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My phone has survived a 10 foot fall many times haha. I almost launched my phone in my high powered rocket last week which hits a max acceleration of 34G's haha

I used a similar app a long time ago that did a similar thing. I might try this out. good idea. thank you

interesting approach. ill give it a go and see what i come up with. thanks for this idea

ok, sounds good. Finding that time is the hardest part i feel. IE, finding the landing force of a parachutist is in the same boat as this problem i have. i use 0.1 for dirt and hard surfaces, and 0.3 seconds for bushes. i wonder if tensile strength will be a good starting point for this problem?

in my original post, i use chain and thats what i am using now is a chain. the title is incorrect because i was going to use a cord until i realized htat was going to add extra stuff i did not want to use at the moment haha i suck at movie quotes. and i have not seen that movie. I have had to...

ok, so using the idea of F=ma. Now what about F=p/t using the idea of momentum?

this is my test bench i made to test hanging weight until failure and now i want to test snatch force or drop force or whatever it is really called. my goal is to find out at what force the eye bolt pulls through the plastic or the part fails by dropping the weight.

yeah, this is almost like the 700 post thread about impact force of a car into a wall haha. Im thinking if i stick with F=mv/t and use a small value of t such as 0.1, it might yield some decent value. But thats a super rough guess

yeah i forgot to edit the title. i was going to go with a nylon webbing cord, but that was adding more complication, so in the original post, i changed it to a chain haha

thank you for your answer. Impact testing i know is super complicated... hence the mass amounts of testing for crash test dummies and cars. The deformation and elasticity of the object, material properties, and the type of ground or object it impacts with. Usually for concrete i use a time of...

im not sure i see why time is important haha? I already have 40Lbs of weights hanging from the chain at the moment. As for the strain gauge or scale, i like that idea. It gives some real numbers to work with. I am looking for some basic calculations as well so i can add those to a report im...

The eyebolt on the 3d printed part is held in place by a washer on both sides and then a single nut. I want to know how much force the washers puts on the 3d printed part when the weights fall and hit the max length of chain. This will tell me how much shock force (thats what i call it) the...

i have a question. I have set up an expierment and need some values. I have a round, 3d printed disk, 0.25 inches thick and 4 inches in diameter. In the center, i have a eye-bolt attached. I have hooked up a chain to the eyebolt and some weight to the chain. I have 40Lbs hanging right now and...

Perfect. I will see if i can test it at home and go from here. this does help me a lot believe it or not. thanks for the information

Ok, i have been searching the google for about an hour and have come up dry. I am trying to either test at home or find some actual numbers for how much inner pressure a thin walled (3mm) cardboard tube take before it bursts? IE like filling it with water and then it pops. Second, im looking...

For those who have built those small model rockets, please help me understand a very very simple physics problem that i for some reason do not fully get. Why is it that the motor tube does not have a retention ring on it, to prevent the motor from flying out the nose cone? i know it cant be...

well, better than me haha. i taught high school math for a few years, but physics was never my strong suite haha.

thank you very much my friend:) your an excellent teacher

thanks for all the help. I really do appreciate it alot. Our actual project is to design an active drag system for our rocket. This ADS will be used to slow the rocket down to achieve as close to 10,000 feet as possible. so far our design is pretty cool.

average terminal velocity is 234.68. IF this was free fall... but we know its not

just making sure but K is the same as Beta right? give me a few moments to get this calculated. might be about an hour or so. had something come up that is very important

with m(dv/dt) = mg-kv^2 and if i set dv/dt = a = 0 therefore 0=mg-kv^2 then solve for v using the correct signs kv^2 = mg v^2=(mg)/k v=sqrt(mg/k) if i am misunderstanding you, im sorry haha:)

i solved for v which then becomes imaginary due to the negative in the sqrt. so i need to solve for v a different way. ok, iw ill write soon

nope, it will be imaginary. i am trying to see which way to solve this.

Vt = sqrt(-mg/k)

just from a basic FBD. so technically it should be kv^2 or the Bv^2.. i did forget about the quadratic drag haha

well, no time like the present. Lets give it a go. ok, we know that F=ma and in free fall, a=0 so i would have 0=mg-kvt where kvt or kv is pointing up on a FBD and mg is down but before velocity is terminal, so just kv we would have ma=mg-kv a=dv/dt m(dv/dt)=mg-kv i can seperate and solve the...

My friend (someone from the rocket forums) used an L1030 motor. You are correct that it is a 54mm rocket. Ok, for the Cd and 200m/s that makes sense. I was thinking it was either that or vt. I have loved every step of this project because i have learned so much from you guys.

i think so haha. I just got home from my 3 hour train ride from school. Which equation should i start with? or is it a whole new equation you have in mind? thanks

ok, thats what i thought. I told that to my professor and he didnt have time to think about my response during the team meeting and he looked confused haha, but i re-assured him that a $80 sensor may not be the most accurate and precise. the "truth" being the consistent beta values that give...

second, i noticed some of my Beta values are negative. technically they should not be negative at all because B=(Cd)(area)(rho)(0.5)

there are other Vt equations as well i could probably use. now to answer your question, i thought Vt would be in my data. but maybe its not since its not completely free fall. If it were free fall, i would expect to see acceleration go to 0 at some point when the object has stopped accelerating...

somewhere along the line we were discussing finding the terminal velocity. We discussed how its exponential and if i integrate the equation we have, i do indeed get the terminal velocity equation inside. Somehow i thought we were finding both the Cd and terminal velocity thats why haha. but...

ok, cool. I have noticed i do not have 0 in my data for acceleration so im guessing its when acceleration is a differnt value. post #40 is that not the DE?

"To do that: What is the acceleration of the rocket at terminal velocity?" Well, my thinking is either 0 or 9.81 since terminal velocity is where Fd and Fg equalize. So probably 0 then

so what is happening at 200m/s with this graph? thats where the system settles too. Is this terminal velocity? I am not sure if it is or not because using the Vt calculation using my 0.62 Cd, density, etc, i get that Vt is close to 600m/s thanks

beta changes alot from 50 to 100m/s but then the system settles around 200m/s. The Drag Coef seems to be an average of 0.6 from 150 to 200m/s.

I agree about the pride. That is why im REALLY grateful that you showed me ALL the steps. I do not expect that from everyone or anyone, and when i have no single clue what i am doing, its so nice to see an example ALL the way through. That is how i learn. step by step and examining each section...

this reminds me of slope field diagrams in a way, im guessing where the lin flattens out is terminal velocity? not sure that is correct thinking on my end because based on a terminal V calculator online, i get 598m/s based on the same numbers i used. I am probably jumping ahead EDIT, i added a...

sorry it has taken so long. I REALLY REALLY REALLY APRECIATE your mass amounts of PAITENCE with me. I am really good at book work and homework, but real world stuff gets to me and it makes me sad and nervous for my future carer.

At 50m/s. At 55m/s a drastic change

i will keep chopping data back until it either changes or i find out of a wrong calculation EDIT: i just saw your post stating the same thing as me.

hmm, its better but not the same. i made sure no crazy values were in the mix either.

yup, something is not right. I selected all the average velocities as X and Beta as Y and this is my plot

me niether. Let me close excel and refresh my RAM.. MAYBE some stupid glitch. I will be back shortly

i get the exact same thing as post 104

i get the exact same thing

then what will the plotted data be? Im not sure i follow, but i will change the axis to match what you have said.

correct, and when i select that data, the velocity is a linear function with negative slope, and B is linear with 0 slope. So somewhere, either excel is trying to be "smart" and adjust the data or i have selected the wrong data haha, which is possible

even plotted against velocity as the x axis or even as a data set, its still looks scattered