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  • FETOAU
    replied
    Bob,

    In answer to your question. The formula was derived from the basic engineering requirements for the maximum wind force generated by a given wind speed. There are a number of modifiers that were generated by consensus that are used for different conditions such as site, exposure etc. This method is included by reference in all current building codes.

    I used the worst case for all conditions so the result of my equation would be the absolute worst loading is in lbs/ft^2. If you want to determine N/m^2 substitute for .00256 and enter V (wind velocity) in M/sec.

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  • jep24601
    replied
    Yes, Francisco Moser chose Mexico City to set the world one hour record on his bicycle in 1984 because of the thinner air (incidentally he also added weights to the wheels) after Eddie Merkx had set the record there in 1972. Moser reset the record in 1994 again in Mexico City.

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  • Willy
    replied
    Well that came as a bit of a shock to me AK, I would have bet that the human engine would have favored the increased air density and oxygen over the decrease in resistance from altitude. Good to know.
    Speaking for myself I know that I won't be setting any records at 12,000 Ft.

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  • A.K. Boomer
    replied
    Good info Willy,
    be it from cold or lack of altitude air density even plays a major role in cycling, and that is generally at a much slower speed - and even though the human machine needs copious amounts of fresh oxygen to help fire off the muscles most all of your major speed records and assaults are set where there is less of it, simply for the fact that both the bike and the rider slip though the higher altitude air with less resistance... even with the huge trade-off - it's still well worth it...

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  • FETOAU
    replied
    Bob,

    The .00256 is the density of air. I don't remember the units. It has been years since I did this modified the equation. There are also four other constants involved that I reduced to the .85. I will look up all and post all units and constants within the next 24 hrs.

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  • Willy
    replied
    Don't forget to include air density in your calculations as it is an important factor.
    Air at 90°F weighs approximately 1.1 kg/cu. meter, at 0°F this same cubic meter weighs about 20% more. The air hitting the side of a structure is significantly heavier during cold weather.

    Just as air density is an important factor in the aviation industry it is also a key factor affecting fuel mileage while operating a vehicle on the road. This is a key factor affecting fuel costs in the transportation industry during the winter months. With large frontal area vehicles the increased air density can mean an additional $100/day between summer and winter fuel consumption.
    Lower fuel density and thus lower BTU figures and the increased rolling resistance encountered during cold weather are factors as well, but the biggest factor affecting fuel consumption is the increase of air density.

    To illustrate how significant this is, I've included a quote from an article that points out how much more additional power is required to push a large truck through cold air vs. warmer air.

    By Jason Rhyno You navigated your way through a wicked whiteout and have just battled winter winds for 12 hours. The load was delivered safely, and on


    How To Calculate Air Drag

    If you are using metric units for speed and area:
    Air drag force ( in pounds) = 0.008678 x air density x coefficient of drag x frontal area x air speed x air speed
    To get HP, multiply the air drag force by road speed (note that speed gets multiplied three times to get to HP and fuel consumption, so HP to push air goes up with speed x speed x speed.)
    Example: Your truck has a 0.5 drag coefficient and a vehicle speed of 100km/h and a 10.7 square meter frontal area, and no oncoming wind on a hot, humid summer day (90F 23C). Air is weighing 1.1013 kg/cu metre, and you are pushing the air with 511lbs or 0.607psi or using 63.1 KW or 84.7 HP of power to do that.
    Drag = 0.008678 x 1.1013 x 0.5 x 10.7 x 100 x 100 = 511.282 lbs
    At the same speed in winter (0F, -18C), it would take 642 lb of push, or 106 HP.
    If you have a 40km/h head wind on that winter day, the force jumps to 1259 lb and 208 HP just for air, not including rolling resistance.
    This formula can be used for a truck, car, motorcycle or the wind blowing on you when you carry a 4 x 8 sheet of plywood on a windy day just by adjusting the measurements and the drag coefficient ( 1.1 for a flat plate, around 0.5 for many vehicles, 0.1 for a smooth air foil).
    — Courtesy of Ray Cambell

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  • Bob Fisher
    replied
    What is the origin of the constants? Why not combine them? Bob.

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  • FETOAU
    replied
    wind pressure in psf is wind speed (mph) squared x .00256 x 0.85

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  • Paul Alciatore
    replied
    Wind loading? You betcha! For small buildings it is built into the building codes. I lived in South Florida and the building codes there took hurricanes into account. CBS (concrete block/stucco) construction is common. Re-bar protrudes up from the slab into the hollows of some of the blocks and re-bar is inserted down those hollows. Concrete is poured down them to form reinforced concrete columns. The top row of blocks are U shaped and are used as forms for more re-bar and concrete to make horizontal beams on top of the block wall. Steel clips ("hurricane clips") are inserted in those beams and stick up to tie to the roof to hold it down. Many of the roofs are covered with concrete shingles or heavy tile.

    Larger buildings will be checked for wind load by the architect. Many of them will sway as much as several feet in heavy winds and they must be built to take not just the force, but the motion into account.

    I worked as a TV engineer for many years and was responsible for the maintenance of tall towers (usually 1000 feet). After making sure they can hold up their own weight as well as the weight of the antennae and transmission lines (6" or 9" diameter copper coax is HEAVY), the next consideration for them is wind loading. They may be essentially hollow, but the wind forces on all those round and angle shaped steel pieces does add up. Wind and ice are the primary causes of tall tower failure.

    And then there are bridges. There have been some well known examples of bridge failure due to winds. A bridge is a complicated structure and each one must be analyzed individually. Bridge design engineers must have nightmares.

    Every city building inspector, every architect, every structural engineer is well aware of wind loading and they do take it into account. There is a lot of experimental data on the effects of wind on elements of various shapes: flat, round, H and I beams, angles, etc. Many of these shapes will have different wind loads on different sides.

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  • darryl
    replied
    Thanks for the links. That looks to be some informative reading- and has already answered my question. Just what I was looking for.

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  • J Tiers
    replied
    Every state building code will have a required minimum wind load for any new building built. It varies by area, with some areas having much higher loads than others. Then also various general national and international building codes also have such standards. Usually the state codes are based on some widely recognized code.

    It's basic stuff for architects and structural engineers, although every so often it seems that some building turns out not to have been built quite up to standards.

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  • boslab
    replied
    Its in the building regulations in the UK , wind loading of walls, they have gotten quite strict about it since numerous deaths have Occurred with walls falling on folk.

    A good guide for smaller stuff, not sure how it applies to big things like hangers though!
    Mark

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  • dp
    replied
    There are several answers depending on the specifics of the situation. Here are some guidelines:

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  • darryl
    started a topic wind pressure

    wind pressure

    We've had some wind here lately, though it has calmed down for now. But it made me wonder if there are any figures or specs for how much force wind can apply to a flat surface. Seems obvious that the higher the wind speed, the more force would be applied. I'm thinking mostly in terms of buildings. Some would be wood framed, some would be concrete block, others would be reinforced concrete structures such as high rises. Yeah I know, some would be cardboard boxes and scavenged tin roofing, etc.

    I was at our local airport cafe today and of course there are some rather large buildings which are hangers for aircraft. These are generally one large open space inside, and off-hand don't seem of a construction that would make them able to withstand that much wind. If 100 mph gusts come along- well I can envision some of these structures simply collapsing.

    My question- purely academic- how much push would be expected from wind blowing against a flat surface?
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