 - Yamaha Grizz 700 W/EPS = 792 Lbs

Cast wheels/ 25" Bighorn Tires / Winch / Skids

#4 - Polaris Sportsman 850 XP W/EPS = 832 Lbs

Bone stock

#5 - Polaris Sportsman 800 X2 = 930 Lbs

26" Bighorns / front cargo box full of misc cargo est. (25 Lbs of cargo in front box)

[timjones]

So, in a totally unscientific comparison:

UTV = 32 lbs/sq foot

ATV = 35 lbs/sq foot

MAN = 112.5 lbs/sq foot

If this was truly the case, wouldn't the man fall through the thin ice before the UTV. I think you need to consider the actual foot print of the vehicle in contact with the ice and not the total square footage of the vehicle. So if the UTV weighs 1200 lbs and each tire has a foot print of say 1 sq. ft then it would be;

1200/4= 300 lbs/sq foot

This wouldn't be totally true because the weight distribution would not be equal to all 4 tires, but should be relatively close, depending on what your carrying and where it is located on the UTV.

[Dave S]

Retired,

you bring up a good point. Although the weight would be dispersed over a larger area, the actual contact area would be concentrated more as you stated. If there was a full contact area for the ATV and UTV, the numbers I presented would be more realistic.

Although comparing machine to machine, there would be little difference as the tires share about the same contact area.

[fishingforester]

I can't help myself, my ANALytical mind has to throw in 2 cents...

Devil's Advocate:

As the said 225 lb angler takes a step, is his/her weight not only on one foot at a given moment? Size of single foot (with pac boot) is roughly 12" X 6" = 1/2 square foot, putting force at 450 lbs/sq ft.

In reality, ice is not entirely static; it does have some elasticity, so the amount of force exerted by a certain amount of weight is spread out into a larger area than the exact area of contact.

I don't have the math equations to crack the code on this at hand, nor the ambition to track them down, but I have heard stories of a person on a snowmobile that stops, steps off, foot goes thru ice but snowmo doesn't sink. Little different weight distribution than ATV's and UTV's, but I think the principle is similar enough.

[Dave]

That sounds logical, ff.

Track manufacturers claim even though you're adding 400 pounds of tracks on a machine, the footprint impact on each corner is less than an ATV with tires on it; since the weight is spread out over a wider surface area.

[swedishpimple]

Thank God for Retired on Osakis

Lep7MM....your reasoning was extremely flawed. The weight of the machine is not distributed over the the 35 s.f. footprint, only over the surface area of the 4 tires.

Tracks vs. tires may be a better comparison.

[Dave S]

Thank God for Retired on Osakis

Lep7MM....your reasoning was extremely flawed.

Tracks vs. tires may be a better comparison.

Thus my claim of it being totally unscientific.

[lookingforluck]

So, as a guide with no ice being safe, what do you atv users consider to be a "safe" amount or ice for an ATV vs. UTV? I am thinking about getting an atv or utv and have thought it might not be a good idea to get a UTV because I will not be able to use it early/late ice like I could with an atv. What are your guys thoughts?

Thanks, LFL

[Dave S]

The DNR will say 6", I wouldn't go on anything thinner than 8" of consistant ice.

[lookingforluck]

The distance between the tires has to make some kind of difference, doesn't it? If you have the the same contact area (4 tires) and same weight but have them in real close, say a foot apart as compared to 10 feet this would change how the ice underneth is effected right?

[Dave S]

Yes.

It was mentioned in an earlier post about ice being somewhat pliable (I think that was the word).

One way to think about it is pushing your finger into an inflated balloon. The area around the contact point will flex and play a role in the strength and support of the actual contact patch.

In the case of something as small as an ATV/UTV, the area inbetween all 4 tires will allow the ice to absorb some of the weight of the machine by allowing some give.

Without having the software to show an example how the ice area is affected by the weight of a vehicle, it's tough to really explain other than the balloon example. The higher stress areas will become lighter and the farther you get away from the stress, it will be darker.

If you've ever received a video from any of the mfgr's showing frame flex and the stress areas of their machines, you'll see the color change in the high stress areas. It's really the same concept with ice. It's not just the contact area that's affected, but the surrounding areas as well.

This says it best:

In reality, ice is not entirely static; it does have some elasticity, so the amount of force exerted by a certain amount of weight is spread out into a larger area than the exact area of contact.

[fishingforester]

It wasn't easy, but I did find an online text version of an engineering book (Introduction to Cold Regions Engineering By Dean R. Freitag, Terry T. McFadden). For a textbook, it was actually an interesting read. Wading through the variable-rich equations, I did find an easy-to-interpret graph, indicating that size of the FOOTPRINT affects the load that a specifically thick sheet of ice can support.

Besides thicker ice being able to hold more weight (duh!), the graph shows that size of the radius of a vehicle's footprint has an increased effect as the ice gets thicker.

By far, footprint is not the only variable when figuring out how much ice can support. Elasticity, temperature, depth of water below ice are a few of the other variables covered; but in comparing 2 objects on the same piece of ice, radius of the footprint, and not contact area, is the determining factor.

[Dave S]

since the chart is in metric untis of measure, I did a quick check to see what 1 kN equals in lbs.

1 kN = 224.8 lbs

I do question how much water, if any, they're factoring in in this chart. The top measurement for a vehicle with a 10 foot radius figures out to approx. 180K lbs on 37.9 inches.

[fishingforester]

I know, I head-scratched about that too. I resigned myself to accept the graph for the relationship illustrated rather than the raw numbers. But I also brought to mind pictures of the Brainerd fishing contest, and that ice is honeycombed with Strikemaster signatures.

[rowdyf]

We must remember that when given "safe" ice numbers these are for clear ice.

Generally ice that is white or "honey-combed" has trapt air in it which reduces load bearing capabilities. Some estimates I've seen are that when calculating ice strength for white ice it is 50% strength of clear ice.

Also a static load (your 4 wheeler parked) will need more ice than your wheeler moving across ice at speed.

If comparing frozen slush and water between that and ice below, only the slush should be used as a safe estimate of effective ice depth.

Stress/cracking of ice will also cause load bearing losses.

We could get into a heavily based engineering discussion, but let's face it, are you going to take your $$ wheeler out on 2" of ice? No. Stick by the recommended guidelines.

[rowdyf]

One more thing, according to the Army Corps of Engineers (Ice Engineering, Safe loads on Ice Sheets bulletin) , a formula used for decades to estimate the minimum ice thickness required to support a load is:

k = 4(P)^1/2

where k is ice thickness in inches

P is the load or gross weight in tons

Plug and chug

[fishingforester]

I do question how much water, if any, they're factoring in in this chart. The top measurement for a vehicle with a 10 foot radius figures out to approx. 180K lbs on 37.9 inches.

[Dave S]

rowdyf and fishingforester,

thanks to both of you for providing the information.

[lookingforluck]

Thanks guys, this has been very informative.

LFL