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sure! you can contact me on sales@physicsgizmos.com.au

I'm so pleased it was helpful! Your comment made my day :-)

Glad it was helpful!

Hi @kazzdevlin5339. Its not the speed (m/s) that determines the g-force you feel, its the acceleration (m/s/s or metres per second squared). This is, how quickly your speed is changing. If you went from 0m/s (speed) to 100 m/s (also a speed) in, say, 3 seconds, that means your speed is changing at 33.3 metres per second every second. In other words, your acceleration is 33.3 metres per second per second. So that woule be a bit over 3 "g"s of accleration. (33.3/9.8). If you went from 0m/s to 100m/s in 2 seconds, your acceleration would be 50 m/s/s (ie: speed changes 50m/s every second). That would be approx 5 g of acceleration. (50/9.8=5.10 g to be more accurate. Does that help?

@ianthompson4882 So if i state 1 M/S^2 would achieve the speed of 5 meters a second in 5 seconds or 10 meters in 10 seconds or I'm I way off? The time factor is critical, @ 100 m/s in 2 seconds, feels like 5g, but at 8 seconds, it may only feel like 1.3g? So, in the game, these ships are going 1000 m/s. Let's say it takes 15 seconds to get to that speed that's close to 7 Gs whereas two seconds, nets you 51Gs and a really big mess? To further complicate this, let's say we're in space no air, no friction. Nothing to slow us. If I were to fire the thrusters only on the left side of the ship, inducing us to spin right, does this exacerbate the G-force effect? Our initial heading still remains the same, but we're effectively facing in a new direction.

@@kazzdevlin5339 You're on the right track with your 'translational' (straight line) motion examples. ...and you said Maths wasn't your strong suit! ;-) Spinning complicates matters. When you spin, centripetal acceleration is involved. Turning is a form of acceleration, because your velocity is changing (even though its the direction of your velocity, not the 'speed' of your velocity that changes. But ts still a change in velocity, and still involves acceleration) I'm sure you can imagine when you're on one of those amusement rides that spins really fast and you feel pushed outward. That outward 'push' you feel is not actually a force, it's your inertia trying to resist the centripetal (rotational) acceleration. Your centripetal acceleration depends on how your tangential (rotational) speed AND your distance from the centre of rotation. the formula is a=mv^2/r So if you're accelerating in a straight line AND spinning, you need to vector-add the translational (straight line) acceleration to the centripetal (rotational) acceleration. The divide your total acceleration by 9.8 to get the 'g's you'd experience. Does that make sense?

@ianthompson4882 Moral of the story: If you went from 0 to 1000 m/s doing 7Gs, turning is very bad. Wanna thank you for all the help much appreciated

Thank you kindly

Each line needs to have only equation on it. If there are any other characters on that line (even a space or a tab) on the line, it stuffs up the alignment feature.

Students and teachers can register for a free license. www.autodesk.com/education/edu-software/overview?sorting=featured&filters=individual If you're NOT a student or teacher, then its actually REALLY expensive. Like, thousands of dollars each year. But it's very, very powerful and heavily used in industry. I use Inventor for all my educational stuff (using my Teacher license) and FreeCAD for anything commercial I do.

btw. hope you're well mate!

Thanks Phil, much appreciated. I hope you find them really useful. We're all well, hope you are too! Enjoy teaching Physics!

Hi Richard, Unfortunately this is probably not the sensor you're after. It's designed for high school Physics experiments, not for industrial applications. But thanks for watching!

Thank Rohan - hope you're well mate!

Thanks! I'm very pleased they are useful.

It's a lot like the ping pong ball launcher! Hope you're well :-)

Hey Kieran, good to hear from you. I hope you're well!

Btw... your bike powered ipad charger was pretty cool!