Parachute Physics
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- Опубликовано: 25 июн 2024
- This LabRat video discusses the basic physics behind non-gliding parachutes. The video includes insights on the physics, provides some mathematical analysis, and presents parachute drop tests to determine the drag coefficients of small test parachutes. This video is geared towards high school physics teachers and students and beginning engineering students. More information of this topic can be found at www.labratscientific.com.
I'm doing research for an assesment and I was really stressed out until I found this video. Even my friend, who was not doing homework, stayed to watch the whole thing. On behalf of tired students, thank you for your service.
I'm that friend, very good video
Great video! I'm working on launching a high altitude balloon and wanted to learn more about parachutes. Thank you for putting this together.
this video single handily saved my physics paper that i did on how mass of a parachute effects its terminal velocity, thanks legend
Very interesting video on parachute. Your explanations are clear and simple to understand. Good job, I hope you will continue to teach us cool things like that !
Studying for A Level Physics Unit 1 for june 2025. This video is great as i am teaching myself lab practicals from youutube vids like this one.
Thanks so very much.
The radius mislabelled cause me to invent new equations to get results
Great video! I'm working on launching myself into the air and wanted to learn more about parachutes. Thank you for putting this together.
Great Video! good break down of the topic!
Thankyou, your video helped me with my thesis
This video was very helpful, thank you for putting it together.
Thanks! this is a great video!
Thank you.. Wonderful videos
Great video!
it was really helpful sir...keep up with such content. Thank You
Great video.
thank you sir
this was really helpful♥
Great video! I'm working on launching children into the air in order to learn more about parachutes. Thank you for creating this video.
Is there calculus in the process?
I am working on something related to calc and I am interested with doing something with parachutes
Hey, awesome video. I have a question.
I'm using the exact same sensor to measure the velocity of a toy parachute, however my readings are all over the place. They're spikes that go from negative to positive continuously. My values are also way off, for context if I drop my parachute (ignoring the spikes) the maximum velocity was around 15 m/s which is quite high. I've checked that it's not a problem with the sensor since I've tried many others and they all seem to have the same problem. I was wondering if you did any calibration or changed any settings for it to be such a smooth curve with accurate values.
ALSO, can you explain why you're velocity starts at 0 but then goes down in the negative direction, and then proceeds to be positive at the end?
Hmmm, I'm not sure what to say... Are you using a guide string to keep the parachute falling in a straight and orderly manner? If the chute is swinging and drifting off to the edge of the sensor "cone" there could be noisy data. It's been a while, but I don't recall changing the default settings on the sensor. It is my experience that if the drop object is falling too fast the sensor data can be choppy. Have you tried using the "old school" stop watch method.
Nice tutorial. Appreciate the extensive details. One question, does the hole in the parachute to allow for the guide line impact the Cd?
A small hole wont affect the Cd noticably. However, larger holes will allow more air to pass through the canopy and thus reduce the mass of air the canopy must try to accelerate. This will reduce the CD. The air passing through the larger hole also increases the pressure behind the canopy, further reducing the drag.
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Well taught, but how to calculate for self weight of the canopy of radius 34 meter long ?
Please i have a reaserch this year and i hope that you can help me how i can do asimilar experience about What surface area of parachute fabric should be provided to prevent the jumper from breaking his leg upon landing?
So when the mass is increased with a constant parachute area and shape the drag coefficient should decrease? the greater the mass the higher the terminal velocity since terminal velocity is squared and is inversely proportional the drag coefficient should decrease right?
Sir,
At 15.48 mins, area of the canopy when laid out flat is to be measured but how to calculate the area when the canopy is of Aeroconical and or parabolic shaped which cannot be laid out flat on the surface.
Regards
Great video, a bit confused about how u calaculated the area of the canopy?
Simply lay out the parachute flat on a table and measure its area. For a square chute, Area = Side 1 x Side 2. For a circular chute, Area = Pi x Radius2. In other words, it just the geometric area of the canopy material.
where that formuler of pressur come from?
May I ask as to where did you get your values for your drag coefficient as I would like to reference it in my paper! If anyone also has an idea as to where I can find it, please pitch in:)
My drag numbers are based on data provided in the reference book known as Horner Drag. The book has been around for a long time and includes a lot of experimental data with many hand drawn graphs and figures.
@@labratscientific1127 Thankyou so much!
@@labratscientific1127 what page is it
how effective is a parachute in .088 psi? How big would the parachute have to be for 1000lbs 100lbs how long would it take to slow down from 20,000 mph?
At 0.088 psi (12.67 lb/ft2) a 5 ft radius circular chute would produce about 700 lbs of drag. Parachutes are generally designed to give a certain impact velocity. The chute size is calculated by rearranging the Drag Equation to solve for Area. Assume terminal velocity where Drag = Weight (in your case 1000 lbs) and you get an area of about 3,000 ft2, which is a 31 radius circular chute. That's 62 feet across. 20,000 mph is about Mach 25 which is far to fast for deploying a parachute. Parachutes can only survive Mach numbers on the order of 2.0 to 2.5... The time it takes to slow down is much more difficult to calculate because the velocity changes as the parachute descends due to changing atmospheric density... If the chute provided a constant 1g (32.2 ft/sec2) deceleration it would take about 15 minutes to slow down from 20,000 MPH (29,333 ft/sec).
@@labratscientific1127 so a parachute will still work in low density atmospheres... that's weird. Maybe earth atmosphere denser than I think then. Thanks for taking the time to reply I appreciate it.
for the first example problem, at the end, you have mistaken radius with the diameter, ( at10.08) 34.4 m should be the radius of the area we got which is 3725m.sq
You are correct! Good catch. Its nice to know that someone is really looking at the details. I'll correct the video in the near future. Thanks.
There was no correction to date. I had a sleepless night trying to calcukte the diameter 34.4m. Now if radius is 34,.4 then it makes sense. My head hurts less now.
I am trying to help my daughter with a physics experiment. The experiment is to drop a 4 pound pumpkin from 30 feet. we are attempting to make a parachute for this project. I was wondering if you can help me with the parachute size we would need for the pumpkin to fall and not break? thank you
Ideally the first step is to determine the maximum impact velocity that a pumpkin can survive. This would be done with drop tests (my guess is 2-3 feet…). The fall time and associated impact velocity for a given drop height can be calculated as follows: Fall Time (in sec.) = Square Root (fall distance (in feet) / 16.1) and Impact Velocity (in Ft/sec) = Fall Time x 32.2. This is the necessary terminal velocity of the pumpkin & parachute system. The drag equation (Drag = 1/2 x Air Density x (Vel x Vel) x Cd x S) is used to determine the necessary flat area of the parachute. Applying some algebra: S = (2 x Drag) / ((Vel x Vel) x air density x Cd). Where S = flat area of the chute in Square Feet, Air Density = 0.00237 lb*sec3/ft4, Cd = 0.6, and Drag = Pumpkin Weight (in lbs). We can set Drag = Weight if we assume the pumpkin/parachute system has reached terminal velocity before hitting the ground. Terminal Velocity means the acceleration is zero (not 32.2 ft/sec2), which is achieved when the drag perfectly counteracts the weight. EXAMPLE: For a 3 ft free fall drop. Fall Time = 0.43 sec. Impact Velocity = 13.9 ft/sec. For a 4 lb pumpkin: Parachute Area (S) = 29.6 ft2, which is a 3 Ft diameter round parachute (S = 3.1416 x Radius Squared). For this to work the height from which the pumpkin is dropped needs to be 20 ft (?) to make sure the chute opens and terminal velocity is achieved. Kind of hard to give a full lesson in this forum, but maybe this will help...
@@labratscientific1127 thank you I understand what you are saying about this being hard to explain on this platform. We made a 32 sq ft parachute and did a 25 foot drop test with a 4 pound dumbbell and it thermal velocity was reached at about 20 feet where the parachute was able to counter act the fall. It was pretty successful we did not try with the pumpkin yet.
Can you please show us the calculations to get the drag coefficient, it’s a bit confusing
The drag coefficient for a parachute is best determined by conducting tests (i.e. wind tunnel or drop tests). For a wind tunnel test - the measured force (the drag force) is entered into the drag equation: Drag = 1/2 x Air Density x Velocity2 x Cd x Canopy Area. In my equation above, "Velocity2" is "Velocity Squared". Rearrange the equation to solve for Cd... You will need to know the air density and and wind tunnel air velocity. For a drop test, the Drag = Weight once the parachute has reached Terminal Velocity (maybe a second or two after the release for a model rocket parachute). The average velocity can be determined by timing how long it takes the parachute to fall a certain distance - after it has reached terminal velocity.
Can you build a parachute out of metal? not worrying about deployment
Yes, metal could be used to make a parachute. Aeroshells used on Mars entry vehicles are a sort of "parachute" - they are just not big enough to slow down the lander sufficiently. A non-porous material like metal does not have the best drag characteristics since "porosity" (the amount of air that passes through the material) can affect the drag. Some porosity is good - especially when trying to control opening loads.
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@@labratscientific1127 Thanks for taking the time to answer
Well, the shape of these parachutes are closer to a hollow hemisphere, for which the drag coefficient opposite stream is 1.42 (see Engeneering toolbox here: www.engineeringtoolbox.com/drag-coefficient-d_627.html)
So, imho there's no need to search for errors, your measurement is close to what others did.
The T-10 military parachute has a Cd around 1.3 also...