Help with calculating vertical jump with kinematic equations

In summary: Yep, thanks for the help guys just needed to talk through it with somebody I guess. I added in a potentiometer to adjust for that height difference between standing flat footed and going up on your toes.
  • #1
mjohnk11
3
0
Ok, so I'm using an arduino with an led matrix and a homemade pressure sensor to figure out the height of a vertical jump and then display the info back. First I get the amount of time between when the person left the board and when they land back on it. I then divide the time in half and use the equation:

Yf = 1/2 * g * t^2 substituting g for 386.088 inches per second^2 giving
Yf = 193.044 * t^2

My problem now comes in here. I'm doing this inside a room with an 8 foot ceiling for easy reference on how high I am jumping. I'm 6' 1" so when I jump so that my head just barely touches the ceiling my vertical is roughly 23". The arduino, though, is reading a little over 16". My first thought was that the pressure pad was staying down for some time after I jumped, but I recorded a jump and it's working right starting the time right when I leave the pad. In fact I even used the time I got from comparing when I left the ground on the jump and when I landed in the video and got .58 seconds. Using the equations before that gives a little over 16 inches max height (like what the arduino measured). So, does anyone have any idea what's going on here?
 
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  • #2
Your center of mass might be in different position relative to your feet when you separate from ground, and when you touch back. If you post the video and mark where you get 23" from on some screen captures, we could probably analyze it further.
 
  • #3
Yes, it's most likely to do with posture, but you're looking for a reason why the time recorded might be too short. So bending your legs ready to land could not account for it. Maybe it's your arms - if you raise them when descending you'll land sooner.
 
  • #4
I think it's the other way around. He was trying not to jump too high, so he might have took off with legs still bent, but landed on straight legs.
 
  • #5
I thought about it being posture related also, but when going through the video it doesn't look like that. I'd upload it but it just a recording of my legs and a light on the arduino for comparison of where the algorithms at. On the jump I start with legs bent and jump up--legs extending and toes eventually pointing down. The arduino doesn't recognize me as leaving the ground until my toes come off the board (although I'm not pushing down anymore: this may be where the error is). I go up then land in the exact same manor, feet pointed down, until they touch the board then the rest of my body comes down to absorb the impact. When I begin to touch the board again is when the algorithm gets the total time. It takes very little force (probably about 10 pounds) to trigger the pressure pad.

The measurement for comparison is the top of my head, which should be a fairly accurate recording of how high I'm getting off the ground since that would be highest point I can reach.

The two errors I'm thinking of right now is that the difference between being flat footed and having feet extended is causing the error, or it's the one I mentioned earlier where I'm already decelerating before leaving the pressure pad. Thanks for the responses guys, let me know if either of those or anything else seems like a more likely cause. It just threw me off a lot when I started getting these consistently differing results.

Edit: Actually looking back at the posts now, the first reason seems like what you were saying k^2
 
  • #6
mjohnk11 said:
Edit: Actually looking back at the posts now, the first reason seems like what you were saying k^2
Not exactly. K^2 and I have both been thinking about differences in posture between take-off and landing, but it looks like the explanation is much simpler. You take off and land with toes extended, so your effective height is some inches more than you thought.
 
  • #7
haruspex said:
Not exactly. K^2 and I have both been thinking about differences in posture between take-off and landing, but it looks like the explanation is much simpler. You take off and land with toes extended, so your effective height is some inches more than you thought.

Yep, thanks for the help guys just needed to talk through it with somebody I guess. I added in a potentiometer to adjust for that height difference between standing flat footed and going up on your toes.
 

What is a kinematic equation?

A kinematic equation is a mathematical formula that describes the motion of an object without considering the forces that cause the motion.

How do you calculate vertical jump using kinematic equations?

The formula for calculating vertical jump using kinematic equations is: Vf² = Vi² + 2ad, where Vf is final velocity, Vi is initial velocity, a is acceleration, and d is displacement. You will need to measure the height of the jump (d) and the time it takes to complete the jump (t) to solve for Vf.

What is the acceleration of a vertical jump?

The acceleration of a vertical jump can vary depending on the individual's strength and technique. However, in most cases, the acceleration will be approximately 9.8 m/s², which is the acceleration due to gravity.

What factors can affect the accuracy of calculating vertical jump using kinematic equations?

The accuracy of calculating vertical jump using kinematic equations can be affected by factors such as air resistance, variations in technique, and measurement errors. It is important to take multiple measurements and use proper techniques to minimize these factors.

Can kinematic equations be used to calculate other types of jumps or movements?

Yes, kinematic equations can be used to calculate various types of jumps and movements, as long as the motion is one-dimensional and can be described using the equations of motion (Vf = Vi + at and d = Vit + 1/2at²). These equations can also be applied to other types of linear motion, such as running or throwing a ball.

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