Thanks. Glad it made it all the way to you. If you ever have video ideas send them my way!

Hey thanks very much, I appreciate that!

You're welcome, I'd like to watch how different tire pressure causes to speed on certain engine RPM

Thanks, that means a lot 👍& I'll take it 🤠

Well I ain't monetized so I guess I'm not sure if it is required...

Everything seems obvious once we learn it:)

I have not included horsepower in any of my videos as of yet; partially because, in my opinion, it’s harder to utilize in calculations. The second reason being that it scares me. :) I absolutely agree with you about horsepower being useless when excessive rpm is required. I tend to agree with those who say torque has much more to do with 0 to 60 time than HP. I think this is illustrated by the fact that electric cars have vastly less torque by the time they reach max horsepower rpm. Besides, horsepower is just a function of torque! Maybe I shouldn’t of said that then I also have to say torque is just a function of HP. :( Give me a little time to read a couple hundred physics forums and I will try to get a video out ;) - thanks

@@redneckphysics Math is one thing (I am a mathematician by trade), but for me it’s always been about the feel of pull between 1500 and 4000 rpm when driving on the street/road… 6k doesn’t come into play very often between traffic lights 😄 usually 🚔 ☹️ I do my builds to be vehicles that I can actually drive and have fun, not for track specs. Anyway, thanks again and Merry Christmas 🎄

That is, total weight able to be pulled, not worrying about how fast.

Absolutely. If you had 25,000lbs trailer behind a truck that weighed 7,000lbs. If you had 500lb-ft of crankshaft torque And you multiply it by your Trany ratio(let’s say it’s a manual trany) and then by your differential ratio you will probably have somewhere around 10k lb-ft of wheel torque. If you have 29 inch tires then you would have 8200ish lbs of wheel force available at the rear axle and assuming that you have somewhere near 5000 pound King pin weight you should be able to accelerate with at least that much force without losing traction… Now if we’re going up a 10° slope… we will need the sin(10°)*32000(combined weight) lbs of force to overcome gravity. This comes out to 6080lbs then add on 600lbs of rolling resistance and we get 6700lbs of force needed. We had over 8K available with 29inch tires, but if we throw 42inch monsters on the truck we would only have 5.8k available thus we could not budge the load! At that point we would just go into 4L:)

By and large, a bigger tire, will give you worse MPG. That being said, it is possible for a larger diameter tire that is much lighter to give you better MPG. It’s all about the tires moment of inertia( resistance to spinning) and obviously about how heavy you are on the pedal. When a figure skater holds their arms out, they’re spinning at a certain speed, but when they pull their arms in, they start spinning faster, even though there was no force supplied to them because they brought their mass closer to the center of rotation. that’s an example of changing a moment of inertia. If you have an extremely large diameter ring of wood(or some relatively light material), it will have the same moment of inertia(resistance to rotating) as a much smaller diameter, solid lead disc. The point being the farther out the masses, the harder it is to spin. When you get into the weeds of rotational dynamics, you actually find out that increasing tire size decreases the inertial mass of your car, although this is typically negated by other stuff. If you’d like a deeper dive into it let me know, and I’ll give you some more detail. Otherwise, I don’t wanna boar you:)

Hi there, here’s my best shot. More mass (more weight assuming your near earths surface) does not change the wheel torque or wheel force but it does change the equivalent mass (see a few paragraphs down) of your car. If your new wheel and tire setup is such that more mass is farther out than the moment of inertia of your wheel will be higher. Moment of inertia is just a bodies resistance to a change in rotational motion. The formula for moment of inertia of a solid disk(your tire wheel) is 1/2MR^2. This tells us that while more mass in general is bad how far out that mass is (R) matters way, way, way more. I still need to make a video on moment of inertia so in the meantime watch this kzfaq.info/get/bejne/ed6leLVh1q-7ep8.html. Most people would answer your question as follows... If you want to find out how much power your car has you compare the energy present in your car at a standstill (basically zero) to the amount of energy present in your car at the end of a certain time of gunning it down the track. At the latter of these two measurements, we would find that your car would have kinetic energy(1/2mv^2) and rotational kinetic energy(1/2Iw^2). The less rotating mass your car's drivetrain had the more your engine could focus on translating that power into translational energy(speed of some part of your car with respect to the ground) whereas if you had 42" rock crawling tires than your engine would have to redirect more of its resources into getting all that rotating mass moving. So to find out how much power your losing to the task of "spinning up this higher moment of inertia" you can balance some energy equations. But in my opinion it still gives you a very incomplete if not inaccurate picture.. Because we can’t derive the wheel force from those calculations! And all we want is F so we can plug it into F=ma and find out our cars linear acceleration comes in. As it turns out a change in a tire(or any rotating drive components) mass or mass distribution will not effect the wheel torque and hence, assuming no change in rolling diameter, no change in wheel force. What a change in mass or distribution of that mass(moment of inertia) will affect is the equivalent mass of your car. This is what people mean when they say "a pound of tire is worth 5 pounds of frame." Check out this link for a great explanation as to what equivalent mass is. hpwizard.com/rotational-inertia.html For real give that page a read it is well worth it even if not all the formulas make sense you can catch the drift of the concept. You are right that you probably don't have way more traction from your new tires( since always keep in mind that contact patch size has no direct correlation to traction available out side of giving you a higher probability that the rubber will be in contact with some good clean tarmac.) That is to say that given the same frictional coefficients, a truck on wheels or on tracks has the same amount of possible traction. I suppose the loss of power might be attributed to your wheel swap if the new wheels have more mass farther out than the old ones...

Oh boy....! here I go. You asked the right question if you want a really long reply :).... The short answer is that power is unaffected by gearing or tire diameter (as long as the total moment of inertia of the drive train is no different). I still need to make a video on moment of inertia so in the meantime watch this kzfaq.info/get/bejne/ed6leLVh1q-7ep8.html. So first off, power is work done with respect to how long it took to do that work, and work is just the how far you were able to move something times the force required to make it move. Power is just how this work is done... here is my video on that...kzfaq.info/get/bejne/kNGmnMV9ys2mqX0.html If you want to find out how much power your car has you compare the energy present in your car at a standstill (basically zero) to the amount of energy present in your car at the end of a certain time of gunning it down the track. At the latter of these two measurements, we would find that your car would have kinetic energy and rotational kinetic energy. The less rotating mass your car's drivetrain had the more your engine could focus on translating that power into translational energy(speed of some part of your car with respect to the ground) whereas if you had 42" rock crawling tires than your engine would have to redirect more of its resources into getting all that rotating mass moving. I will continue answering your question after I get some of my homework done :)

In almost all cases a bigger tire will lower your dyno readings. This does not directly relate to the diameter of the tire but rather to "how hard" the tire is to turn (the tire's moment of inertia or distribution of mass). The engine will still be giving out the same power it's just that it has to spend more of it getting the tires moving. Now, if you measure power like the big three do (I believe)...then you get the car up to the RPM - which you know your HP peak is at - and then let it sit at that speed before reading your dyno you will find there is no difference between big or little tires. Because the tires (and rest of drivetrain) is already up to speed they stop stealing the engines resources. As far as load moving ability goes... Under acceleration (when your car is speeding up) the following equation describes how tire size and mass distribution fits in. F=T/(R+I/M*R). I = moment of inertia, F = force propelling your car down the road, T = torque at the wheel, M= mass of the tire, R= radius of the tire.

I'm looking at the yanmar 325 vs 425 and your video helped so much. Ty

It all depends on how you drive... if you are doing a lot of rapid acceleration you will see less than zero improvement in fuel economy. If you are cruising at highway speeds you may see a tiny, teeny, tiny improvement... also its possible to get a larger diameter tire that still has a lower resistance to rotation(moment of inertia). It all about how far out the weight is in your tire. So look for a good lightweight tire.

Your absolutely correct. I definitely want to do a video on dynamic rotational losses. I suppose the only time that we could actually ignore these losses would be when the vehicle is no longer accelerating but maintaining a velocity.

I found this formula very helpful F=T/(R+I/M*R). I = moment of inertia. Do you know of any equations that describe the dynamic losses in terms of energy?