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The force on each car is purely longitudinal because the restraining force of necessity runs thru the central point (origin). For example, the East and West forces cancel so far as the origin is concerned, so no lateral force exists from the viewpoint of the North or South car. The restraining force on each car, then, is the same as if it were tied to an ...


4

No, positive acceleration, alone, does not need to imply positive velocity. Another term for "acceleration in the opposite direction of the velocity" is just "deceleration". The negative value is being rejected based on physical context. That is, it is being rejected based on the fact you are starting from rest along with the fact that ...


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Alhtough this answer should be awarded to Pete, because he has spotted it first. The accelerometer has a sensitivity of 5%, which corresponds to 100[mV/g]. what you are measuring is 0.1[m/s^2], which is about 0.01g. So the actual signal you should be receiving from the piezoelectric accelerometer is about 1[mV], which is quite low. Now, if you see the ...


3

You are unlikely to get any useful results from this, for several reasons. The accelerometer "x,y,z" directions are relative to the accelerometer. Since the runner's leg is rotating, they are constantly changing orientation relative to the ground. You don't know how to split the "1g" gravitational acceleration into the correct components at each measured ...


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While the acoustic analysis idea may* work, I think the accellerometer data would be much easier to analyze. If you are looking for something that can give you better rates of data, try looking into other accellerometers. This one from Adafruit has a maximum rate of 800 Hz(datasheet). According to your calculations, this should be enough. You could log this ...


3

The complication here is that the simple relationship between wight (or downforce assisted equivalent) coefficient of friction and traction is only true up to a point and in performance tyre that point is where the thermal and mechanical loads on a tyre start to change its mechanical properties. Performance tyres are generally designed to work best with a ...


2

One way might be to try to model and understand the physical problem. You are however clear in your question that you would prefer an empirical approach where constants are tuned to experience data. I would suggest that the easiest and most general way is to use a piece-wise approximation of the velocity curve. Generically this may be made by using basis ...


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You actually don't need to know the force on the wheel to work on the power. Just look at the kinetic energy of the powder. Kinetic energy = $(1/2)mv^2$. For 0.013 kg of powder, the kinetic energy is $(1/2)(0.013)(104.72)^2=71$ Joules. You need to provide that much Joules every second, so the power is just 71 Joules/second = 71 Watts. Of course, this ...


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I assume you will be doing this in a software, matlab or Abaqus for example, it’s not particularly easy thing to do with your given info, but. Perhaps this example method will help Abaqus Infos at csiamerica Matlab function I am unsure how directly your data types will fit, but these are some links to get you started


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As posed, you have three unknowns - $a1$, $a2$, and $a3$. The tension on the cable is everywhere uniform, but the accelerations are different because the static force components affecting the tension are different. $F=ma$ at each mass. So for mass $m1$, the tension is $m1*(g*cos53 +a1)$. Likewise $m2*(g*cos37 + a2)$ and $1/2*m3*(g*cos90 +a3)$. These are all ...


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In principle yes, it can be done. It would be enough to put a MEMS accelerometer inside anything with an MCU and the rf component. The device should work like an Inertial Measurement Unit (IMU) which might not be the simplest thing to do. Still you need to keep track of movements just for a few seconds and this can be done easily through simple integration, ...


2

When working with tangential and normal directions/coordinate system in the circular motion, you can thing of: tangential acceleration: the change in the magnitude of velocity normal acceleration: the change in the velocity direction In a case of uniform circular motion, the magnitude of the tangential acceleration is zero, because the magnitude of the ...


2

As suggested I would try to use accelerometer. Below you will find a link to source code that might be helpful. I used a ADXL 345 accelormeter in combination with a TIVA ARM Cortex M4 Microcontroller. You are welcome to fork the source code TIVA - ADXL345 Accelerometer Demo Using I2C Tiva_I2C_Nokia_ADXL345 source code


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I will rephrase DKNGuyen's answer (which is correct - I upvoted ). I assume that the positive (+) sign is for position, velocities and acceleration is to the right. The object is starting from rest. Therefore its velocity is zero. We don't explicitly know the initial position, but we can assume that its a very small value different that zero (any value ...


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You are almost there. In order to calculate the tangential and the normal acceleration you need to take them as vectors. The angles are $\theta= \arctan(16/12)=-53.13 deg$ for velocity $\phi= \arctan(16/12)=-26.51 deg$ for accelaration Therefore the vectors for acceleration and velocity would be And then you can find the component of tangential and ...


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You are using a piezo electric accelerometer. There are always always always offsets and low frequency noise on a piezoelectric accel. Depending on the charge amp, anything below about 3 hz is meaningless. Just high pass filter the signal to get rid of it. If you need to read below 3 hz then you shouldn't use a piezoelectric. Use mems or piezo resistive in ...


1

Point $B$ is the system's initial instantaneous center. So the acceleration of point $A$ is equal and opposite to the acceleration of $G$ at $t=0$. This is the trick to solving this problem. There is a force couple of $2452.5\,N$ at slip inception with an arm of $0.8\,m$ to the instantaneous center $B$. There is some additional tension on the rope needed to ...


1

Since the conveyor belt is resting, the last equation does hold's true. The fact that it doesn't, suggests to me that there is some numerical error in the calculations. I've run through your equations and they seem ok. I couldn't follow all your conventions from your sketch, so I tried solving the problem myself. The results I got are slightly different. i.e....


1

First of all you need to be about which reference frame you are estimating the acceleration of A. I assume you need the accelation of A wrt to the inertial frame of reference. The most generic way is to link the acceleration of A to B. Then find the acceleration of B with respect to O. See the following link for a starting point The bottom line is that if ...


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You are fine. I'm not checking your work, but if you know engine torque, the using gear ratios you can calculate torque at the wheel. Force at the road is torque divided by wheel radius, and then A = F/M


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I know how much the frame of reference is rotated along the x and y axes and I know the position of the vector in this frame of reference. If you know how much, then you must know the order. How much is dependent on the order. The problem is, there are several ways to get from one reference frame to another via rotations. You need to look at what you mean ...


1

First, a bit of background. I spent 6 years running a packaging test lab here in the U.S. and part of my job was collecting shock and vibration data from the field for different transportation modes (semi tractor/trailers, TOFC, railcar, delivery trucks, ships...etc) around the world, then using that data to develop and/or validate lab test protocols. I've ...


1

I couldn't add a comment (new to the forum) so posting in the answer section. Bosch has since acknowledged this to be an error and updated the datasheet's errata. They have added a 3/2 factor to the original formula. I am wondering, do we really need to do this way? The LSB$/$g sensitivity just needs to be divided by the register value and with that, if we ...


1

Unfortunately, the arrangement you show won't work very well. There are a whole raft of reasons, but they boil down to two classes - gimbal lock and singularities. I'll try to illustrate the problem with an example of an entire class of motion that this setup can't handle. Based on your sketch, I'm assuming the IMU output is acceleration, and that you are ...


1

The first thing you need to note is that when your engine is rotating freely, i.e. your clutch is disengaged, then it "produces" little to no torque, regardless of the current RPM. The torque rating of the engine at a certain RPM is the torque you need to apply to the output shaft to bring it to an instantaneous standstill. Thus, to mobilise the torque, you ...


1

If you insist on using the accelerometer data, there are probably libraries, searching "accelerometer tool" (on the google play store returns several results that allow exporting of data etc), depending on your platform that do most of the maths for you. but seeing as acceleration is rate of change of velocity which in turn is rate of change of distance. you ...


1

Let's put aside this specific game, and just say we want to calculate the speed of some generic vehicle. I think that will answer what you are really trying to get at. Our vehicle is going to obey Newtons\'s second law $F=ma$, force = mass times acceleration. So, let's add up all of the forces that act on the train. The main ones will be the driving ...


1

Accelerometers are capable of measuring the acceleration they experience relative to free-fall. Accelerometers are used to measure the upwards acceleration that counters gravity when at rest, its a hoax that it measures the acceleration due to gravity. This acceleration is measured as 1 g (g = 9.8 m/s2) on the z-axis, when both pitch and ...


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The factor here is mass, not weight - weight is the force produced by mass and the effect of gravity. The same kinds of forces are produced by acceleration. In "weightless space" (zero gravity) a spacecraft accelerating at 1g (9.81 m/s2) will impose the same force on its occupants as earth's gravity. For a high speed spacecraft, a lot of the time will be ...


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For the first question, you can split up the equations of motion into normal and tangential direction: $$ma_n=F_n$$ $$ma_t=F_t.$$ The tangential force $F_t=12/\pi \text{ N}$. Hence, $a_t=\frac{12}{10\pi} \frac{\text{m}}{\text{s}^2}$. For the normal force (centripetal force) you have to remember that $F_n=m\frac{v^2}{r}$. Hence, $a_n=\frac{v^2}{r}=1\frac{\...


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