# Home project help please - max wind speed to tip over object

• Lawn/Garden
• johndalessandro
In summary, if the wind speed is 25 mph and the weight of the footprint is 2.5 lbs, the footprint must be made of a material that can withstand a force of 3 lbs per square foot.
johndalessandro
Hi guys, thanks for helping me! I am working on a home project that includes a privacy shade for my backyard. I have a 6'x15' vertical shade attached to two portable posts that are held up right by four legs ("X" shaped footprint). the surface footprint of the legs covers an area of 46"x70". I am trying to figure out how heavy my two posts have to be to keep my vertical shade from tipping over in the wind at 25mph. This will determine what kind of material I use for the post (probably steel) and if needed how heavy of a sand bag I would need to add. I calculated that at 25mph the shade will experience 1.6lb of force per sqft (not sure if this helps, not an engineer or student).

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The force that I have for 1 sqft at 25mph is 2.5 lbs so yo might want to check your number to start.

Next you might want to consider gust which are about 30% larger than the average wind speed.

Your problem is basically one of counteracting torques. In this case the tendency of the screen to rotate about the ends of your "footprint" due to the torque produced by the wind. Torque is a twisting force defined as applied force (in this case the wind hitting the screen) Fw times the perpendicular distance of the line of action L from the pivot point P (in this case the end of the footprint". This torque must be counteracted by the weight of the footprint, W time the perpendicular distance of the line of action l from the pivot point P.

Now there is another problem since the feet are not connected by a ridged structure. If the wind force is not dead center on the screen the load on the footprints will not be equal. Also, the net wind force is probably not exactly 3 feet above the ground. So, you must include a safety factor. How much? That is just a guess. Perhaps one of our engineers can suggest one. Also, you need to expect in any situation of deployment that the wind might come from either direction so the positioning of the weigh should take this into consideration.

russ_watters, johndalessandro and Lnewqban
johndalessandro said:
im working on a home project that includes a privacy shade for my backyard.
Have you thought about material options that let some of the wind through but still provide adequate privacy? Perhaps a slightly open dark mesh material?

Or even better would be to use a material or assembly almost like Venician Blinds, where the angled pieces overlap enough that you can't see through them, but they are much more transparent to wind...

johndalessandro, Lnewqban and dlgoff
In retrospect I think you need a rigid beam connecting the two mounting posts as the wind will create a tension in the fabric that will tend to pull the supports inward toward one another seemingly fairly easily.

russ_watters, berkeman and Lnewqban
gleem said:
The force that I have for 1 sqft at 25mph is 2.5 lbs so yo might want to check your number to start.

Next you might want to consider gust which are about 30% larger than the average wind speed.

Your problem is basically one of counteracting torques. In this case the tendency of the screen to rotate about the ends of your "footprint" due to the torque produced by the wind. Torque is a twisting force defined as applied force (in this case the wind hitting the screen) Fw times the perpendicular distance of the line of action L from the pivot point P (in this case the end of the footprint". This torque must be counteracted by the weight of the footprint, W time the perpendicular distance of the line of action l from the pivot point P.

View attachment 295523

Now there is another problem since the feet are not connected by a ridged structure. If the wind force is not dead center on the screen the load on the footprints will not be equal. Also, the net wind force is probably not exactly 3 feet above the ground. So, you must include a safety factor. How much? That is just a guess. Perhaps one of our engineers can suggest one. Also, you need to expect in any situation of deployment that the wind might come from either direction so the positioning of the weigh should take this into consideration.
Thanks so much for your insight, so the way I am thinking is, anything under 3ft won't make it tip but any force above 3ft would.. in that case would it acceptable to calculate Fw at 5 ft? when I do the math I divide the max wind force to the total sqft of the shade or total sqft of the shade above 3ft? Also, wouldn't "I" be from the center/post to the edge of the footprint. If this is the case half of 70 or 35in?

berkeman said:
Have you thought about material options that let some of the wind through but still provide adequate privacy? Perhaps a slightly open dark mesh material?

Or even better would be to use a material or assembly almost like Venician Blinds, where the angled pieces overlap enough that you can't see through them, but they are much more transparent to wind...

View attachment 295524
I thought about that but I wanted to keep the build simple, something I can easily roll up and stow in the shed when I don't want it up.

You could also think about staking or perhaps better) planting the posts with concrete footings.

Or better yet concrete footings with receptacles to temporarily receive the supporting poles. This might even reduce the need of a cross beam between the poles, but you might still want a quick release for the canvas if the wind becomes too strong. Additionally, you could also make "U" shaped cuts in the canvas to relieve some of the wind pressure.

## 1. What is the maximum wind speed that can cause an object to tip over?

The maximum wind speed that can cause an object to tip over depends on various factors such as the size, shape, weight, and stability of the object. Generally, objects with a larger surface area and higher center of gravity are more prone to tipping over. However, it is difficult to determine an exact wind speed as it can also vary based on the direction and duration of the wind.

## 2. How can I calculate the maximum wind speed for a specific object?

To calculate the maximum wind speed for a specific object, you can use the formula: V = (M x g)/(C x A), where V is the maximum wind speed, M is the mass of the object, g is the acceleration due to gravity, C is the drag coefficient, and A is the projected area of the object. The drag coefficient and projected area can be determined experimentally or by using mathematical models.

## 3. Can a strong foundation prevent an object from tipping over in high winds?

Yes, a strong foundation can help prevent an object from tipping over in high winds. A sturdy and well-anchored base can increase the stability of the object and make it more resistant to wind forces. However, it is still important to consider the size and shape of the object as well as the wind speed and direction.

## 4. Are there any design strategies to reduce the risk of objects tipping over in high winds?

Yes, there are various design strategies that can help reduce the risk of objects tipping over in high winds. These include using a wider base, lowering the center of gravity, and adding wind-resistant features such as aerodynamic shapes or wind deflectors. Additionally, using materials with high strength and durability can also increase the stability of the object.

## 5. What are some common objects that are prone to tipping over in high winds?

Objects with a high center of gravity and a narrow base are more likely to tip over in high winds. Some common examples include tall structures like cranes and towers, outdoor furniture, and vehicles such as trailers or RVs. Trees and power poles can also be susceptible to tipping over in strong winds if they have weak or shallow root systems.

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